29:I[613436,["/_next/static/immutable/chunks/0att3g9h0h9m7.js","/_next/static/immutable/chunks/0n_ejf1w61anu.js","/_next/static/immutable/chunks/1g-8_acy_u6v_.js","/_next/static/immutable/chunks/1_wyo_rtcon-8.js","/_next/static/immutable/chunks/2cei39krol38n.js","/_next/static/immutable/chunks/2w601m8wsm1ht.js","/_next/static/immutable/chunks/0b9145ckv2cou.js","/_next/static/immutable/chunks/17r05ydov_hpt.js","/_next/static/immutable/chunks/1dtj60b48sxo1.js","/_next/static/immutable/chunks/2cs5ldqqqz25p.js"],"default"] 2a:Tf87,A brief note on **Waves and Sound**. Read it before a test. Try each question at the end before you open its answer. ## The big ideas - $v = f\lambda$. The **medium** sets the speed; the **source** sets the frequency. On a string, $v = \sqrt{\dfrac{T}{\mu}}$. - Two waves in step: whole wavelengths of path difference are **loud**, odd half wavelengths are **quiet**. **Beats:** $|f_1 - f_2|$ each second. - **Standing waves:** a string fixed at both ends or an open pipe gives $f_n = \dfrac{nv}{2L}$; a pipe closed at one end gives only odd multiples of $\dfrac{v}{4L}$. - **Doppler:** motion towards each other raises the pitch: $f' = f\,\dfrac{v \pm v_o}{v \mp v_s}$. ## The facts at a glance - **Wave speed:** $v = f\lambda$ - **Beats:** $\lvert f_1 - f_2 \rvert$ each second - **String, open pipe:** $f_n = \dfrac{nv}{2L}$ - **Closed pipe:** $\dfrac{v}{4L}, \dfrac{3v}{4L}, \dfrac{5v}{4L}, \ldots$ - **Resonance tube:** $\lambda = 2(l_2 - l_1)$ - **Moving source:** $f' = f\,\dfrac{v}{v \mp v_s}$ ## A worked example A string $0.75\ \text{m}$ long has a tension of $81\ \text{N}$ and a mass of $0.01\ \text{kg}$ for each metre. Find the wave speed and its first three harmonics. 1. **The speed.** $v = \sqrt{\dfrac{81\ \text{N}}{0.01\ \text{kg/m}}} = 90\ \text{m/s}$. 2. **The fundamental.** $f_1 = \dfrac{90\ \text{m/s}}{2 \times 0.75\ \text{m}} = 60\ \text{Hz}$. 3. **The harmonics.** $120\ \text{Hz}$ and $180\ \text{Hz}$. ## Practice with new numbers ```practice { "title": "A source moving away", "vars": { "f": [ 500, 800 ], "vs": [ 17, 34 ] }, "values": { "fh": "f*340/(340 + vs)", "wrong": "f*340/(340 - vs)" }, "question": "A siren of {{f}} Hz moves away from you at {{vs}} m/s. Sound travels at $340\\ \\text{m/s}$. What frequency do you hear, in hertz?", "answer": { "value": "fh", "unit": "Hz", "digits": 1 }, "hint": "Moving away: $f' = f\\,\\dfrac{v}{v + v_s}$.", "mistakes": [ { "value": "wrong", "say": "That is for a source coming towards you. Moving away, the pitch is lower." } ], "solution": "$f' = {{f}}\\ \\text{Hz} \\times \\dfrac{340\\ \\text{m/s}}{ (340 + {{vs}})\\ \\text{m/s} } = {{fh:1}}\\ \\text{Hz}$." } ``` ## A program to try ```python import math L, T, mu = 0.75, 81, 0.01 # m, N, kg/m # Step 1: the wave speed on the string v = math.sqrt(T / mu) # Step 2: the harmonics, f_n = n v / 2L print("Speed:", v, "m/s") for n in [1, 2, 3]: print(f"Harmonic {n}: {n * v / (2 * L):.0f} Hz") ``` !try Read it, then run it. It repeats the worked example. 1. Change `T` to `324`. What happens to every harmonic? ## Mistakes to avoid - **Thinking a higher frequency travels faster.** It has a shorter wavelength instead. - **Taking the average frequency as the beats.** The beats are the difference. - **Using $\dfrac{v}{2L}$ for a closed pipe.** It is $\dfrac{v}{4L}$, with odd harmonics only. - **Putting the Doppler signs the wrong way.** Towards each other always raises the pitch. ## Check yourself 1. A pipe $0.17\ \text{m}$ long is closed at one end. Sound travels at $340\ \text{m/s}$. What is its lowest note? !hint Closed at one end: $f_1 = \dfrac{v}{4L}$. !answer $f_1 = \dfrac{340\ \text{m/s}}{4 \times 0.17\ \text{m}} = 500\ \text{Hz}$. 2. Forks of $384\ \text{Hz}$ and $381\ \text{Hz}$ sound together. How many beats do you hear each second? !hint Beats each second $=$ the difference of the frequencies. !answer $384\ \text{Hz} - 381\ \text{Hz} = 3$ beats each second. 3. Why does an approaching car's horn sound higher than it really is? !hint What does the moving horn do to the spacing of its crests in front of it? !answer The horn chases its own crests, so they reach you closer together: a shorter wavelength and a higher frequency. 4. What is the wavelength of a $1360\ \text{Hz}$ note in air, where sound travels at $340\ \text{m/s}$? !hint $\lambda = \dfrac{v}{f}$. !answer $\lambda = \dfrac{340\ \text{m/s}}{1360\ \text{Hz}} = 0.25\ \text{m}$.2b:T1215,A brief note on **Entropy, Gibbs Energy and Spontaneity**. Read it before a test. Try each question at the end before you open its answer. ## Entropy - A spontaneous change happens on its own. That says nothing about how fast. - A fall in enthalpy helps, but some endothermic changes are spontaneous too. - Entropy, $S$, measures how spread out matter and energy are. Gas $>$ liquid $>$ solid. - $\Delta S > 0$ for melting, boiling, dissolving and reactions that make more moles of gas. - $\Delta S = q_{rev}/T$; for a change of state, $\Delta S = \Delta H / T$. - Second law: $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$ for a spontaneous change, with $\Delta S_{surr} = -\Delta H/T$. - Third law: a perfect crystal has $S = 0$ at $0\ \text{K}$. $\Delta_r S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$. ## Gibbs energy - $G = H - TS$, and $\Delta G = \Delta H - T\Delta S$. - $\Delta G < 0$: spontaneous. $\Delta G = 0$: equilibrium. $\Delta G > 0$: not spontaneous. - $\Delta H < 0$, $\Delta S > 0$: always. $\Delta H > 0$, $\Delta S < 0$: never. - Same signs: $\Delta G$ changes sign at $T = \Delta H / \Delta S$. ## Gibbs energy and K - $\Delta G^\circ = -RT\ln K = -2.303\,RT\log K$. - $\Delta G^\circ < 0$: $K > 1$. $\Delta G^\circ = 0$: $K = 1$. $\Delta G^\circ > 0$: $K < 1$. - At $298\ \text{K}$, $2.303\,RT \approx 5.71\ \text{kJ/mol}$. ## A worked example A change has $\Delta H = +40\ \text{kJ/mol}$ and $\Delta S = +120\ \text{J/(K mol)}$. Is it spontaneous at $298\ \text{K}$, and where does it turn? 1. **Units.** $\Delta S = 0.120\ \text{kJ/(K mol)}$. 2. **At 298 K.** $\Delta G = 40 - 298 \times 0.120 = +4.24\ \text{kJ/mol}$. Not spontaneous. 3. **The turning point.** $T = \dfrac{40\,000}{120} = 333\ \text{K}$. 4. **Meaning.** Both signs are positive, so it is spontaneous above $333\ \text{K}$. ## Practice with new numbers ```practice { "title": "Where delta G changes sign", "vars": { "dh": [ 25, 45, 80 ], "ds": [ 90, 125, 200 ] }, "values": { "tx": "1000*dh/ds", "wrong": "dh/ds" }, "question": "A change has delta H = +{{dh}} kJ/mol and delta S = +{{ds}} J/(K mol). Above what temperature, in K, is it spontaneous?", "answer": { "value": "tx", "unit": "K", "digits": 0 }, "hint": "Delta G is zero where T = delta H / delta S. Put delta H in J/mol first.", "mistakes": [ { "value": "wrong", "say": "Change delta H to J/mol before dividing by delta S in J/(K mol)." } ], "solution": "$T = \\dfrac{ {{dh}} \\times 1000 }{ {{ds}} } = {{tx:0}}\\ \\text{K}$." } ``` ## A program to try The program tries each of the four sign cases at a cold and a hot temperature. ```python # The four sign cases of delta G = delta H - T delta S, tried at a cold and a hot temperature. cases = [(-50, 100), (50, -100), (-50, -100), (50, 100)] # delta H in kJ/mol, delta S in J/(K mol) for dh, ds in cases: cold = dh - 200 * ds / 1000 hot = dh - 1000 * ds / 1000 print(f"dH {dh:+} kJ/mol, dS {ds:+} J/(K mol): at 200 K {cold:+.0f}, at 1000 K {hot:+.0f} kJ/mol") ``` !try Run it. Which case changes sign between $200\ \text{K}$ and $1000\ \text{K}$, and in which direction? 1. Add a fifth case with $\Delta S = 0$. Does temperature matter for it? ## Mistakes to avoid - **Only exothermic changes are spontaneous.** Melting ice takes in heat and still happens. - **Mixing kJ and J.** Divide $\Delta S$ by $1000$ before combining it with $\Delta H$ in kJ/mol. - **Celsius in the formulas.** Every $T$ must be in kelvin. - **$\Delta G$ and $\Delta G^\circ$ confused.** At equilibrium $\Delta G = 0$; $\Delta G^\circ$ fixes $K$. - **A negative $\Delta G$ means fast.** It says the change can happen, not how quickly. ## Check yourself 1. Predict the sign of $\Delta S$ when ammonia gas dissolves in water. !hint Is the gas becoming more or less spread out? !answer Negative. Gas molecules become trapped in the liquid, so they are less free. 2. Why is a reaction with $\Delta H > 0$ and $\Delta S < 0$ never spontaneous? !hint Look at the sign of each part of $\Delta H - T\Delta S$. !answer Both $\Delta H$ and $-T\Delta S$ are positive, so $\Delta G$ is positive at every temperature. 3. A reaction has $K = 1000$ at $298\ \text{K}$. Is $\Delta G^\circ$ positive or negative, and about how large? !hint $\log K = 3$. Each $5.71\ \text{kJ/mol}$ is one unit of $\log K$. !answer Negative, about $-3 \times 5.71 = -17.1\ \text{kJ/mol}$. 4. What is $\Delta S_{total}$ for a system at equilibrium? !hint Is there a push in either direction? !answer Zero. Neither direction raises the total entropy.2c:Tf85,A brief note on **Combinations**. Read it before a test. Try each question at the end before you open its answer. ## The formula - A combination is a selection: order does not matter. - $^{n}C_{r} = \dfrac{n!}{r!\,(n - r)!} = \dfrac{n(n - 1)\cdots(n - r + 1)}{r!}$. - $^{n}P_{r} = {}^{n}C_{r} \times r!$: choose, then arrange. - $^{n}C_{0} = {}^{n}C_{n} = 1$, $^{n}C_{1} = n$ and $^{n}C_{2} = \dfrac{n(n - 1)}{2}$. ## Properties | Rule | In symbols | | --- | --- | | symmetry | $^{n}C_{r} = {}^{n}C_{n - r}$ | | equal values | $^{n}C_{a} = {}^{n}C_{b}$ means $a = b$ or $a + b = n$ | | Pascal's rule | $^{n}C_{r} + {}^{n}C_{r - 1} = {}^{n + 1}C_{r}$ | | neighbours | $\dfrac{^{n}C_{r}}{^{n}C_{r - 1}} = \dfrac{n - r + 1}{r}$ | ## Geometry - $n$ points, no three in a line: $^{n}C_{2}$ lines and $^{n}C_{3}$ triangles. - A polygon of $n$ sides: $^{n}C_{2} - n = \dfrac{n(n - 3)}{2}$ diagonals. - If $k$ of the points lie on one line: subtract $^{k}C_{3}$ triangles, and replace $^{k}C_{2}$ lines by $1$. ## Selections with conditions - Must be in: put them in, choose the rest from the others. - Must be out: choose everything from the others. - Separate groups: multiply the counts. - At least one: total minus none. - Words and codes: select, then arrange. ## A worked example A quiz team of $4$ is chosen from $9$ girls and $5$ boys. How many teams have at least one boy? **Step 1. All teams.** $^{14}C_{4} = 1001$. **Step 2. No boy.** $^{9}C_{4} = 126$. **Step 3. Subtract.** $1001 - 126 = 875$. ## Practice with new numbers ```practice { "title": "Lines through points", "vars": { "n": [ 9, 11, 12, 16, 17, 19 ] }, "values": { "c": "n*(n - 1)/2", "p": "n*(n - 1)", "m": "n - 1" }, "question": "${{n}}$ points lie in a plane, and no three of them are on one straight line. How many lines can be drawn through two of the points?", "answer": { "value": "c", "digits": 0 }, "hint": "Two points fix one line, and the pair has no order.", "mistakes": [ { "value": "p", "say": "That counts each line twice, once from each end. Divide by $2$." } ], "solution": "$^{ {{n}} }C_{2} = \\dfrac{ {{n}} \\times {{m}} }{2} = {{c}}$." } ``` ## A program to try The program lists every quiz team of the worked example, keeps those with a boy, and checks the count against the formula. ```python # A quiz team of 4 from 9 girls and 5 boys, with at least one boy. from itertools import combinations from math import comb girls = ["G" + str(k) for k in range(1, 10)] boys = ["B" + str(k) for k in range(1, 6)] teams = [t for t in combinations(girls + boys, 4) if any(p in boys for p in t)] print("By listing:", len(teams)) print("By formula:", comb(14, 4) - comb(9, 4)) print("One of them:", teams[0]) ``` !try Run it. Do the listing and the formula give the same number? 1. Change the rule to teams with at least two boys. Predict the count first. ## Mistakes to avoid - **$^{n}P_{r}$ for a team.** A team has no order. Use $^{n}C_{r}$. - **Only $a = b$.** $^{n}C_{a} = {}^{n}C_{b}$ also allows $a + b = n$. - **Sides as diagonals.** Take the $n$ sides away from $^{n}C_{2}$. - **One chosen first.** Picking one woman first counts some committees more than once. - **Selecting without arranging.** Words need the $r!$ as well. ## Check yourself 1. Work out $^{17}C_{15}$. !hint Use the symmetry first. !answer $^{17}C_{2} = \dfrac{17 \times 16}{2} = 136$. 2. If $^{n}C_{2} = 28$, what is $n$? !hint $n(n - 1) = 56$. !answer $n = 8$, since $8 \times 7 = 56$. 3. How many diagonals does a polygon with $9$ sides have? !hint $\dfrac{n(n - 3)}{2}$. !answer $\dfrac{9 \times 6}{2} = 27$. 4. Choose $2$ of $6$ singers and $1$ of $4$ drummers for a band. In how many ways? !hint Two separate choices multiply. !answer $^{6}C_{2} \times {}^{4}C_{1} = 15 \times 4 = 60$. 