29:I[613436,["/_next/static/immutable/chunks/2e-_pntwh4kb0.js","/_next/static/immutable/chunks/0n_ejf1w61anu.js","/_next/static/immutable/chunks/23grczsvieze4.js","/_next/static/immutable/chunks/21nt0qfuunlyw.js","/_next/static/immutable/chunks/2cei39krol38n.js","/_next/static/immutable/chunks/3qdlim0u4lvjl.js","/_next/static/immutable/chunks/0b9145ckv2cou.js","/_next/static/immutable/chunks/17r05ydov_hpt.js","/_next/static/immutable/chunks/1dtj60b48sxo1.js","/_next/static/immutable/chunks/1o5nhfcb2s9jp.js"],"default"] 2a:Tefe,A brief note on **Beats, Resonance and the Doppler Effect**. Read it before a test. Try each question at the end before you open its answer. ## Beats - Two close frequencies drift in and out of step. - **Beat frequency:** $f_b = |f_1 - f_2|$. **Beat period:** $T_b = 1/f_b$. - **Pitch heard:** $\dfrac{f_1 + f_2}{2}$. - **Wax** lowers a fork's frequency. **Filing** raises it. **Tightening** raises a string's. - Beats faster after a change: the two frequencies moved apart. ## The resonance tube - Node at the water, antinode a distance $e$ above the open end. - $l_1 + e = \lambda/4$, $l_2 + e = 3\lambda/4$: only odd quarter waves. - **Speed of sound:** $v = 2f(l_2 - l_1)$. The end correction cancels. - **End correction:** $e = \dfrac{l_2 - 3l_1}{2} \approx 0.3d$. ## The Doppler effect - $f' = f\,\dfrac{v \pm v_o}{v \mp v_s}$. Upper signs when that body moves towards the other. - **Source moving:** the wavelength changes. **Listener moving:** the rate of meeting crests changes. - Speeds are measured relative to the air. - **Echo** from a reflector coming at $u$: $f\,\dfrac{v + u}{v - u}$. ## A worked example A fire engine with a $600\ \text{Hz}$ siren passes you at $15\ \text{m/s}$. 1. **Coming.** $f' = 600\ \text{Hz} \times \dfrac{340\ \text{m/s}}{325\ \text{m/s}} = 628\ \text{Hz}$. 2. **Going.** $f' = 600\ \text{Hz} \times \dfrac{340\ \text{m/s}}{355\ \text{m/s}} = 575\ \text{Hz}$. 3. **The drop.** About $53\ \text{Hz}$, heard the moment it passes. ## Practice with new numbers ```practice { "title": "A whistle moving away", "vars": { "ff": [ 420, 560, 630, 840 ], "s": [ 12, 18, 24 ] }, "values": { "fh": "ff*340/(340 + s)", "wrong": "ff*340/(340 - s)" }, "question": "A whistle of {{ff}} Hz moves straight away from you at {{s}} m/s. The speed of sound is 340 m/s. What frequency do you hear, in Hz?", "answer": { "value": "fh", "unit": "Hz", "digits": 0 }, "hint": "A source moving away: add its speed to $v$ on the bottom.", "mistakes": [ { "value": "wrong", "say": "That is for a source coming towards you." } ], "solution": "$f' = {{ff}} \\times \\dfrac{340}{340 + {{s}}} = {{fh:0}}$ Hz." } ``` ## A program to try The program finds the two notes of the fire engine. ```python # A fire engine's siren: what you hear before and after it passes. v = 340.0 # m/s f = 600.0 # Hz s = 15.0 # speed of the fire engine, in m/s coming = f * v / (v - s) going = f * v / (v + s) print("Coming:", round(coming), "Hz") print("Going:", round(going), "Hz") ``` !try Run it. Check both lines against the worked example. 1. Change `s = 15.0` to `s = 30.0`. Does the drop double? ## Mistakes to avoid - **Adding frequencies for beats.** Take the difference. - **Taking $l_2 - l_1$ as $\lambda$.** It is $\lambda/2$. - **Swapping the Doppler signs.** Towards always raises the note. - **A one-step echo.** The reflector is a listener, then a source. ## Check yourself 1. Forks of $315\ \text{Hz}$ and $318\ \text{Hz}$ sound together. What is the beat period? !hint First find the beat frequency. !answer $f_b = 3\ \text{Hz}$, so $T_b = \tfrac13\ \text{s}$, about $0.33\ \text{s}$. 2. A tube resounds at $l_1$ and $l_2$. How much is $l_2 - l_1$, in wavelengths? !hint Each resonance adds half a wave. !answer Half a wavelength. 