5. Write $^{7}C_{2} + {}^{7}C_{3}$ as one term, and find its value. !hint Pascal's rule with $n = 7$ and $r = 3$. !answer $^{8}C_{3} = 56$.2d:T1102,A brief note on **Chordates: Fishes to Mammals**. Read it before a test. Try each question at the end before you open its answer. ## The phylum - Four features at some stage: **notochord**, **dorsal hollow nerve cord**, **paired pharyngeal gill slits**, **post-anal tail**. - Non-chordates: no notochord, a **ventral, solid, double** nerve cord, no gill slits, a dorsal heart, no post-anal tail. - **Urochordata**: notochord only in the larval tail (*Ascidia*, *Salpa*, *Doliolum*). - **Cephalochordata**: notochord from head to tail, for life (*Branchiostoma*). - **Vertebrata**: notochord in the embryo, replaced by a vertebral column. - Vertebrates add a ventral muscular heart, kidneys and paired fins or limbs. ## The seven classes - **Cyclostomata**: no jaws, a sucking mouth. Heart 2, cold-blooded. - **Chondrichthyes**: cartilage, separate gill slits, no air bladder. Heart 2, cold-blooded. - **Osteichthyes**: bone, an operculum, an air bladder. Heart 2, cold-blooded. - **Amphibia**: moist skin, a cloaca, a tympanum. Heart 3, cold-blooded. - **Reptilia**: dry skin with scales or scutes. Heart 3 (4 in crocodiles), cold-blooded. - **Aves**: feathers, pneumatic bones, air sacs. Heart 4, warm-blooded. - **Mammalia**: hair, mammary glands, pinnae. Heart 4, warm-blooded. - Cartilaginous fishes: placoid scales, claspers, internal fertilisation, often viviparous. - Bony fishes: four pairs of gills, external fertilisation, mostly oviparous. - Amphibians breathe with gills, lungs and skin; development goes through a larva. - *Ornithorhynchus* is the egg-laying mammal. ## A worked example Name the class of an animal with hair, a four-chambered heart and wings. 1. **Hair** is found only in mammals. 2. **Wings** are not enough to make a bird: a bird must have feathers. 3. **Class:** Mammalia. The flying fox, *Pteropus*, fits. ## Practice with new numbers ```practice { "title": "Gill openings of a hagfish", "vars": { "p": [ 6, 8, 10, 12 ] }, "values": { "t": "2*p" }, "question": "A hagfish has {{p}} pairs of gill openings. How many gill openings does it have in all?", "answer": { "value": "t", "unit": "openings", "digits": 0 }, "hint": "A pair is two, one on each side.", "mistakes": [ { "value": "p", "say": "That counts the pairs. Double it for both sides." } ], "solution": "$2 \\times {{p}} = {{t}}$ openings." } ``` ## A program to try The program prints one line for each class of vertebrates: its mark, its heart and its blood. ```python # One line on each class of vertebrates. classes = [ ("Cyclostomata", "no jaws", 2, "cold"), ("Chondrichthyes", "cartilage", 2, "cold"), ("Osteichthyes", "bone, operculum", 2, "cold"), ("Amphibia", "moist skin, cloaca", 3, "cold"), ("Reptilia", "dry scaly skin", 3, "cold"), ("Aves", "feathers, hollow bones", 4, "warm"), ("Mammalia", "hair, mammary glands", 4, "warm"), ] for name, mark, chambers, blood in classes: print(f"{name:15} {mark:22} {chambers} chambers, {blood}-blooded") warm = [c[0] for c in classes if c[3] == "warm"] print("warm-blooded classes:", warm) ``` !try Run it. Which two classes are warm-blooded? 1. A crocodile's heart has 4 chambers. Why does the Reptilia line still say 3? !hint Look at what most reptiles have. !answer Most reptiles have a three-chambered heart. The crocodile is the exception. ## Mistakes to avoid - **Calling every chordate a vertebrate.** Sea squirts and lancelets have no backbone. - **Giving a shark an operculum.** Its gill slits open separately. - **Linking four chambers with warm blood.** Crocodiles have four and are cold-blooded. - **Saying every mammal gives birth to live young.** The platypus lays eggs. ## Check yourself 1. In which subphylum is the notochord lost when the adult forms? !hint Think of the sea squirt. !answer Urochordata. The notochord lies only in the tail of the larva. 2. Give two features that set a bony fish apart from a cartilaginous one. !hint Look at the gills and the swim organ. !answer A bony fish has an operculum over its gills and an air bladder. A cartilaginous fish has neither. 3. Which two classes of vertebrates are homoiothermous? !hint They keep a steady body temperature. !answer Aves and Mammalia. 4. Name an amphibian with no limbs. !hint It lives in damp soil. !answer *Ichthyophis*.2e:Tfcb,A brief note on **Genetics by Computer**. Read it before a test. Try each question at the end before you open its answer. ## One gene - A gamete is one letter: `list("Tt")` gives `['T', 't']`. - A child is `"".join(sorted(a + b))`; the capital, dominant letter comes first. - Two loops over the parents' letters fill the Punnett square. - `Counter(kids)` counts genotypes. Tt with Tt gives 1 TT : 2 Tt : 1 tt, which is 3 tall to 1 short. - Tt with tt, a test cross, gives 1 tall to 1 short. ## Two genes - `product("Rr", "Yy")` gives the four gametes RY, Ry, rY and ry. - Four gametes meet four in 16 boxes. Sort each gene of the child on its own. - RrYy with RrYy gives 9:3:3:1. Each gene alone still gives 3:1, and $\dfrac{3}{4} \times \dfrac{3}{4} = \dfrac{9}{16}$. - A cross with rryy reveals a plant's hidden small letters. ## Chance - `random.choice(parent)` picks one gamete; one choice from each parent makes a child. - `random.seed(k)` makes every run repeat the same numbers. - A few plants wander far from the ratio; thousands settle near it. ## A worked example What does this print? ```python from collections import Counter kids = ["".join(sorted(a + b)) for a in "Ww" for b in "ww"] print(Counter("round" if "W" in k else "wrinkled" for k in kids)) ``` 1. **The boxes.** W with w gives Ww twice; w with w gives ww twice. 2. **The phenotypes.** Ww holds a capital W, so it is round; ww is wrinkled. It prints `Counter({'round': 2, 'wrinkled': 2})`, a 1:1 ratio. ## Practice with new numbers ```practice { "title": "A ratio from counts", "vars": { "a": [ 459, 612, 930 ], "b": [ 151, 207, 298 ] }, "values": { "r": "a/b", "wrong": "b/a" }, "question": "A count of pea plants finds {{a}} with purple flowers and {{b}} with white. What is the ratio of purple to white, to 1?", "answer": { "value": "r", "unit": "", "digits": 2 }, "hint": "Divide the purple count by the white count.", "mistakes": [ { "value": "wrong", "say": "That is white to purple. Put the purple count on top." } ], "solution": "$\\dfrac{ {{a}} }{ {{b}} } = {{r}}$, so about {{r}} to 1. Close to 3 to 1, as Mendel's rule says." } ``` ## A program to try The program crosses Pp with pp and then Pp with Pp, and counts the flower colours each time. ```python # Purple flowers (P) over white (p): cross Pp with pp, then Pp with Pp from collections import Counter def cross(mum, dad): return ["".join(sorted(a + b)) for a in mum for b in dad] for mum, dad in [("Pp", "pp"), ("Pp", "Pp")]: kids = cross(mum, dad) looks = Counter("purple" if "P" in k else "white" for k in kids) print(mum, "x", dad, kids, dict(looks)) ``` !try Run it. Which cross gives 3 purple to 1 white? 1. Add the cross `("PP", "pp")`. What colours appear? ## Mistakes to avoid - **No sort.** Then `tT` and `Tt` are counted as two genotypes. - **Testing for TT only.** One capital letter is enough for the dominant look. - **Two letters of one gene in a gamete.** A gamete has one letter of each gene. - **Both random letters from one parent.** Take one from each. - **Judging a ratio from a few plants.** Chance needs many plants to settle. ## Check yourself 1. Why is `"".join(sorted(a + b))` used instead of `a + b`? !hint What would a + b give for t and T? !answer It would give `tT`, which Counter treats as different from `Tt`. Sorting writes every genotype the same way, capital first. 2. How many boxes does the square of AaBb with AaBb have, and why? !hint Count the gametes of each parent. !answer 16. Each parent makes 4 gametes, and each meets each of the other's: 4 × 4. 3. A round plant crossed with rr gives only round seeds. What is it? !hint Could it pass on an r? !answer RR. A plant with r would give rr, wrinkled, seeds in half the boxes. 4. A simulation of 40 plants gives 30 % short. Is the 3:1 rule wrong? !hint How far can 40 plants wander? !answer No. With so few plants, chance easily moves the share several percent from 25 %. Simulate many more before judging.2f:T12a3,A brief note on **A Data Project from Question to Report**: a whole data project, from the question to the report. Try each question at the end before you open its answer. ## The six steps 1. **Ask** one clear question. 2. **Collect** the data in a table. 3. **Clean** it: check any value that cannot be true. 4. **Show** it: a frequency table or histogram for the shape, a line graph for the change over time. 5. **Summarise** it: the centre, the spread and the chances. 6. **Report** it: the question, the method, the numbers, a graph and one limit. ## Summaries for a month of weather | Want | Use | |---|---| | a typical day | the median, or the mean if there are no extreme days | | how much the days vary | the IQR with the median, the standard deviation with the mean | | the chance of rain | its relative frequency: rainy days $\div$ all days | | rainy days to expect in $N$ days | the relative frequency $\times N$ | Weather comes in **spells**, so the chance of two rainy days in a row is often more than the product of the two chances. ## What a report may claim - Compare groups with their means **and** their sizes. - An **association** is not a **cause**: "the rainy days were cooler" is fine; "rain makes it cooler" is not proved. - One month in one town is a limit; say so. ## A worked example Ten spring days had the highest temperatures $24, 22, 21, 25, 27$, $26, 23, 20, 22, 26\ ^\circ\text{C}$, with rain on days 2, 3, 8 and 9. 1. **Typical day.** Sorted: $20, 21, 22, 22, 23$, $24, 25, 26, 26, 27$. The median is $\dfrac{23 + 24}{2} = 23.5\ ^\circ\text{C}$. 2. **Chance of rain.** $\dfrac{4}{10} = 0.4$. 3. **Two groups.** The 4 rainy days had a mean of $21.25\ ^\circ\text{C}$, and the 6 dry days $25.17\ ^\circ\text{C}$. 4. **The report.** "On these 10 days, the rainy days were cooler on average. Ten days in one town cannot tell us about other seasons." ## Practice with new numbers ```practice { "title": "A mean from two groups of days", "vars": { "a": [ 4, 6 ], "ma": [ 19, 21 ], "mb": [ 25, 27 ] }, "values": { "b": "20 - a", "m": "(a*ma + (20 - a)*mb)/20", "half": "(ma + mb)/2" }, "question": "Over 20 days, {{a}} rainy days had a mean of {{ma}} °C and the other {{b}} days a mean of {{mb}} °C. What was the mean of all 20 days?", "answer": { "value": "m", "unit": "°C", "digits": 2 }, "hint": "Find each group's total first: days $\\times$ mean.", "mistakes": [ { "value": "half", "say": "Halfway is right only for groups of the same size. There were more dry days." } ], "solution": "$\\dfrac{ {{a}} \\times {{ma}} + {{b}} \\times {{mb}} }{20} = {{m}}\\ ^\\circ\\text{C}$." } ``` ## A program to try The program works out the worked example. ```python import statistics # Ten spring days, in degrees Celsius, and the days it rained. temps = [24, 22, 21, 25, 27, 26, 23, 20, 22, 26] rain_days = [2, 3, 8, 9] rainy = [temps[d - 1] for d in rain_days] dry = [t for d, t in enumerate(temps, 1) if d not in rain_days] print("days:", len(temps), " median:", statistics.median(temps), "°C") print("rainy days:", len(rainy), " mean:", statistics.mean(rainy), "°C") print("dry days:", len(dry), " mean:", round(statistics.mean(dry), 2), "°C") print("chance of rain, estimated:", len(rain_days) / len(temps)) ``` !try Run it. Do its numbers match the worked example? 1. Add day 5 to the rainy days. What happens to the two means? ## Mistakes to avoid - **Starting with a graph** before there is a clear question. - **Giving a mean with no spread and no group sizes.** - **Multiplying the chances of linked days** as if they were separate. - **Writing "because"** when the data only shows that two things went together. ## Check yourself 1. Which graph shows how the temperature changed from day to day, and which shows how many days had each temperature? !hint One keeps the days in order; one groups them. !answer A line graph shows the change from day to day. A histogram or bar chart shows how many days had each temperature. 2. It rained on 8 of 40 days. Estimate the chance of rain, and the number of rainy days to expect in 25 days. !hint Relative frequency first. !answer $\dfrac{8}{40} = 0.2$, so about $0.2 \times 25 = 5$ rainy days. 3. A report says: "Foggy mornings make the afternoons cooler." What should it say instead? !hint What did the data really show? !answer For example: "In this record, days with a foggy morning had cooler afternoons on average." That reports the association without claiming a cause. 4. Give two limits for a report on 30 days of weather at one school. !hint How long was the record, and where was it taken? !answer One month may not be like other months, and one school may not be like other places in the town.30:T10eb,A brief note on **Relative Motion**. Read it before a test. Try each question at the end before you open its answer. ## The big ideas - **Relative velocity:** $v_{AB} = v_A - v_B$, with signs. $v_{BA} = -v_{AB}$. - **The same way:** subtract the speeds. **Opposite ways:** add them. - **Meeting and catching up:** time $=$ gap $\div$ closing speed. Or write both positions and set them equal. - **In two dimensions** the velocities add as vectors: rain slants towards a moving walker; a boat moves along the sum of its own and the river's velocity. ## The facts at a glance - **Along a line:** $v_{AB} = v_A - v_B$ - **Km/h to m/s:** multiply by $\dfrac{5}{18}$ - **Catching up:** $t = \dfrac{\text{gap}}{v_{\text{fast}} - v_{\text{slow}}}$ - **Rain on a walker:** $\tan\theta = \dfrac{v_{\text{walker}}}{v_{\text{rain}}}$ - **Boat heading straight across:** $t = \dfrac{d}{u}$, drift $= wt$ - **Boat landing opposite:** $u\sin\theta = w$, $t = \dfrac{d}{u\cos\theta}$ ## A worked example A car $120\ \text{m}$ behind a bus goes at $23\ \text{m/s}$, and the bus at $15\ \text{m/s}$, the same way. When does the car reach the bus? Then a boat that moves at $3\ \text{m/s}$ in still water heads straight across a river $60\ \text{m}$ wide that flows at $1.2\ \text{m/s}$. How long does it take, and how far downstream does it land? 1. **The car.** $t = \dfrac{120\ \text{m}}{23\ \text{m/s} - 15\ \text{m/s}} = 15\ \text{s}$. 2. **The boat's time.** $t = \dfrac{60\ \text{m}}{3\ \text{m/s}} = 20\ \text{s}$. 3. **The drift.** $1.2\ \text{m/s} \times 20\ \text{s} = 24\ \text{m}$. ## Practice with new numbers ```practice { "title": "Tilt the umbrella", "vars": { "r": [ 6, 8, 10 ], "w": [ 3, 4.5 ] }, "values": { "a": "atan2(w, r)*180/pi", "wrong": "atan2(r, w)*180/pi" }, "question": "Rain falls straight down at {{r}} m/s. A cyclist rides at {{w}} m/s. At what angle from the vertical, in degrees, should the umbrella be tilted forward?", "answer": { "value": "a", "unit": "°", "digits": 1 }, "hint": "Relative to the rider, the rain comes at $\\theta$ from the vertical with $\\tan\\theta = \\dfrac{v_{\\text{rider}}}{v_{\\text{rain}}}$.", "mistakes": [ { "value": "wrong", "say": "That is the angle from the horizontal. Measure from the vertical." } ], "solution": "$\\tan\\theta = \\dfrac{ {{w}}\\ \\text{m/s} }{ {{r}}\\ \\text{m/s} }$, so $\\theta = {{a:1}}^\\circ$." } ``` ## A program to try ```python # A car 120 m behind a bus closes the gap; then a boat crosses a river. gap, v_car, v_bus = 120, 23, 15 # m, m/s, m/s print("Car catches the bus after", gap / (v_car - v_bus), "s") width, u, w = 60, 3.0, 1.2 # m, m/s, m/s t = width / u # heading straight across print("Boat crosses in", t, "s and lands", w * t, "m downstream") ``` !try Read it, then run it. It repeats the worked example. 1. Change `u = 3.0` to `u = 4.0`. How do the time and the drift change? ## Mistakes to avoid - **Adding speeds of things that go the same way.** Subtract them. - **Mixing km/h and m/s** in one calculation. - **Mixing up A relative to B with B relative to A.** They have opposite signs. - **Using the boat's speed over the ground** for the crossing time. Only the part across the river carries it over. ## Check yourself 1. Two trains come towards each other at $72\ \text{km/h}$ and $36\ \text{km/h}$. How fast does the gap close, in m/s? !hint Towards each other: add, then multiply by $\dfrac{5}{18}$. !answer $108\ \text{km/h} \times \dfrac{5}{18} = 30\ \text{m/s}$. 2. A boat that moves at $4.5\ \text{m/s}$ in still water heads straight across a river $90\ \text{m}$ wide. How long does it take? !hint Only the boat's own speed carries it across. !answer $t = \dfrac{90\ \text{m}}{4.5\ \text{m/s}} = 20\ \text{s}$, whatever the river does. 3. When does car A seem to stand still to the driver of car B? !hint Relative velocity is the difference of the two velocities. !answer When the two cars have the same velocity: the same speed in the same direction. 4. The positions are $x_A = 6t$ and $x_B = 40 + 4t$, in metres, with $t$ in seconds. When and where do they meet? !hint Set the two positions equal. !answer $6t = 40 + 4t$ gives $t = 20\ \text{s}$, at $6\ \text{m/s} \times 20\ \text{s} = 120\ \text{m}$.31:T1385,A brief note on **Enthalpy and Hess's Law**. Read it before a test. Try each question at the end before you open its answer. ## Enthalpy - $H = U + pV$, a state function. At constant pressure, the heat taken in is $\Delta H$. - $\Delta H = \Delta U + \Delta n_g R T$. Count gases only. $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$. - Exothermic: $\Delta H < 0$. Endothermic: $\Delta H > 0$. - $q = C\Delta T = m c \Delta T$. For an ideal gas, $C_p - C_v = R$. - Bomb calorimeter: constant volume, gives $\Delta U$. Coffee-cup calorimeter: constant pressure, gives $\Delta H$. ## Standard enthalpies - Standard state: the pure substance at 1 bar, at the stated temperature. - Reverse an equation: change the sign of $\Delta H$. Multiply it: multiply $\Delta H$. - $\Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H$. - $\Delta_f H^\ominus$: one mole of a compound from its elements. Elements in their standard states have zero. - $\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) - \sum b_i \Delta_f H^\ominus(\text{reactants})$. - Combustion is always negative, atomisation always positive. $\Delta_{sol}H = \Delta_{lattice}H + \Delta_{hyd}H$. ## Hess's law and bond enthalpies - The enthalpy change is the same by any route. Add equations like algebra. - For gases, $\Delta_r H \approx \sum(\text{bonds broken}) - \sum(\text{bonds formed})$. - Mean bond enthalpies are averages, so the answer is an estimate. ## A worked example Find $\Delta_f H^\ominus$ of CO from two combustions: carbon gives $-393.5\ \text{kJ mol}^{-1}$, and CO gives $-283.0\ \text{kJ mol}^{-1}$. 1. **Target.** $\text{C} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}$. 2. **Steps.** Burn the carbon, then reverse the burning of CO: $-393.5$ and $+283.0$. 3. **Add.** $\Delta_f H^\ominus = -393.5 + 283.0 = -110.5\ \text{kJ mol}^{-1}$. ## Practice with new numbers ```practice { "title": "Delta H from Delta U", "vars": { "du": [ -500, -300, -100 ], "dn": [ -2, -1, 1, 2 ] }, "values": { "dh": "du + dn*8.314*300/1000", "w1": "du - dn*8.314*300/1000", "w2": "du + dn*8.314*300" }, "question": "At 300 K, a gas reaction has an internal energy change of {{du}} kJ/mol and changes the moles of gas by {{dn}}. Find its enthalpy change, in kJ/mol.", "answer": { "value": "dh", "unit": "kJ/mol", "digits": 1 }, "hint": "Delta H = Delta U + (change in moles of gas) times R times T. Turn joules into kilojoules.", "mistakes": [ { "value": "w1", "say": "The term is added, not taken away." }, { "value": "w2", "say": "R times T comes out in joules. Divide by 1000 first." } ], "solution": "$\\Delta H = {{du}} + \\dfrac{ ({{dn}}) \\times 8.314 \\times 300 }{1000} = {{dh:1}}\\ \\text{kJ/mol}$." } ``` ## A program to try The program writes each equation as a dictionary and adds them with their enthalpy changes. ```python # Hess's law with equations as dictionaries: products positive, reactants negative def combine(*steps): """Each step is (factor, equation, dH). Returns the summed equation and its dH.""" total, dH = {}, 0.0 for factor, eq, h in steps: for s, n in eq.items(): total[s] = total.get(s, 0) + factor * n dH += factor * h return {s: n for s, n in total.items() if abs(n) > 1e-9}, dH burn_C = ({"C": -1, "O2": -1, "CO2": 1}, -393.5) burn_CO = ({"CO": -1, "O2": -0.5, "CO2": 1}, -283.0) # C + 1/2 O2 -> CO: burn the carbon, then undo the burning of CO eq, dH = combine((1, *burn_C), (-1, *burn_CO)) print(eq) print("Delta H =", round(dH, 1), "kJ/mol") ``` !try Run it. Which substances cancel in the sum? 1. Change the factors to `(1, *burn_C)` and `(1, *burn_CO)`. What does the sum describe now? ## Mistakes to avoid - **Liquids in $\Delta n_g$.** Count gases only. - **Joules with kilojoules.** $RT$ comes out in joules. - **Reactants minus products.** Formation enthalpies go products minus reactants. - **Formed minus broken.** Bond enthalpies go broken minus formed. - **A reversed equation with the old sign.** Change the sign. ## Check yourself 1. What is $\Delta_f H^\ominus$ of graphite? !hint Graphite is carbon in its standard state. !answer Zero. 2. Which calorimeter measures $\Delta H$ directly? !hint Which one works at constant pressure? !answer The coffee-cup calorimeter. 3. For $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$, is $\Delta H$ larger or smaller than $\Delta U$? !hint Count the moles of gas made. !answer Larger: one mole of gas is made, so $\Delta H = \Delta U + RT$. 4. Why is breaking a bond always endothermic? !hint Do the atoms attract each other in the bond? !answer The atoms attract each other, so energy must be put in to pull them apart. 5. If $\text{X} \rightarrow \text{Y}$ has $\Delta H = +30\ \text{kJ}$ and $\text{Y} \rightarrow \text{Z}$ has $\Delta H = -50\ \text{kJ}$, what is $\Delta H$ for $\text{X} \rightarrow \text{Z}$? !hint Add the steps. !answer $+30 - 50 = -20\ \text{kJ}$.32:Te42,A brief note on **Permutations**. Read it before a test. Try each question at the end before you open its answer. ## The formulas | Situation | Count | | --- | --- | | $r$ of $n$ different objects in order | ${}^nP_r = \dfrac{n!}{(n - r)!}$ | | all $n$ different objects | $n!$ | | $r$ places, repeats allowed | $n^r$ | | $n$ objects, groups of $p$, $q$ alike | $\dfrac{n!}{p!\, q!}$ | Also $0! = 1$, ${}^nP_0 = 1$ and ${}^nP_r = n \times {}^{n-1}P_{r-1}$. ## Conditions - **Together.** Glue the objects into one block. Multiply by the orders inside it, unless they are alike. - **Apart.** All arrangements minus the together ones. - **Fixed places.** Fill the restricted places first. - **Digits.** A number cannot start with $0$. Split into cases when $0$ meets another condition. - **Rank.** Count the words before it, one place at a time, then add $1$. ## A worked example How many arrangements of BALLOT have the two L's side by side? **Step 1. All arrangements.** $6$ letters with L twice: $\dfrac{6!}{2!} = 360$. **Step 2. Glue the L's.** The block and B, A, O, T make $5$ units, all different: $5! = 120$. **Step 3. Inside the block.** The L's look alike, so there is only one order. **Answer:** $120$ of the $360$ arrangements keep the L's together. ## Practice with new numbers ```practice { "title": "Three letter codes", "vars": { "n": [ 8, 9, 11, 12 ] }, "values": { "P": "n*(n - 1)*(n - 2)", "rep": "n*n*n", "first": "n", "last": "n - 2" }, "question": "A code is $3$ different letters chosen in order from the first {{n}} letters of the alphabet. How many codes are there?", "answer": { "value": "P", "digits": 0 }, "hint": "The letters must be different. How many choices does each place have?", "mistakes": [ { "value": "rep", "say": "That allows a letter to be used twice. Each place has one choice fewer." } ], "solution": "${}^{ {{n}} }P_3 = {{first}} \\times ({{first}} - 1) \\times {{last}} = {{P}}$." } ``` ## A program to try The program lists the arrangements of BALLOT and keeps those where the L's stand together. ```python # Check the worked example of the note: the letters of BALLOT with the two L's together. from itertools import permutations word = "BALLOT" rows = set(permutations(word)) together = [r for r in rows if "LL" in "".join(r)] print("Different arrangements:", len(rows)) print("With the two L's side by side:", len(together)) ``` !try Run it. Does it find $120$? 1. Change `word` to `"BALLOON"`. Predict both counts first. ## Mistakes to avoid - **$n^r$ when objects cannot repeat.** Use ${}^nP_r$. - **$\dfrac{n!}{r!}$ for ${}^nP_r$.** The bottom is $(n - r)!$. - **Dividing by $p$ for a repeated letter.** Divide by $p!$. - **Forgetting the orders inside a block.** Different objects in a block can swap. - **A rank one too small.** Add $1$ to the words before it. ## Check yourself 1. Find ${}^9P_2$. !hint Two factors, starting at $9$. !answer $9 \times 8 = 72$. 2. How many arrangements does the word ADDED have? !hint D three times, A once, E once. !answer $\dfrac{5!}{3!} = 20$. 3. Four friends and a teacher stand in a row, and the teacher must stand at one end. How many rows are possible? !hint Fill the teacher's place first: two ends to choose from. !answer $2 \times 4! = 48$. 4. How many $2$ digit numbers have two different digits, each from $1$ to $9$? !hint No $0$ is used, so each place is free. !answer ${}^9P_2 = 72$. 5. In dictionary order, which arrangement of D, O, G comes first, and which comes last? !hint Put the letters in alphabetical order: D, G, O. !answer DGO comes first and OGD comes last.33:T123e,A brief note on **Animals Without a Backbone**. Read it before a test. Try each question at the end before you open its answer. ## The simplest phyla - **Porifera (sponges):** cellular level; water in by **ostia**, through the **spongocoel**, out by the **osculum**; **collar cells**; spicules or spongin; hermaphrodite. *Sycon*, *Spongilla*, *Euspongia*. - **Cnidaria:** **cnidoblasts** with nematocysts; diploblastic; one opening to the gut; **polyp** and **medusa**; **metagenesis** in *Obelia*. *Physalia*, *Adamsia*, *Pennatula*, *Gorgonia*, *Meandrina*, *Aurelia*. - **Ctenophora (comb jellies):** **eight rows of comb plates**; **bioluminescence**; only sexual reproduction. *Pleurobrachia*, *Ctenoplana*. ## The worms - **Platyhelminthes:** flat, **acoelomate**, hooks and suckers, **flame cells**. *Taenia*, *Fasciola*, *Planaria*. - **Aschelminthes:** round, **pseudocoelomate**, muscular pharynx, **sexes separate**. *Ascaris*, *Wuchereria*, *Ancylostoma*. - **Annelida:** **metameres**, true coelom, **nephridia**, closed circulation, double ventral nerve cord; **parapodia** in *Nereis*. *Pheretima*, *Hirudinaria*. ## The higher non-chordates - **Arthropoda:** the largest phylum; **chitin** exoskeleton, **jointed appendages**, open circulation, **malpighian tubules**. *Apis*, *Bombyx*, *Laccifer*, *Anopheles*, *Culex*, *Aedes*, *Locusta*, *Limulus*. - **Mollusca:** head, foot, visceral hump, **mantle**, **radula**. *Pila*, *Pinctada*, *Sepia*, *Loligo*, *Octopus*. - **Echinodermata:** spiny skin of ossicles, **water vascular system**, **radial adult and bilateral larva**, no excretory system. *Asterias*, *Echinus*, *Antedon*, *Cucumaria*, *Ophiura*. - **Hemichordata:** **proboscis, collar and trunk**; proboscis gland. *Balanoglossus*, *Saccoglossus*. ## A worked example A tray from a field trip holds $13$ insects, $4$ spiders and $3$ prawns. How many legs are on the tray? 1. **Insects.