3. A siren comes towards you, then goes away. When does its note drop? !hint The note depends on the direction of motion, not the distance. !answer At the moment it passes you. Before that it is steady and high, after that steady and low. 4. Why does an echo from a car coming towards you sound higher than the sound that was sent? !hint Think of two steps. !answer The car meets the crests faster, as a listener. Then it sends them back as a source that moves towards you. Both steps raise the frequency.2b:T1076,A brief note on **Buffers, Salt Hydrolysis and Solubility**. Read it before a test. Try each question at the end before you open its answer. ## Salt hydrolysis - An ion from a weak acid or a weak base reacts with water. - Strong acid and strong base: neutral. Weak acid and strong base: basic. Strong acid and weak base: acidic. - $K_h = K_w / K_a$ (or $K_w / K_b$) and $h = \sqrt{K_h / C}$. - Weak acid and strong base: $\text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a + \log C)$. - Strong acid and weak base: $\text{pH} = 7 - \tfrac{1}{2}(\text{p}K_b + \log C)$. - Weak acid and weak base: $\text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a - \text{p}K_b)$. ## Buffers - Common ion effect: a shared ion pushes back the ionisation of a weak electrolyte. - Acidic buffer: weak acid and its salt. Basic buffer: weak base and its salt. - $\text{pH} = \text{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$; $\text{pOH} = \text{p}K_b + \log\dfrac{[\text{salt}]}{[\text{base}]}$. - Best within one pH unit of $\text{p}K_a$. Blood: $\text{H}_2\text{CO}_3/\text{HCO}_3^-$. ## Solubility product - $\text{A}_x\text{B}_y$: $K_{sp} = [\text{A}]^x[\text{B}]^y = x^x y^y s^{x+y}$. - AB: $s^2$. $\text{AB}_2$: $4s^3$. $\text{A}_2\text{B}_3$: $108s^5$. - A common ion lowers $s$: for AB in $c$ mol/L of a common ion, $s \approx K_{sp} / c$. - $Q_{sp} > K_{sp}$: a precipitate forms. ## A worked example Silver carbonate, $\text{Ag}_2\text{CO}_3$, has $K_{sp} = 8.1 \times 10^{-12}$. How much dissolves in a litre of water? 1. **Type.** It is $\text{A}_2\text{B}$: $[\text{Ag}^+] = 2s$ and $[\text{CO}_3^{2-}] = s$. 2. **Ksp.** $K_{sp} = (2s)^2 s = 4s^3$. 3. **Solve.** $s^3 = \dfrac{8.1 \times 10^{-12}}{4} = 2.03 \times 10^{-12}$, so $s = 1.27 \times 10^{-4}\ \text{mol/L}$. ## Practice with new numbers ```practice { "title": "Solubility of a salt of type AB", "vars": { "m": [ 1.6, 2.5, 4.9, 8.1 ] }, "values": { "s": "sqrt(m)", "wrong": "m/2" }, "question": "A salt of the type AB has Ksp = {{m}} × 10^-10. Its molar solubility is S × 10^-5 mol/L. What is S?", "answer": { "value": "s", "digits": 2 }, "hint": "For AB, Ksp = s², so s = √Ksp. The square root of 10^-10 is 10^-5.", "mistakes": [ { "value": "wrong", "say": "Take the square root of Ksp; do not halve it." } ], "solution": "$s = \\sqrt{ {{m}} \\times 10^{-10} } = {{s:2}} \\times 10^{-5}$ mol/L." } ``` ## A program to try The program finds the molar solubility of three salts of different types from their $K_{sp}$. ```python # The molar solubility of a salt AxBy from its Ksp: Ksp = x^x * y^y * s^(x + y). def solubility(ksp, x, y): return (ksp / (x ** x * y ** y)) ** (1 / (x + y)) salts = [("AgCl", 1.8e-10, 1, 1), ("CaF2", 5.3e-11, 1, 2), ("Ag2CO3", 8.1e-12, 2, 1)] for name, ksp, x, y in salts: print(f"{name:7} Ksp = {ksp:.1e} s = {solubility(ksp, x, y):.2e} mol/L") ``` !try Run it. Which salt is the most soluble? 1. Add aluminium hydroxide, $\text{Al(OH)}_3$, with $K_{sp} = 1.3 \times 10^{-33}$. What are its x and y? ## Mistakes to avoid - **Every salt is neutral.** Only a salt of a strong acid and a strong base is. - **The Henderson fraction upside down.** The salt goes on top for an acidic buffer. - **$K_{sp} = s^2$ for every salt.** Count the ions in the formula. - **Comparing $K_{sp}$ of different salt types.** Compare their solubilities instead. - **Forgetting dilution on mixing.