** Three pairs each: $13 \times 6 = 78$ legs. 2. **Spiders.** Four pairs each: $4 \times 8 = 32$ legs. 3. **Prawns.** Five pairs each: $3 \times 10 = 30$ legs. 4. **All the legs.** $78 + 32 + 30 = 140$ legs. All of them are arthropods: their legs are jointed. ## Practice with new numbers ```practice { "title": "Legs in a jar", "vars": { "i": [ 8, 14, 21 ], "s": [ 3, 6, 7 ] }, "values": { "n": "6*i + 8*s", "wrong": "6*(i + s)" }, "question": "A jar from a field trip holds {{i}} insects and {{s}} spiders. How many legs are there in the jar?", "answer": { "value": "n", "unit": "legs", "digits": 0 }, "hint": "An insect has three pairs of legs. A spider has four pairs.", "mistakes": [ { "value": "wrong", "say": "A spider has 8 legs, not 6. Count the two groups apart." } ], "solution": "$6 \\times {{i}} + 8 \\times {{s}} = {{n}}$ legs." } ``` ## A program to try The program names the phylum of three animals from one telling feature each. ```python # A key: one telling feature names the phylum. key = { "collar cells": "Porifera", "stinging cells": "Cnidaria", "comb plates": "Ctenophora", "flame cells": "Platyhelminthes", "jointed legs": "Arthropoda", "water vascular system": "Echinodermata", } animals = {"Spongilla": "collar cells", "Planaria": "flame cells", "Antedon": "water vascular system"} for name, feature in animals.items(): print(name, "->", key.get(feature, "look again")) ``` !try Run it. Which phylum does it give for *Planaria*? 1. Add `"Pila": "radula"` to `animals`. What does the key say? ## Mistakes to avoid - **A jellyfish, a starfish and a cuttlefish are not fish.** None has a backbone. - **Water enters a sponge by the ostia**, not by the osculum. - **Roundworms are pseudocoelomate**, not coelomate. - **Earthworms and leeches are monoecious**; *Nereis* is dioecious. - **Echinoderm larvae are bilateral**; only the adults are radial. ## Check yourself 1. Which phylum has a gastro-vascular cavity with only one opening? !hint It also has stinging cells. !answer Cnidaria. 2. Name the organs of excretion in flatworms, annelids and insects. !hint One is a cell, one a coiled tube, one a set of fine tubes. !answer Flame cells, nephridia and malpighian tubules. 3. Which mollusc is the pearl oyster? !hint Its name begins with P. !answer *Pinctada*. 4. Give the three uses of the water vascular system. !hint Think of what tube feet do. !answer Movement, catching and passing food, and breathing. 5. Why is *Balanoglossus* no longer placed among the chordates? !hint It now has a phylum of its own among the non-chordates. !answer It is now placed in the separate phylum Hemichordata, among the non-chordates.34:Tfc5,A brief note on **Growth and Populations by Computer**. Read it before a test. Try each question at the end before you open its answer. ## Exponential growth - A loop doubles with `N = 2 * N`; the formula `N0 * 2 ** (t / T)` jumps to any time and works on numpy arrays. - With a growth rate: $N = N_0 e^{rt}$, and $T = \dfrac{\ln 2}{r}$. Use `np.log` and `np.exp`, not `np.log10`. - On a log axis, `plt.yscale("log")`, exponential growth is a straight line. ## Logistic growth - $\dfrac{dN}{dt} = rN\left(1 - \dfrac{N}{K}\right)$: nearly exponential while $N$ is small, stopping at the carrying capacity $K$. - An Euler step: `N = N + r * N * (1 - N / K) * dt`. Keep `dt` much smaller than $\dfrac{1}{r}$. - The S-curve grows fastest at $\dfrac{K}{2}$, at $\dfrac{rK}{4}$, and reaches it at $t = \dfrac{\ln(K/N_0 - 1)}{r}$. ## Fitting - `np.polyfit(t, np.log(N), 1)` gives the slope $r$ and the intercept $\ln N_0$ of exponential counts. - `curve_fit(model, t, N, p0=guess)` returns the best parameters in the order of the model's arguments after `t`. - A fit makes the sum of squared misses smallest. ## A worked example What does this print? ```python N = 40 for hour in range(3): N = N + 0.5 * N * (1 - N / 400) * 1 print(round(N)) ``` 1. **Hour 1.** $40 + 0.5 \times 40 \times 0.9 = 58$. 2. **Hour 2.** $58 + 0.5 \times 58 \times 0.855 = 82.795$. 3. **Hour 3.** $82.795 + 0.5 \times 82.795 \times 0.793 = 115.6$. It prints `116`. ## Practice with new numbers ```practice { "title": "Growth with e", "vars": { "n": [ 50, 120 ], "r": [ 0.7, 1.1 ], "h": [ 3, 5 ] }, "values": { "N": "n*exp(r*h)", "wrong": "n*r*h" }, "question": "A population of {{n}} cells grows as $N = N_0 e^{rt}$ with $r = {{r}}$ per hour. What does `n * np.exp(r * t)` give for $t = {{h}}$ h?", "answer": { "value": "N", "unit": "cells", "digits": 0 }, "hint": "Work out $rt$ first, then $e^{rt}$.", "mistakes": [ { "value": "wrong", "say": "The rate goes into the power of $e$; it does not multiply the count directly." } ], "solution": "${{n}} \\times e^{ {{r}} \\times {{h}} } = {{N}}$ cells." } ``` ## A program to try The program grows the same population exponentially and logistically, side by side, and prints both every two hours. ```python # Exponential and logistic growth side by side, hour by hour. import numpy as np N0, r, K = 30, 0.6, 900 # cells, per hour, cells dt = 0.01 # hours N_exp, N_log = N0, N0 for step in range(1, round(12 / dt) + 1): N_exp = N_exp + r * N_exp * dt N_log = N_log + r * N_log * (1 - N_log / K) * dt t = step * dt if abs(t - round(t)) < 1e-9 and round(t) % 2 == 0: print(f"{round(t):2d} h exponential {N_exp:9.0f} logistic {N_log:5.0f} cells") print("doubling time while small:", round(np.log(2) / r, 2), "h") ``` !try Run it. After how many hours do the two columns clearly part? 1. Change `K` to `3000` cells. When do they part now? ## Mistakes to avoid - **`^` for a power.** In Python a power is `**`. - **Hours and minutes mixed.** Keep $t$, $T$ and $dt$ in the same unit as $r$. - **No `dt` in the step.** Without it, each step is one whole unit of time. - **A line fitted to N.** Fit the line to $\ln N$. - **The time not first.** curve_fit puts the times into the model's first argument. ## Check yourself 1. A culture of $25$ cells doubles every $12$ min. How many cells are there after one hour? !hint How many doublings are in 60 minutes? !answer $5$ doublings: $25 \times 32 = 800$ cells. 2. What is the doubling time for $r = 0.2$ per hour? !hint $T = \dfrac{\ln 2}{r}$. !answer $\dfrac{0.693}{0.2} = 3.47$ h. 3. A logistic population has $K = 600$ cells. At what count is it growing fastest? !hint Where is the S-curve steepest? !answer At $\dfrac{K}{2} = 300$ cells. 4. Why fit a straight line to $\ln N$ and not to $N$? !hint What is $\ln$ of $N_0 e^{rt}$? !answer $\ln N = \ln N_0 + rt$ is a straight line in $t$, with slope $r$. $N$ itself curves upward.35:Tf4e,A brief note on **Probability**. Try each question at the end before you open its answer. ## The basics - A **probability** is a number from $0$, impossible, to $1$, certain. - With equally likely outcomes: $P = \dfrac{\text{outcomes we want}}{\text{all outcomes}}$. - **Not:** $P(\text{not } A) = 1 - P(A)$. ## When we cannot count - **Relative frequency** = successes divided by tries. It estimates the probability. - A short run jumps about; a long run settles near the true probability. - A coin or a die has **no memory**: past results do not change the next chance. ## Two events together | Question | How | |---|---| | Two dice, two coins | list all the equally likely pairs and count | | Both, when the events do not affect each other | multiply the chances | | At least one | $1 - P(\text{none})$ | | Two draws without putting back | the second chance uses what is left in the bag | ## A worked example A coin is tossed and a die is rolled. What is the chance of a head and a 5 or more? 1. **All the outcomes.** $2 \times 6 = 12$ equally likely pairs. 2. **The ones we want.** (H, 5) and (H, 6): $2$ pairs. 3. **The chance.** $\dfrac{2}{12} = \dfrac{1}{6}$. 4. **A check.** The events do not affect each other, so multiply: $\dfrac{1}{2} \times \dfrac{2}{6} = \dfrac{1}{6}$. ## Practice with new numbers ```practice { "title": "At least one hit", "vars": { "p": [ 0.2, 0.3, 0.4 ], "n": [ 2, 3, 5 ] }, "values": { "one": "1 - pow(1 - p, n)", "none": "pow(1 - p, n)", "q": "1 - p" }, "question": "A player scores from a free throw with a chance of {{p}}, and each throw is separate. What is the chance of scoring at least once in {{n}} throws?", "answer": { "value": "one", "digits": 3 }, "hint": "Find the chance of missing every throw first: $(1 - {{p}})^{ {{n}} }$.", "mistakes": [ { "value": "none", "say": "That is the chance of missing every throw. Take it from 1." } ], "solution": "$P(\\text{no score}) = {{q}}^{ {{n}} } = {{none}}$, so $P(\\text{at least one}) = 1 - {{none}} = {{one}}$." } ``` ## A program to try The program lists all 12 outcomes of the worked example and counts the ones we want. ```python from itertools import product # Every equally likely way that a coin and a die can land together. outcomes = list(product("HT", [1, 2, 3, 4, 5, 6])) print(len(outcomes), "outcomes") want = [o for o in outcomes if o[0] == "H" and o[1] >= 5] print("head and 5 or more:", want, len(want), "/", len(outcomes)) print("probability:", round(len(want) / len(outcomes), 3)) ``` !try Run it. Does it agree with the worked example? 1. Change `o[1] >= 5` to `o[1] % 2 == 0`, for a head and an even number. ## Mistakes to avoid - **A probability above 1 or below 0.** It cannot be. - **Treating the 11 sums of two dice as equally likely.** The 36 pairs are; the sums are not. - **Adding for "both".** Multiply the chances of events that do not affect each other. - **Forgetting that the bag changes** when the first item is not put back. ## Check yourself 1. A letter is picked at random from the word STATISTICS. What is the probability that it is a T? !hint How many letters are there, and how many are T? !answer There are 10 letters, and 3 of them are T, so $\dfrac{3}{10}$. 2. The chance of a delay is $0.12$. What is the chance of no delay? !hint The two chances add up to 1. !answer $1 - 0.12 = 0.88$. 3. A cup lands upright 54 times in 150 throws. About how many upright landings do you expect in 500 throws? !hint First find the relative frequency. !answer $\dfrac{54}{150} = 0.36$, so about $0.36 \times 500 = 180$. 4. Two cards are drawn from five cards numbered 1 to 5, without putting the first back. What is the chance that both are odd? !hint Three of the five cards are odd. After one odd card is taken, how many odd cards are left among how many? !answer $\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{3}{10}$.36:Tfe9,A brief note on **Modern Physics**. Read it before a test. Try each question at the end before you open its answer. ## The big ideas - Light comes in **photons**: $E = h\nu = \dfrac{hc}{\lambda}$, with $hc = 1240\ \text{eV nm}$. - **Photoelectric effect:** $KE_{\max} = h\nu - \phi = eV_0$. Below the threshold no electrons leave; brightness changes how many, not how fast. - **Bohr's hydrogen:** $E_n = -\dfrac{13.6\ \text{eV}}{n^2}$ ($Z^2$ times that for an ion with one electron). A jump gives a photon of the energy difference. - **Series:** Lyman ends on $n = 1$ (ultraviolet), Balmer on $n = 2$ (visible), Paschen on $n = 3$ (infrared). - **Matter waves:** $\lambda = \dfrac{h}{mv}$; for an electron through $V$ volts, $\lambda = \dfrac{1.227}{\sqrt{V}}\ \text{nm}$. ## The facts at a glance - Photon: $E = \dfrac{1240\ \text{eV nm}}{\lambda}$ - Einstein: $KE_{\max} = h\nu - \phi$, $eV_0 = KE_{\max}$ - Graph of $V_0$ against $\nu$: slope $\dfrac{h}{e}$, the same for every metal - Hydrogen level: $E_n = -\dfrac{13.6\ \text{eV}}{n^2}$ - Lines from level $n$: up to $\dfrac{n(n - 1)}{2}$ - De Broglie: $\lambda = \dfrac{h}{p}$ ## A worked example Light of $300\ \text{nm}$ falls on a metal whose work function is $3.4\ \text{eV}$. Find the largest kinetic energy and the stopping potential. Then find the de Broglie wavelength of an electron speeded up by $81\ \text{V}$. 1. **The photon.** $\dfrac{1240\ \text{eV nm}}{300\ \text{nm}} = 4.13\ \text{eV}$. 2. **Einstein.** $KE_{\max} = 4.13\ \text{eV} - 3.4\ \text{eV} = 0.73\ \text{eV}$, so $V_0 = 0.73\ \text{V}$. 3. **The electron.** $\lambda = \dfrac{1.227}{\sqrt{81}}\ \text{nm} = 0.136\ \text{nm}$. ## Practice with new numbers ```practice { "title": "Work function from the threshold", "vars": { "lam0": [ 275, 520, 680 ] }, "values": { "phi": "1240/lam0", "wrong": "lam0/1240" }, "question": "A metal's threshold wavelength is {{lam0}} nm. What is its work function, in electronvolts? Use $hc = 1240\\ \\text{eV nm}$.", "answer": { "value": "phi", "unit": "eV", "digits": 2 }, "hint": "At the threshold, $\\phi = \\dfrac{hc}{\\lambda_0}$.", "mistakes": [ { "value": "wrong", "say": "Divide $hc$ by the wavelength." } ], "solution": "$\\phi = \\dfrac{1240\\ \\text{eV nm}}{ {{lam0}}\\ \\text{nm} } = {{phi:2}}\\ \\text{eV}$." } ``` ## A program to try ```python import math hc = 1240 # eV nm phi = 3.4 # eV # Step 1: light of 300 nm on the metal ke = hc / 300 - phi print(f"KE max: {ke:.2f} eV, stopping potential {ke:.2f} V") # Step 2: an electron speeded up by 81 V print(f"de Broglie wavelength: {1.227 / math.sqrt(81):.3f} nm") ``` !try Read it, then run it. It repeats the worked example. 1. Change `phi = 3.4` to `phi = 4.5`. What does a negative answer mean? ## Mistakes to avoid - **Thinking brighter light gives faster electrons.** Only the frequency sets $KE_{\max}$. - **Dividing by $n$ instead of $n^2$** in $E_n = -\dfrac{13.6\ \text{eV}}{n^2}$. - **Forgetting $Z^2$** for $\text{He}^+$ and $\text{Li}^{2+}$. - **Forgetting the square root** in $\lambda = \dfrac{1.227}{\sqrt{V}}\ \text{nm}$. ## Check yourself 1. What is the energy of a photon of $207\ \text{nm}$? !hint $E = \dfrac{1240\ \text{eV nm}}{\lambda}$. !answer $E = \dfrac{1240\ \text{eV nm}}{207\ \text{nm}} = 5.99\ \text{eV}$, about $6.0\ \text{eV}$. 2. What is the energy of the $n = 3$ level of $\text{He}^+$? !hint $E_n = -\dfrac{13.6\,Z^2}{n^2}\ \text{eV}$ with $Z = 2$. !answer $E_3 = -\dfrac{13.6 \times 4}{9}\ \text{eV} = -6.04\ \text{eV}$. 3. Why does a dim violet light eject electrons from a metal that a bright red light cannot? !hint One photon frees one electron. Which photon has more energy? !answer Each violet photon has more energy than the work function, while each red photon has less, however many there are. 4. What is the de Broglie wavelength of an electron speeded up from rest by $400\ \text{V}$? !hint $\lambda = \dfrac{1.227}{\sqrt{V}}\ \text{nm}$. !answer $\lambda = \dfrac{1.227}{20}\ \text{nm} = 0.0614\ \text{nm}$.37:Tf16,A brief note on **Systems, Heat, Work and the First Law**. Read it before a test. Try each question at the end before you open its answer. ## Systems - **System:** the part we study. **Surroundings:** everything else. The **boundary** separates them. - **Open:** energy and matter cross. **Closed:** only energy. **Isolated:** neither. - **State variables:** $p$, $V$, $T$ and $n$. - **State functions** ($p$, $V$, $T$, $U$) depend only on the state. **Heat and work** depend on the path. - **Extensive** (mass, volume, $U$) grow with the amount. **Intensive** ($T$, $p$, density) do not. ## Heat, work and U - $U$ is the total energy stored in the system. We measure only $\Delta U$. - **Sign rule:** energy into the system is positive. $q > 0$ for heat absorbed, $w > 0$ for work done on the system. - **Adiabatic:** $q = 0$, so $\Delta U = w$. ## The first law - $\Delta U = q + w$. The energy of an isolated system is constant. - **Constant outside pressure:** $w = -p_\text{ex}\Delta V$. $1$ L bar $= 100$ J, $1$ L atm $= 101.3$ J. - **Free expansion:** $w = 0$. - **Reversible, isothermal, ideal gas:** $w = -2.303\,nRT\log\dfrac{V_f}{V_i}$, the most work possible. - **Isothermal ideal gas:** $\Delta U = 0$, so $q = -w$. ## A worked example A gas absorbs 500 J of heat and expands from 2.0 L to 5.0 L against a constant 1.2 bar. Find $w$ and $\Delta U$. 1. **Volume change.** $\Delta V = 5.0 - 2.0 = 3.0$ L. 2. **Work.** $w = -1.2 \times 3.0 = -3.6$ L bar $= -360$ J. 3. **First law.** $\Delta U = 500 + (-360) = 140$ J. ## Practice with new numbers ```practice { "title": "Heat and work for a gas in a cylinder", "vars": { "q": [ 300, 450, 650 ], "p": [ 1.5, 2.5 ], "dv": [ 0.8, 1.6, 2.4 ] }, "values": { "w": "-p*dv*100", "dU": "q - p*dv*100" }, "question": "A gas absorbs {{q}} J of heat and expands by {{dv}} L against a constant {{p}} bar. What is $\\Delta U$, in J?", "answer": { "value": "dU", "unit": "J", "digits": 0 }, "hint": "First find $w = -p_{ex}\\Delta V$ in J. Then add $q$.", "mistakes": [ { "value": "w", "say": "That is the work alone. Add the heat to get $\\Delta U$." } ], "solution": "$w = -{{p}} \\times {{dv}} \\times 100 = {{w:0}}$ J. $\\Delta U = {{q}} + ({{w:0}}) = {{dU:0}}$ J." } ``` ## A program to try The program works through the example above. ```python # A gas absorbs heat and expands against a constant outside pressure. Find w and dU. q = 500.0 # heat absorbed, J p_ext = 1.2 # bar V1, V2 = 2.0, 5.0 # L w = -p_ext * (V2 - V1) * 100 # 1 L bar = 100 J dU = q + w print(f"Work: w = {w:.0f} J") print(f"Change in internal energy: dU = {dU:.0f} J") ``` !try Run it. Compare each line with the worked example. 1. Change `p_ext` to `2.0`. Does the internal energy still rise? ## Mistakes to avoid - **Dropping the minus sign** in $w = -p_\text{ex}\Delta V$. - **Leaving work in L bar** when $q$ is in J. - **Writing $\Delta q$ or $\Delta w$.** Heat and work are not state functions. - **Thinking $q = 0$ at constant temperature.** For an ideal gas it is $\Delta U$ that is zero. ## Check yourself 1. Is a sealed can of soda in a fridge open, closed or isolated? !hint Can matter cross? Can heat cross? !answer Closed. Heat leaves through the metal, but nothing gets in or out. 2. A gas gives out 30 J of heat and has 70 J of work done on it. What is $\Delta U$? !hint Heat out is negative. Work on the gas is positive. !answer $\Delta U = -30 + 70 = +40$ J. 3. A gas expands by 0.50 L against 4.0 bar. What is $w$? !hint $w = -p_\text{ex}\Delta V$, then 1 L bar = 100 J. !answer $w = -4.0 \times 0.50 = -2.0$ L bar $= -200$ J. 4. An ideal gas expands into a vacuum. Why is $w$ zero? !hint What does the gas push against? !answer There is no outside pressure, so the gas does no work.38:T1125,A brief note on **The Fundamental Principle of Counting**. Read it before a test. Try each question at the end before you open its answer. ## The multiplication principle - An event in $m$ ways, then another in $n$ ways: $m \times n$ ways in all. It extends to any number of stages. - A tree diagram shows why. Every branch splits into the same number of branches. - **With repetition**, each place keeps all its choices: $k$ places with $n$ choices give $n^k$. - **Without repetition**, each place has one choice fewer: $n \times (n - 1) \times (n - 2) \times \dots$ - A number cannot start with $0$. ## The addition principle - A task in $m$ ways OR in $n$ other ways: $m + n$ ways, if no way is in both. - Stages one after another: multiply. Separate cases: add. - Split a hard count into cases that cannot overlap, count each, then add. - Fill the place that has a rule first, such as an even or odd last digit. ## Factorials - $n! = 1 \times 2 \times \dots \times n$, and $0! = 1$. - $n! = n \times (n - 1)!$, so each factorial comes from the one before. - Cancel before multiplying: $\dfrac{n!}{(n - 2)!} = n(n - 1)$. - A product of $k$ numbers counting down from $n$ is $\dfrac{n!}{(n - k)!}$. - $(2n)! \neq 2 \times n!$, and $(a + b)! \neq a! + b!$. ## A worked example How many $4$-digit numbers with four different digits are greater than $6000$? 1. **The first digit.** It must be $6$, $7$, $8$ or $9$: $4$ ways. 2. **The other places.** Nine digits are left for the second place, then $8$ and $7$. 3. **Multiply.** $4 \times 9 \times 8 \times 7 = 2016$ numbers. ## Practice with new numbers ```practice { "title": "Codes of letters and digits", "vars": { "a": [ 3, 4, 5 ], "b": [ 2, 3, 6 ] }, "values": { "ways": "a*a*b", "wrong": "a + a + b" }, "question": "A code is ${{a}}$ choices of letter, then another letter from the same ${{a}}$ (repeats allowed), then one of ${{b}}$ digits. How many codes are there?", "answer": { "value": "ways", "digits": 0 }, "hint": "Three places, one after another: multiply.", "mistakes": [ { "value": "wrong", "say": "The places follow each other, so the counts multiply." } ], "solution": "${{a}} \\times {{a}} \\times {{b}} = {{ways}}$ codes." } ``` ## A program to try The program checks the three rules of this note by listing: a product of choices, a sum of separate cases, and the factorial rule. ```python # Check the rules of the note by listing every case. from itertools import product import math # Multiplication: a 2-letter code from A, B, C, D, then one digit from 1 to 3. codes = list(product("ABCD", "ABCD", "123")) print("codes listed:", len(codes), " 4 x 4 x 3 =", 4 * 4 * 3) # Addition: a snack is either one of 5 fruits or one of 3 biscuits. fruits = ["apple", "banana", "guava", "pear", "plum"] biscuits = ["salt", "sweet", "cream"] print("snacks:", len(fruits + biscuits), " 5 + 3 =", 5 + 3) # Factorials: n! = n x (n - 1)! for n in [1, 4, 7]: print(f"{n}! =", math.factorial(n), " n x (n - 1)! =", n * math.factorial(n - 1)) ``` !try Run it. Do the listed counts agree with the rules? 1. Add the digit `4` to the code's last place. Predict the new count first. ## Mistakes to avoid - **Adding stages.** Choices one after another multiply. - **Keeping every choice without repetition.** Each place loses one choice. - **Overlapping cases.** Add only cases with nothing in common. - **$0! = 0$.** It is $1$. - **Cancelling inside a factorial.** $\dfrac{8!}{4!}$ is $8 \times 7 \times 6 \times 5$, not $2!$. ## Check yourself 1. A password is one letter from A to Z followed by $2$ digits, which may repeat. How many passwords are there? !hint Three places: $26$, $10$ and $10$ ways. !answer $26 \times 10 \times 10 = 2600$ passwords. 2. Dev chooses either one of $6$ films or one of $4$ plays. How many choices does he have? !hint A film or a play: the cases do not overlap. !answer $6 + 4 = 10$ choices. 3. How many $3$-digit numbers with different digits end in $0$? !hint Fill the units place with $0$ first. !answer Hundreds: $9$ ways, tens: $8$ ways, so $72$ numbers. 4. Find $\dfrac{14!}{12!}$. !hint Cancel $12!$. !answer $14 \times 13 = 182$. 5. Why is $0! = 1$? !hint Use $n! = n \times (n - 1)!$ with $n = 1$. !answer $1! = 1 \times 0!$ and $1! = 1$, so $0!$ must be $1$. Also, there is exactly one way to arrange nothing.39:T10be,A brief note on **How Animals Are Classified**. Read it before a test. Try each question at the end before you open its answer. ## Levels and systems - **Cellular** (Porifera), **tissue** (coelenterates, ctenophores), **organ** (Platyhelminthes), **organ system** (annelids to chordates). - **Gut**: incomplete, one opening (flatworms); complete, a mouth and an anus. - **Circulation**: open, blood bathes the tissues (arthropods, molluscs); closed, blood in vessels (annelids, chordates). ## Symmetry - **Asymmetrical**: no plane gives equal halves (most sponges). - **Radial**: any plane through the central axis (coelenterates, ctenophores, adult echinoderms). - **Bilateral**: one plane only, left and right (most other animals). ## Germ layers and the body cavity - **Diploblastic**: ectoderm and endoderm, with mesoglea between (coelenterates). - **Triploblastic**: ectoderm, mesoderm and endoderm (flatworms to chordates). - **Coelomate**: a cavity lined by mesoderm (annelids, molluscs, arthropods, echinoderms, hemichordates, chordates). - **Pseudocoelomate**: mesoderm in scattered pouches (Aschelminthes). **Acoelomate**: no cavity (Platyhelminthes). ## Segments and the notochord - **Metameric segmentation**: segments outside and inside, organs repeated (annelids, arthropods, chordates). - **Notochord**: a rod from the mesoderm on the dorsal side of the embryo; chordates have it, non-chordates do not. ## A worked example A marine animal has true tissues, bilateral symmetry, three germ layers, a true coelom and no notochord. 1. **Tissues.** It is not Porifera. 2. **Bilateral and triploblastic.** It is not a coelenterate or a ctenophore. 3. **A true coelom.** It is not a flatworm or a roundworm. 4. **No notochord.** It is a coelomate non-chordate: Annelida, Arthropoda, Mollusca, Echinodermata or Hemichordata. ## Practice with new numbers ```practice { "title": "Non-chordates in a drawer", "vars": { "n": [ 25, 40, 50 ], "c": [ 6, 10, 12 ] }, "values": { "pct": "100*(n - c)/n", "wrong": "100*c/n" }, "question": "A drawer holds {{n}} specimens, of which {{c}} are chordates. What per cent are non-chordates?", "answer": { "value": "pct", "unit": "%", "digits": 1 }, "hint": "First find how many are not chordates.", "mistakes": [ { "value": "wrong", "say": "That is the share of chordates. Subtract them from all the specimens first." } ], "solution": "${{n}} - {{c}}$ non-chordates, so $\\dfrac{ {{n}} - {{c}} }{ {{n}} } \\times 100 = {{pct}}\\%$." } ``` ## A program to try The program prints the basic features of six groups as a table. ```python # The basic features of the main animal groups, one line each. groups = [ ("Porifera", "cellular", "mostly none", "-", "none"), ("Coelenterata", "tissue", "radial", "2", "none"), ("Platyhelminthes", "organ", "bilateral", "3", "none"), ("Aschelminthes", "organ system", "bilateral", "3", "pseudocoelom"), ("Annelida", "organ system", "bilateral", "3", "coelom"), ("Chordata", "organ system", "bilateral", "3", "coelom"), ] print(f"{'group':<16}{'level':<14}{'symmetry':<13}{'layers':<8}cavity") for name, level, symmetry, layers, cavity in groups: print(f"{name:<16}{level:<14}{symmetry:<13}{layers:<8}{cavity}") ``` !try Run it. Which group in the table has a pseudocoelom? 1. Add a line for Arthropoda. What would you write in each column? ## Mistakes to avoid - **Calling the mesoglea a germ layer.** Coelenterates stay diploblastic. - **Giving flatworms a pseudocoelom.** They have no cavity; roundworms have the pseudocoelom. - **Thinking every coelomate is a chordate.** Only animals with a notochord are. - **Calling a gut with one opening complete.** It needs a mouth and an anus. ## Check yourself 1. Which level of organisation do flatworms reach? !hint Their tissues are grouped into organs. !answer The organ level. 2. Name the kind of circulation of a mollusc. !hint Does the blood stay in vessels? !answer Open circulation. 3. What does a pseudocoelomate lack that a coelomate has? !hint Think of what lines the cavity. !answer A cavity lined all round by mesoderm. 4. Which structure separates chordates from all other animals? !hint It is a rod in the embryo. !answer The notochord.3a:T11aa,A brief note on **The Periodic Table as Data**. Read it before a test. Try each question at the end before you open its answer. ## The table - `pd.read_csv(io.StringIO(text), sep=" ", names=cols)` turns typed text into a DataFrame. `nan` is a missing value. - `df["ie"]` is a column; `df[["symbol", "ie"]]` is a smaller table; `df.set_index("symbol").loc["Na", "ie"]` is one value. - `df.sort_values("radius", ascending=False)` sorts; store the result to keep it. - Filter with `df[condition]`; join conditions with `&` or `|`, each in brackets. - A computed column: `df["ie_eV"] = df["ie"] / 96.49`, for every row at once. ## The plots - `plt.plot(df["z"], df["ie"], "o-")`; label each axis with its unit. - `plt.scatter(..., c=df["block"].map(colours))` colours points by block. - Ionisation enthalpy peaks at the noble gases and dips at the alkali metals after them. - Atomic radius falls across a period and rises down a group. ## The patterns - `df.groupby("block")["ie"].mean()`: the mean of each block. - `df["radius"].corr(df["ie"])`: $r$ from $-1$ to $1$; here strongly negative. - `df.groupby("period")["ie"].diff()`: the change from the element before, inside each period. A negative value is a break: boron and oxygen in period 2. ## A worked example What does this print? ```python import pandas as pd t = pd.DataFrame({"sym": ["Na", "Mg", "Al", "Si"], "ie": [496, 738, 577, 786]}) t["rise"] = t["ie"].diff() print(list(t[t["rise"] < 0]["sym"])) ``` 1. **The changes.** Magnesium $+242$, aluminium $-161$, silicon $+209$; sodium has none. 2. **The filter.** Only aluminium is below zero. 3. **The reason.** Aluminium's outer electron is in a 3p orbital, higher in energy than magnesium's full 3s. It prints `['Al']`. ## Practice with new numbers ```practice { "title": "A change between neighbours", "vars": { "p": [ 1402, 1012 ], "q": [ 1314, 1000 ] }, "values": { "d": "q - p", "wrong": "p - q" }, "question": "In the column `ie`, one row holds {{p}} kJ/mol and the next row holds {{q}} kJ/mol. What does `diff()` put in the second row?", "answer": { "value": "d", "unit": "kJ/mol", "digits": 0 }, "hint": "diff() takes each value minus the one above it.", "mistakes": [ { "value": "wrong", "say": "It is the new value minus the old one, so a fall comes out negative." } ], "solution": "${{q}} - {{p}} = {{d}}$ kJ/mol: a fall, so this element breaks the rise." } ``` ## A program to try The program finds the break in the rise across the p-block of period 4. ```python # Where does the rise of ionisation enthalpy break in period 4's p-block? import pandas as pd ie = pd.Series([579, 762, 947, 941, 1140, 1351], index=["Ga", "Ge", "As", "Se", "Br", "Kr"]) rise = ie.diff() print(rise) print("Breaks at:", list(rise[rise < 0].index)) ``` !try Run it. Which element breaks the rise? 1. Print the mean of the series as well. Is it closer to gallium's value or to krypton's? ## Mistakes to avoid - **A capital letter in a column name.** `df["IE"]` stops with `KeyError: 'IE'`. - **`and` between conditions.** It stops with `ValueError: The truth value of a Series is ambiguous`; use `&` and brackets. - **diff() across periods.** Group by period first, or a new shell looks like a break. - **Missing values in a plot.** Filter with `notna()` before plotting the radius. - **Reading a correlation as a cause.** It says that two columns move together, no more. ## Check yourself 1. Which line keeps the d-block elements whose first ionisation enthalpy is above $700$ kJ/mol? !hint Two conditions, each in brackets. !answer `df[(df[\"block\"] == \"d\") & (df[\"ie\"] > 700)]`. 2. In a plot coloured by block, what does the flat green stretch of period 4 tell you? !hint Where do the added electrons go in the d-block? !answer The added electrons go into the inner 3d orbitals, so the outer 4s electrons feel almost the same pull and the values hardly change. 3. `df["radius"].corr(df["ie"])` gives a value near $-0.8$. What does that mean in words? !hint What does the minus sign say about the two columns? !answer Large atoms tend to have low first ionisation enthalpies, and small atoms high ones. The link is strong but not perfect. 4. Why is magnesium's value higher than aluminium's, though aluminium comes later in the period? !hint Which orbital holds the electron that is removed? !answer Aluminium's outer electron is in a 3p orbital, higher in energy and easier to remove than magnesium's paired 3s electron.3b:T1064,A brief note on **Range, IQR and Standard Deviation**. An average says where the data is; the spread says how far apart the values are. Try each question at the end before you open its answer. ## Three measures of spread | Measure | How to find it | Good to know | |---|---|---| | Range | largest $-$ smallest | one extreme value makes it huge | | IQR | $Q_3 - Q_1$ | the width of the middle half; outliers hardly move it | | Standard deviation | square root of the mean of the squared distances from the mean | uses every value; same units as the data | - **Quartiles.** Sort the data. $Q_1$ is the median of the lower half, $Q_3$ of the upper half. With an odd count, leave the median out of both halves. - **Box plot.** The smallest value, $Q_1$, the median, $Q_3$ and the largest value. The box is the middle half. - **Pairs.** The mean goes with the standard deviation; the median with the IQR. ## When every value changes - **Add a number to every value:** the averages move, the spread does not. - **Multiply every value by $k$:** the averages and the spread are all multiplied by $k$. ## A worked example Nine students need $11, 13, 14, 14, 16, 17, 19, 22$ and $41$ minutes to reach school. 1. **Range.** $41 - 11 = 30$ minutes. 2. **Quartiles.** The median, $16$, is left out. The lower half $11, 13, 14, 14$ gives $Q_1 = 13.5$; the upper half $17, 19, 22, 41$ gives $Q_3 = 20.5$. 3. **IQR.** $20.5 - 13.5 = 7$ minutes, much smaller than the range, because one student needs 41 minutes. 4. **Which to report.** With that outlier, give the median, 16 minutes, with the IQR, 7 minutes. ## Practice with new numbers ```practice { "title": "The standard deviation of three values", "vars": { "m": [ 10, 20, 30 ], "d": [ 2, 3, 5 ] }, "values": { "sd": "sqrt(2*d*d/3)", "lo": "m - d", "hi": "m + d", "var": "2*d*d/3", "ss": "2*d*d" }, "question": "What is the standard deviation of {{lo}}, {{m}} and {{hi}}?", "answer": { "value": "sd", "digits": 2 }, "hint": "The mean is {{m}}. The distances are $-{{d}}$, $0$ and ${{d}}$.", "mistakes": [ { "value": "d", "say": "That is the largest distance. Average the squares of all three distances, then take the square root." } ], "solution": "The squares add up to ${{ss}}$, so the variance is $\\dfrac{ {{ss}} }{3} = {{var}}$ and the standard deviation is $\\sqrt{ {{var}} } = {{sd}}$." } ``` ## A program to try The program works out the worked example, and the standard deviation too. ```python import statistics def quartiles(data): data = sorted(data) half = len(data) // 2 return statistics.median(data[:half]), statistics.median(data[-half:]) times = [11, 13, 14, 14, 16, 17, 19, 22, 41] # minutes q1, q3 = quartiles(times) print("range:", max(times) - min(times)) print("Q1:", q1, " Q3:", q3, " IQR:", q3 - q1) print("standard deviation:", round(statistics.pstdev(times), 2)) ``` !try Run it. Which measure is pulled up most by the 41? 1. Change the `41` to `24`. Which measures change, and which stay? ## Mistakes to avoid - **Forgetting to sort** before finding the quartiles. - **Adding the distances from the mean.** They always make 0; square them first. - **Stopping at the variance.** The standard deviation is its square root. - **Thinking that adding to every value changes the spread.** It only moves the data. ## Check yourself 1. What is the range of $8, 3, 15, 6$ and $11$? !hint Largest minus smallest. !answer $15 - 3 = 12$. 2. A box plot has the five numbers $4, 9, 13, 20, 31$. What is the IQR? !hint The box runs from Q1 to Q3. !answer $20 - 9 = 11$. 3. Some marks have a standard deviation of $7$. Every mark is multiplied by $3$, and then $5$ is added. What is the new standard deviation? !hint Which of the two changes stretches the gaps? !answer $3 \times 7 = 21$. Adding 5 does not change it. 4. Two shops sell the same mean number of loaves a day. Shop A's daily sales have a standard deviation of 4 loaves and shop B's 15 loaves. Which shop can plan its baking more easily? !hint Which shop's sales stay closer to the mean? !answer Shop A. Its daily sales stay much closer to the mean, so each day is easier to predict.3c:T10f9,A brief note on **Electric Circuits**. Read it before a test. Try each question at the end before you open its answer. ## The big ideas - **Ohm's law:** $V = IR$. The resistance of a wire is $R = \rho\dfrac{L}{A}$; power is $P = VI = I^2R = \dfrac{V^2}{R}$. - **In series** the same current flows and the resistances add. **In parallel** the same voltage is across every branch and the reciprocals add. - **Kirchhoff:** the current into a junction equals the current out; round any loop the rises and falls of potential add to zero. - **A real cell:** $V = E - Ir$. A load takes the most power when $R = r$. - **A balanced Wheatstone bridge:** $\dfrac{P}{Q} = \dfrac{R}{S}$. ## The facts at a glance - **Series:** $R = R_1 + R_2 + \ldots$ - **Parallel:** $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots$ - **Junction rule:** $\sum I_{\text{in}} = \sum I_{\text{out}}$ - **Loop rule:** $\sum \Delta V = 0$ - **Terminal voltage:** $V = E - Ir$ - **Most power to a load:** $R = r$, $P_{\text{max}} = \dfrac{E^2}{4r}$ ## A worked example A cell of emf $10\ \text{V}$ and internal resistance $0.5\ \Omega$ feeds a $3\ \Omega$ and a $6\ \Omega$ resistor in parallel. Find the current from the cell, its terminal voltage, and the current in the $3\ \Omega$ resistor. 1. **The pair.** $\dfrac{3\ \Omega \times 6\ \Omega}{3\ \Omega + 6\ \Omega} = 2\ \Omega$. 2. **The current.** $I = \dfrac{10\ \text{V}}{2\ \Omega + 0.5\ \Omega} = 4\ \text{A}$. 3. **The terminal voltage.** $V = 10\ \text{V} - 4\ \text{A} \times 0.5\ \Omega = 8\ \text{V}$. 4. **The $3\ \Omega$ branch.** $I_3 = \dfrac{8\ \text{V}}{3\ \Omega} = 2.67\ \text{A}$. ## Practice with new numbers ```practice { "title": "Equal resistors in parallel", "vars": { "N": [ 3, 4, 5 ], "R": [ 12, 30 ], "V": [ 6, 9 ] }, "values": { "I": "N*V/R", "one": "V/R" }, "question": "{{N}} resistors of {{R}} Ω each are joined in parallel across {{V}} V. What current does the supply give, in amperes?", "answer": { "value": "I", "unit": "A", "digits": 2 }, "hint": "Each branch has the full voltage and takes $\\dfrac{V}{R}$; the branch currents add.", "mistakes": [ { "value": "one", "say": "That is one branch. Add the current of every branch." } ], "solution": "$I = {{N}} \\times \\dfrac{ {{V}}\\ \\text{V} }{ {{R}}\\ \\Omega } = {{I:2}}\\ \\text{A}$." } ``` ## A program to try ```python # A cell of emf 10 V with 0.5 ohm inside feeds 3 ohm and 6 ohm in parallel. E, r = 10.0, 0.5 # V, ohm # Step 1: the parallel pair pair = 3 * 6 / (3 + 6) # Step 2: the current, with the internal resistance in the loop I = E / (pair + r) # Step 3: the terminal voltage, and the current in the 3 ohm branch V = E - I * r print("Pair:", pair, "ohm current:", I, "A terminal voltage:", V, "V") print("Current in 3 ohm:", round(V / 3, 2), "A") ``` !try Read it, then run it. It repeats the worked example. 1. Change `r = 0.5` to `r = 0`. How much higher is the terminal voltage now? ## Mistakes to avoid - **Adding resistors in parallel.** The total is always smaller than the smallest branch. - **Forgetting the internal resistance** in $I = \dfrac{E}{R + r}$. - **Fearing a negative current.** It only means the current flows against your arrow. - **Taking the emf as the terminal voltage** while a current flows. ## Check yourself 1. Two cells, each $1.5\ \text{V}$ with $0.5\ \Omega$ inside, are joined in series to a $5\ \Omega$ lamp. What current flows? !hint In series both the emfs and the internal resistances add. !answer $I = \dfrac{3\ \text{V}}{5\ \Omega + 1\ \Omega} = 0.5\ \text{A}$. 2. A current of $7\ \text{A}$ flows into a junction, and two wires take current out. One carries $2.5\ \text{A}$. What does the other carry? !hint The current in equals the current out. !answer $7\ \text{A} - 2.5\ \text{A} = 4.5\ \text{A}$. 3. A Wheatstone bridge has $P = 50\ \Omega$, $Q = 75\ \Omega$ and $R = 40\ \Omega$. What $S$ balances it? !hint At balance, $\dfrac{P}{Q} = \dfrac{R}{S}$. !answer $S = \dfrac{75\ \Omega \times 40\ \Omega}{50\ \Omega} = 60\ \Omega$. 4. A battery has an internal resistance of $0.8\ \Omega$. Which load takes the most power from it? !hint The power in the load is largest when the load matches the battery. !answer A load of $0.8\ \Omega$, equal to the internal resistance.3d:T1270,A brief note on **Molecular Orbitals and Hydrogen Bonding**. Read it before a test. Try each question at the end before you open its answer. ## Molecular orbitals - Atomic orbitals combine into the same number of molecular orbitals. - Bonding: $\psi_A + \psi_B$, lower energy. Antibonding: $\psi_A - \psi_B$, higher energy, starred. - They combine only with similar energy, the same symmetry and good overlap. - Up to $\text{N}_2$: $\sigma 2s$, $\sigma^* 2s$, $\pi 2p$, $\sigma 2p$, $\pi^* 2p$, $\sigma^* 2p$. - For $\text{O}_2$ and $\text{F}_2$: $\sigma 2p$ comes before $\pi 2p$. - Fill by the Aufbau, Pauli and Hund rules. ## Bond order and magnetism - Bond order $= \dfrac{1}{2}(N_b - N_a)$. - Zero: no stable molecule ($\text{He}_2$, $\text{Be}_2$, $\text{Ne}_2$). - Higher bond order: stronger and shorter bond. - Unpaired electrons: paramagnetic ($\text{O}_2$, $\text{B}_2$). All paired: diamagnetic ($\text{N}_2$, $\text{C}_2$). - $\text{O}_2^+$, $\text{O}_2$, $\text{O}_2^-$ and $\text{O}_2^{2-}$ have bond orders $2.5$, $2$, $1.5$ and $1$. ## The hydrogen bond - $\text{H}$ joined to $\text{F}$, $\text{O}$ or $\text{N}$, attracted to a lone pair on another such atom. - Intermolecular: HF chains, the water network. Intramolecular: ortho-nitrophenol. - It lifts the boiling points of $\text{H}_2\text{O}$, HF and $\text{NH}_3$, and makes ice less dense than water. ## A worked example Find the bond order and the magnetism of the superoxide ion, $\text{O}_2^-$. 1. **Electrons.** Oxygen has $16$, plus one: $17$. 2. **Fill.** $\sigma 1s^2$ $\sigma^* 1s^2$ $\sigma 2s^2$ $\sigma^* 2s^2$ $\sigma 2p^2$ $\pi 2p^4$ $\pi^* 2p^3$. 3. **Count.** Bonding $2 + 2 + 2 + 4 = 10$; antibonding $2 + 2 + 3 = 7$. 4. **Bond order.** $\dfrac{10 - 7}{2} = 1.5$. 5. **Magnetism.** Three electrons in two $\pi^*$ orbitals leave one unpaired: paramagnetic. ## Practice with new numbers ```practice { "title": "Bond order of an oxygen species", "vars": { "q": [ -2, -1, 1, 2 ] }, "values": { "bo": "(4 + q)/2", "wrong": "(4 - q)/2" }, "question": "A species made of two oxygen atoms carries a charge of {{q}}. What is its bond order?", "answer": { "value": "bo", "digits": 1 }, "hint": "Neutral oxygen has 10 bonding and 6 antibonding electrons. A charge changes the antibonding count.", "mistakes": [ { "value": "wrong", "say": "A positive charge removes antibonding electrons, so the bond order rises." } ], "solution": "Antibonding electrons $= 6 - ({{q}})$, so the bond order is $\\dfrac{4 + ({{q}})}{2} = {{bo:1}}$." } ``` ## A program to try The program lists the bond order, bond length and bond enthalpy of $\text{N}_2$, $\text{O}_2$ and $\text{F}_2$, and sorts them by bond order. ```python # Bond order, bond length and bond enthalpy for three diatomic molecules (NCERT data) molecules = {"N2": (3, 110, 946), "O2": (2, 121, 498), "F2": (1, 144, 155)} for name, (order, length, enthalpy) in molecules.items(): print(f"{name}: bond order {order}, length {length} pm, enthalpy {enthalpy} kJ/mol") ordered = sorted(molecules, key=lambda m: molecules[m][0], reverse=True) print("From the highest bond order:", ordered) ``` !try Run it. As the bond order falls, what happens to the length and the enthalpy? 1. Which molecule needs the least energy to break its bond? ## Mistakes to avoid - **One energy order for all.** Oxygen and fluorine put $\sigma 2p$ below $\pi 2p$. - **Forgetting to halve the difference.** Two electrons make one bond. - **Pairing oxygen's $\pi^*$ electrons.** They stay unpaired, so $\text{O}_2$ is paramagnetic. - **Every ion weaker than its molecule.** $\text{O}_2^+$ is stronger than $\text{O}_2$. - **Hydrogen bonds with carbon or chlorine.** Strong ones need $\text{F}$, $\text{O}$ or $\text{N}$. ## Check yourself 1. What is the bond order of $\text{Li}_2$, and is it paramagnetic? !hint Six electrons fill $\sigma 1s$, $\sigma^* 1s$ and $\sigma 2s$. !answer Bond order $\dfrac{4 - 2}{2} = 1$; every electron is paired, so it is diamagnetic. 2. Why does $\text{N}_2^+$ have a weaker bond than $\text{N}_2$, while $\text{O}_2^+$ has a stronger bond than $\text{O}_2$? !hint Which kind of orbital loses the electron in each case? !answer $\text{N}_2$ loses a bonding electron, but $\text{O}_2$ loses an antibonding one. 3. Why does ammonia dissolve so well in water? !hint What can ammonia and water molecules form with each other? !answer Ammonia molecules form hydrogen bonds with water molecules. 4. Which is the more volatile of the two nitrophenols, and why? !hint Where does each form its hydrogen bond? !answer Ortho-nitrophenol: its hydrogen bond is inside the molecule, so its molecules are not held to each other.3e:T100a,A brief note on **Special Angles and Trigonometric Identities**. Read it before a test. Try each question at the end before you open its answer. ## The special angles - Half a square, sides $1, 1, \sqrt{2}$: $\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}}$ and $\tan 45^\circ = 1$. - Half an equilateral triangle, sides $1, \sqrt{3}, 2$: $\sin 30^\circ = \cos 60^\circ = \dfrac{1}{2}$ and $\sin 60^\circ = \cos 30^\circ = \dfrac{\sqrt{3}}{2}$. - $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ and $\tan 60^\circ = \sqrt{3}$. - $\sin 0^\circ = \cos 90^\circ = 0$ and $\sin 90^\circ = \cos 0^\circ = 1$. - $\tan 90^\circ$ is not defined, because $\cos 90^\circ = 0$. - The sines of $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ are $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$. ## The identities - $\sin^2 A + \cos^2 A = 1$, for every angle. - $1 + \tan^2 A = \sec^2 A$, for $A < 90^\circ$. - $1 + \cot^2 A = \text{cosec}^2 A$, for $A > 0^\circ$. - For an acute angle, take the positive square root. - $\sin(90^\circ - A) = \cos A$ and $\tan(90^\circ - A) = \cot A$. ## Proving an identity - Start with the harder side and work on it alone. - Write everything in $\sin A$ and $\cos A$ when stuck. - Look for $1 - \sin^2 A$, $\sec^2 A - 1$ and $\text{cosec}^2 A - 1$: each is a single term. - One angle where a claim fails shows it is not an identity. ## A worked example Prove that $\dfrac{\cos A}{1 - \sin A} = \sec A + \tan A$. 1. **Start with the right side.** $\sec A + \tan A = \dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A} = \dfrac{1 + \sin A}{\cos A}$. 2. **Multiply top and bottom by $1 - \sin A$.** $\dfrac{(1 + \sin A)(1 - \sin A)}{\cos A (1 - \sin A)} = \dfrac{1 - \sin^2 A}{\cos A(1 - \sin A)}$. 3. **Use the identity.** $\dfrac{\cos^2 A}{\cos A(1 - \sin A)} = \dfrac{\cos A}{1 - \sin A}$, the left side. ## Practice with new numbers ```practice { "title": "An angle from its cosine", "vars": { "k": [ 2, 3, 4 ] }, "values": { "A": "60/k", "wrong": "30/k" }, "question": "If $2\\cos({{k}}A) = 1$ and ${{k}}A$ is acute, find $A$.", "answer": { "value": "A", "unit": "°", "digits": 1 }, "hint": "$\\cos({{k}}A) = \\dfrac{1}{2}$. Which angle has a cosine of $\\dfrac{1}{2}$?", "mistakes": [ { "value": "wrong", "say": "$\\dfrac{1}{2}$ is the sine of $30^\\circ$ but the cosine of $60^\\circ$." } ], "solution": "${{k}}A = 60^\\circ$, so $A = {{A}}^\\circ$." } ``` ## A program to try The program prints the sine and cosine of $30^\circ$, $45^\circ$ and $60^\circ$, and checks the first identity for each. ```python # The special angles and one identity, in a few lines import math for deg in [30, 45, 60]: A = math.radians(deg) s, c = math.sin(A), math.cos(A) print(deg, "degrees: sin =", round(s, 3), " cos =", round(c, 3), " sin^2 + cos^2 =", round(s * s + c * c, 3)) ``` !try Run it. Does the last column always show $1$? 1. Add $15^\circ$ and $75^\circ$ to the list. Does the identity still hold? ## Mistakes to avoid - **Swapping $\sin 30^\circ$ and $\cos 30^\circ$.** - **Writing $\tan 90^\circ = 0$.** - **Writing $\sin A + \cos A = 1$.** Only the squares add to $1$. - **Pairing $\tan$ with $\text{cosec}$.** It is $\tan$ with $\sec$, and $\cot$ with $\text{cosec}$. ## Check yourself 1. Find $\sin^2 45^\circ + \tan^2 30^\circ$. !hint $\sin^2 45^\circ = \dfrac{1}{2}$ and $\tan^2 30^\circ = \dfrac{1}{3}$. !answer $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$. 2. If $\cot A = \dfrac{7}{24}$ for an acute angle, find $\text{cosec}\,A$. !hint $\text{cosec}^2 A = 1 + \cot^2 A$. !answer $\text{cosec}^2 A = \dfrac{576 + 49}{576} = \dfrac{625}{576}$, so $\text{cosec}\,A = \dfrac{25}{24}$. 3. Simplify $\tan^2 A \cos^2 A + \cos^2 A$. !hint $\tan^2 A \cos^2 A = \sin^2 A$. !answer $\sin^2 A + \cos^2 A = 1$. 4. Is $\tan A + \cot A = \sec A\,\text{cosec}\,A$ an identity? !hint Write the left side over the common denominator $\sin A \cos A$. !answer Yes. $\dfrac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \dfrac{1}{\sin A\cos A} = \sec A\,\text{cosec}\,A$.3f:Tfaf,A brief note on **How Traits Are Expressed, and Sex Determination**. Read it before a test. Try each question at the end before you open its answer. ## From gene to trait - **DNA** is the information source for making **proteins**. - A **gene** is a section of DNA for one protein. - An **enzyme** is a protein. In a pea plant, one enzyme makes a **growth hormone**. - Efficient enzyme, much hormone: a **tall** plant. Poor enzyme, little hormone: a **short** plant. - A body cell has **two copies** of every gene, one from each parent. So both parents shape each trait. ## Chromosomes and inheritance - Genes lie on **chromosomes**, which come in **pairs** in body cells. - Human body cells: $46$ chromosomes, $23$ pairs. - A **gamete** carries one chromosome of each pair: $23$ in humans. - Fertilisation gives the full number again, so it stays **constant**. - Genes on different chromosomes are passed on **independently**, which gives new mixtures of traits. ## How sex is determined - **Environment:** in some reptiles the temperature of the eggs decides; some snails change sex. - **Humans:** $22$ matched pairs and one pair of sex chromosomes. - Women are **XX**; men are **XY**, with a short Y. - Every egg carries **X**. Half the sperm carry X, half carry Y. - X sperm: a girl (XX). Y sperm: a boy (XY). The **father's sperm** decides. - Each birth has an **equal chance** of a girl or a boy. ## A worked example A body cell of an onion has $16$ chromosomes. How many does its gamete carry, and how many does the zygote have? 1. **The gamete.** One chromosome of each pair: $\dfrac{16}{2} = 8$. 2. **The zygote.** One gamete from each parent: $8 + 8 = 16$. 3. **The point.** The number returns to $16$, so it stays constant. ## Practice with new numbers ```practice { "title": "Chromosomes in a zygote", "vars": { "g": [ 7, 12, 19, 23 ] }, "values": { "z": "2*g", "wrong": "g/2" }, "question": "An egg and a sperm of a species each carry {{g}} chromosomes. How many chromosomes does the zygote have?", "answer": { "value": "z", "unit": "chromosomes", "digits": 0 }, "hint": "The zygote gets one set from each gamete.", "mistakes": [ { "value": "wrong", "say": "Fertilisation adds the two gametes together. It does not halve the number." } ], "solution": "${{g}} + {{g}} = {{z}}$ chromosomes." } ``` ## A program to try The program prints the pairs and the gamete number for three organisms. ```python # Chromosome numbers in body cells and gametes. body_cells = {"cat": 38, "horse": 64, "potato": 48} for name, n in body_cells.items(): print(name, ":", n, "in a body cell,", n // 2, "pairs,", n // 2, "in a gamete") ``` !try Run it. How many chromosomes does a gamete of a horse carry? 1. Add `"rice": 24` to the dictionary. What does it print for rice? ## Mistakes to avoid - **A gene is not the trait.** It is information that works through a protein. - **A gamete does not carry full pairs.** It carries one of each pair. - **The mother does not decide the sex.** All her eggs carry X. - **Earlier births do not change the chance.** It is one half each time. - **Not all animals use genes to decide sex.** Some reptiles use temperature. ## Check yourself 1. Name the kind of protein that speeds up the making of a growth hormone. !hint It speeds up a chemical reaction. !answer An enzyme. 2. How many chromosomes does a human egg carry? !hint One chromosome of each of the 23 pairs. !answer $23$ chromosomes. 3. Which sex chromosomes does a woman have? !hint They are a matched pair. !answer Two X chromosomes, XX. 4. Why is the chance of a girl one half at every birth? !hint Think about the two kinds of sperm. !answer The X sperm and the Y sperm are made in equal numbers, and either may reach the egg. 5. Why does a child of a tall parent and a short parent get genes from both? !hint Where did the zygote come from? !answer The zygote formed from a sperm and an egg, so it has one copy of every gene from each parent.40:T112b,A brief note on **Moles and Reactions by Computer**. Read it before a test. Try each question at the end before you open its answer. ## Formulas as data - Keep atomic masses in a dictionary: `atomic = {"H": 1.008, "C": 12.01, "O": 16.00}`. - Read a formula with `re.findall(r"([A-Z][a-z]?)(\d*)", formula)`: one pair for each symbol, and an empty count means one. - Open brackets first: the number after a bracket multiplies every atom inside. - Molar mass: `sum(atomic[el] * n for el, n in counts.items())`. Mass percents add up to $100$. ## Reactions as functions - Keep an equation as two dictionaries of coefficients. - Rounds $=$ moles $\div$ coefficient. `min(rounds, key=rounds.get)` gives the limiting reagent. - Products: extent $\times$ coefficient; excess left: moles $-$ extent $\times$ coefficient. - Percent yield $= \dfrac{\text{actual}}{\text{theoretical}} \times 100$; at STP a mole of gas fills $22.7\ \text{L}$. ## Solutions - Moles $=$ molarity $\times \dfrac{V \text{ in mL}}{1000}$. - Dilution: $M_1V_1 = M_2V_2$. Mixing: add the moles and the volumes. - End point: moles of base $=$ moles of acid $\times$ the ratio of the equation; a default argument `ratio=1` keeps the common case short. ## A worked example What does this print? ```python moles = {"P4": 0.10, "O2": 0.60} coef = {"P4": 1, "O2": 5} rounds = {s: moles[s] / coef[s] for s in coef} extent = min(rounds.values()) print(min(rounds, key=rounds.get), round(extent, 2)) ``` 1. **Rounds.** Phosphorus $\dfrac{0.10}{1} = 0.10$; oxygen $\dfrac{0.60}{5} = 0.12$. 2. **Limiting.** Phosphorus has fewer rounds, so `min` returns `'P4'`. 3. **Product.** In $\text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}$ the product's coefficient is $1$, so $0.1$ mol of $\text{P}_4\text{O}_{10}$ forms. It prints `P4 0.1`. ## Practice with new numbers ```practice { "title": "Molarity from grams", "vars": { "g": [ 2.4, 4.8, 7.2 ], "v": [ 100, 250, 500 ] }, "values": { "mol": "g/120.37/(v/1000)" }, "question": "{{g}} g of magnesium sulphate, $\\text{MgSO}_4$ ($120.37$ g/mol), is dissolved to make {{v}} mL of solution. What molarity does the program print?", "answer": { "value": "mol", "unit": "mol/L", "digits": 3 }, "hint": "Change grams to moles, and millilitres to litres.", "solution": "$\\dfrac{ {{g}} / 120.37}{ {{v}} / 1000} = {{mol}}$ mol/L." } ``` ## A program to try The program reads a formula, finds its molar mass and the molarity of a solution made from it. ```python import re atomic = {"K": 39.10, "N": 14.01, "O": 16.00} def molar_mass(formula): return sum(atomic[el] * int(n or 1) for el, n in re.findall(r"([A-Z][a-z]?)(\d*)", formula)) f, grams, volume_mL = "KNO3", 10.1, 250 M = molar_mass(f) moles = grams / M print(f"{f}: {M:.2f} g/mol, {moles:.4f} mol") print(f"Molarity: {moles / (volume_mL / 1000):.3f} mol/L") ``` !try Run it. Check the molar mass of potassium nitrate by hand. 1. Change the mass to `20.2` g. What happens to the molarity? ## Mistakes to avoid - **A missing element.** `KeyError: 'K'` means the dictionary has no potassium. - **`int('')`.** An empty count raises `ValueError`; write `int(n or 1)`. - **Comparing moles instead of rounds.** Divide by the coefficient first. - **Millilitres as litres.** Divide by $1000$. - **Adding molarities when mixing.** Add the moles, then divide by the total volume. ## Check yourself 1. How many atoms of each element does a correct reader count in `Zn(NO3)2`? !hint The 2 multiplies everything in the bracket. !answer One zinc, two nitrogen and six oxygen. 2. For $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ with $1$ mol of each gas, which gas does the function name as limiting? !hint Rounds: nitrogen 1, hydrogen one third. !answer Hydrogen, with $\dfrac{1}{3}$ of a round against $1$ for nitrogen. 3. A function returns `6000.0` mol for $0.3\ \text{mol/L}$ in $20\ \text{L}$. What is wrong, and what is the right answer? !hint Was the volume in litres or in millilitres? !answer It multiplied the volume by $1000$ instead of using litres as they are. With the volume in litres, moles $= 0.3 \times 20 = 6$ mol. 4. Why should you test a molar mass function on a formula with brackets? !hint Which part of the code handles brackets? !answer The bracket code is the part most likely to be wrong, and a formula without brackets never runs it.41:T122f,A brief note on **Data Tables, Kinds of Data and Fair Samples**. Read it before a test. Try each question at the end before you open its answer. ## Tables - **Data** is a set of facts. We keep it in a **table**. - A **row** is one **case**, such as one student or one day. A **column** is one **variable**, one fact about every case. - The number of facts is rows $\times$ columns. - A good table has one fact in a cell, one unit in a column, and a blank or NA for a missing value. Never a zero. ## The kinds of data | Kind | What it is | Examples | |---|---|---| | Categorical, nominal | labels with no order | colour, city, PIN code | | Categorical, ordinal | labels with an order | shirt size, poor to good | | Numerical, discrete | counted, whole steps | siblings, goals | | Numerical, continuous | measured, any value | height, time, mass | A number used as a label, such as a roll number, is categorical. Ask: would you ever add two of them? ## Samples - The **population** is everyone we want to know about. The **sample** is the few we ask. - A **fair** sample gives everyone the same chance, so we choose it **by chance**. - A **biased** sample leaves out a kind of person, such as asking only one club or one place. - A **fair question** is neutral. It asks for a fact and does not push the answer. ## A worked example A survey table has 35 rows. Its columns are the student's name, house colour, number of cousins and time to run 50 m. 1. **Facts.** $35 \times 4 = 140$. 2. **Kinds.** The name and the house colour are categorical, nominal. The number of cousins is counted, so it is discrete. The running time is measured, so it is continuous. 3. **Fairness.** If the 35 students were the first 35 to reach the sports ground, the fastest runners may be over-counted. Choosing 35 names by chance from the school list is fairer. ## Practice with new numbers ```practice { "title": "A table that grows", "vars": { "r": [ 12, 15, 21 ], "c": [ 4, 6 ], "a": [ 3, 5 ] }, "values": { "facts": "(r + a)*c", "wrong": "r*c + a" }, "question": "A table has {{r}} rows and {{c}} columns. Then {{a}} more rows are added. How many facts does it hold now?", "answer": { "value": "facts", "unit": "facts", "digits": 0 }, "hint": "Each new row also has {{c}} facts.", "mistakes": [ { "value": "wrong", "say": "Each new row brings {{c}} facts, not 1." } ], "solution": "$({{r}} + {{a}}) \\times {{c}} = {{facts}}$ facts." } ``` ## A program to try The program sorts the columns of a small survey by a rough rule: words, whole numbers or decimals. ```python # A small survey table. Each column is a list of values. survey = { "favourite game": ["chess", "carrom", "ludo", "chess"], "goals scored": [2, 0, 1, 3], "time to school (min)": [12.5, 8.0, 21.75, 15.2], } # A rough rule: words are categorical, whole numbers are counts, decimals are measurements. for column, values in survey.items(): if all(isinstance(v, str) for v in values): kind = "categorical" elif all(isinstance(v, int) for v in values): kind = "numerical, discrete (counted)" else: kind = "numerical, continuous (measured)" print(column.ljust(22), kind) ``` !try Read it, then run it. Does each column get the kind you expect? 1. Add a column `"shoe size": [5, 6, 4, 7]`. What does the rule say? 2. Add a column `"roll number": [11, 12, 13, 14]`. The rule says discrete. Why is that wrong? ## Mistakes to avoid - **Calling every number numerical.** A roll number or a phone number is a label. - **Calling a rounded measurement discrete.** A height of 152 cm was measured, so it is continuous. - **Writing zero for a missing value.** It pulls the mean down. - **Trusting a big sample that is biased.** Many answers from the wrong people still give the wrong picture. ## Check yourself 1. A table has 26 rows and 6 columns. How many facts does it hold? !hint Multiply the rows by the columns. !answer $26 \times 6 = 156$ facts. 2. Which kind of data is the number of stamps in a collection? !hint Do you count stamps, or measure them? !answer Numerical, discrete. Stamps are counted. 3. Which kind of data is a seat on a train: first class, second class or third class? !hint Is it a label? Do the labels have an order? !answer Categorical, ordinal. The classes are labels, and they have an order from first class down. 4. A gym asks its own members how often people in the town exercise. Why is the result too high? !hint Who joins a gym? !answer Gym members exercise more than most people. People who never exercise are left out, so the sample is biased.42:T1115,A brief note on **Electromagnetic Induction and Household Circuits**. Read it before a test. Try each question at the end before you open its answer. ## The big ideas - A current is **induced** in a coil while the field through it **changes**: when a magnet and the coil move relative to each other. - It is larger for **faster motion**, a **stronger magnet** and **more turns**. It reverses when the motion reverses, and it **opposes** the change (Lenz's law). - **Fleming's right-hand rule:** thumb for the motion, forefinger for the field, middle finger for the induced current. - A **generator** turns a coil in a field. **Slip rings** give AC; a **split ring** gives a pulsing DC. - A house has a **live**, a **neutral** and an **earth** wire. Appliances are in **parallel**. A **fuse** in the live wire melts when the current is too large; the **earth wire** protects us from a live metal body. ## The facts at a glance | What | Rule | Unit | |---|---|---| | Frequency | turns each second | hertz (Hz) | | One cycle | $T = \dfrac{1}{f}$ | second (s) | | Changes of direction | $2f$ each second | | | Current of an appliance | $I = \dfrac{P}{V}$ | ampere (A) | | Most power on a circuit | $P = V \times$ fuse current | watt (W) | ## A worked example A generator's coil turns $2700$ times a minute. Find the frequency, the time of one cycle and the changes of direction each second. Then decide whether a $3000\ \text{W}$ geyser on $220\ \text{V}$ is safe on a $15\ \text{A}$ circuit. 1. **Frequency.** $f = \dfrac{2700}{60\ \text{s}} = 45\ \text{Hz}$. 2. **One cycle.** $T = \dfrac{1}{45\ \text{Hz}} = 0.0222\ \text{s} = 22.2\ \text{ms}$. 3. **Changes.** $2 \times 45 = 90$ each second. 4. **The geyser.** $I = \dfrac{3000\ \text{W}}{220\ \text{V}} = 13.6\ \text{A}$, less than $15\ \text{A}$: safe. ## Practice with new numbers ```practice { "title": "Two appliances together", "vars": { "P1": [ 385, 1210 ], "P2": [ 550, 1320, 2090 ] }, "values": { "I": "(P1 + P2)/220", "one": "P1/220" }, "question": "Appliances of {{P1}} W and {{P2}} W are used together on $220\\ \\text{V}$. What is the total current, in amperes?", "answer": { "value": "I", "unit": "A", "digits": 2 }, "hint": "In parallel the powers add; then $I = \\dfrac{P}{V}$.", "mistakes": [ { "value": "one", "say": "Add both powers before you divide by the voltage." } ], "solution": "$I = \\dfrac{ {{P1}}\\ \\text{W} + {{P2}}\\ \\text{W} }{220\\ \\text{V}} = {{I:2}}\\ \\text{A}$." } ``` ## A program to try The program repeats the worked example. ```python # Step 1: a generator that turns 2700 times a minute f = 2700 / 60 print("Frequency:", f, "Hz one cycle:", round(1000 / f, 1), "ms", " reversals each second:", 2 * f) # Step 2: a geyser of 3000 W on 220 V, on a 15 A circuit I = 3000 / 220 print("Geyser:", round(I, 1), "A safe on 15 A:", I <= 15) ``` !try Read it, then run it. Do its answers agree with yours? 1. Change the geyser to `3500` W. Is it still safe on the $15\ \text{A}$ circuit? ## Mistakes to avoid - **Expecting a current from a magnet at rest.** Only a change of field induces a current. - **Counting the changes of direction as $f$.** There are two in every cycle: $2f$. - **Putting the fuse in the neutral wire.** It must be in the live wire. - **Replacing a fuse with a thick wire.** It would never melt, and the wires could catch fire. ## Check yourself 1. A magnet is pushed into a coil and then held still inside it. What does the galvanometer's needle do? !hint A current flows only while the field through the coil changes. !answer It swings while the magnet moves, then goes back to zero when the magnet stops. 2. How long is one cycle of a $60\ \text{Hz}$ supply, in milliseconds? !hint $T = \dfrac{1}{f}$. !answer $T = \dfrac{1}{60\ \text{Hz}} = 0.0167\ \text{s} = 16.7\ \text{ms}$. 3. 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