** Equal volumes halve each concentration. ## Check yourself 1. Is a solution of ammonium chloride acidic or basic? !hint Which parent is weak: hydrochloric acid or ammonia? !answer Acidic. The ammonium ion, from the weak base, gives protons to water. 2. A buffer has twice as much salt as acid. Is its pH above or below $\text{p}K_a$? !hint What is the sign of $\log 2$? !answer Above: $\text{pH} = \text{p}K_a + \log 2 = \text{p}K_a + 0.30$. 3. Why does AgCl dissolve less in potassium chloride solution than in water? !hint Which ion do they share? !answer The chloride ion is common, so the dissolving equilibrium is pushed back. 4. Write $K_{sp}$ of a salt of type $\text{AB}_3$ in terms of $s$. !hint One formula unit gives one A and three B ions. !answer $K_{sp} = s(3s)^3 = 27s^4$.2c:Te01,A brief note on **Geometric Progressions and the AM-GM Inequality**. Read it before a test. Try each question at the end before you open its answer. ## The formulas | What | Formula | | --- | --- | | nth term | $a_n = a r^{n - 1}$ | | sum of $n$ terms | $\dfrac{a(r^n - 1)}{r - 1}$, or $na$ if $r = 1$ | | sum to infinity | $\dfrac{a}{1 - r}$, only if $\lvert r \rvert < 1$ | | geometric mean | $G = \sqrt{ab}$ | - $a$, $b$, $c$ are in GP exactly when $b^2 = ac$. - Three numbers in GP: write them $\dfrac{a}{r}$, $a$, $ar$. - $n$ means between $a$ and $b$: $r = \left(\dfrac{b}{a}\right)^{\frac{1}{n + 1}}$. - For positive numbers, $A \ge G$, with equality only when they are equal. - Two numbers with means $A$ and $G$ are the roots of $x^2 - 2Ax + G^2 = 0$. ## A worked example Find the sum of the first $6$ terms of $5, 15, 45, \ldots$, and the sum to infinity of $5 + \tfrac{5}{3} + \tfrac{5}{9} + \cdots$ **Step 1. The first sum.** $a = 5$ and $r = 3$: $\dfrac{5(3^6 - 1)}{3 - 1} = \dfrac{5 \times 728}{2} = 1820$. **Step 2. The second sum.** $a = 5$ and $r = \tfrac{1}{3}$: $\dfrac{5}{1 - \tfrac{1}{3}} = \dfrac{15}{2}$. **Answer:** $1820$ and $\tfrac{15}{2}$. ## Practice with new numbers ```practice { "title": "A sum to infinity", "vars": { "a": [ 6, 8, 12 ], "q": [ 2, 3, 4 ] }, "values": { "S": "a*q/(q - 1)", "plus": "a*q/(q + 1)", "qm": "q - 1" }, "question": "Find the sum to infinity of the geometric progression with first term {{a}} and common ratio $\\dfrac{1}{ {{q}} }$.", "answer": { "value": "S", "digits": 2 }, "hint": "Use $\\dfrac{a}{1 - r}$. The ratio is less than $1$, so the sum exists.", "mistakes": [ { "value": "plus", "say": "That divides by $1 + r$. The formula has $1 - r$ underneath." } ], "solution": "$\\dfrac{ {{a}} }{1 - \\frac{1}{ {{q}} }} = \\dfrac{ {{a}} \\times {{q}} }{ {{qm}} } = {{S}}$." } ``` ## A program to try The program checks both sums of the worked example with exact fractions. ```python # Check the formulas of a GP on one example: a = 5, r = 3. from fractions import Fraction a, r, n = Fraction(5), Fraction(3), 6 terms = [a * r ** (k - 1) for k in range(1, n + 1)] print("Terms:", [int(t) for t in terms]) print("Sum by adding:", sum(terms), " by the formula:", a * (r ** n - 1) / (r - 1)) b = Fraction(1, 3) print("Sum to infinity of 5 + 5/3 + 5/9 + ...:", a / (1 - b)) ``` !try Run it. Does it find $1820$ and $ frac{15}{2}$? 1. Change `n` to `8`. Predict the new sum first. ## Mistakes to avoid - **$ar^n$ for the nth term.** It is $ar^{n - 1}$. - **The sum formula with $r = 1$.** The sum is $na$. - **A sum to infinity with $|r| \ge 1$.** There is none. - **Forgetting the way up.** A bouncing ball travels each rebound twice. - **AM-GM with negative numbers.** It needs positive numbers. ## Check yourself 1. Find the 4th term of $10, 5, \tfrac{5}{2}, \ldots$ !hint $r = \tfrac{1}{2}$, and the 4th term carries $r^3$. !answer $10 \times \tfrac{1}{8} = \tfrac{5}{4}$. 2. Find the sum to infinity of $1 - \tfrac{1}{2} + \tfrac{1}{4} - \cdots$ !hint $a = 1$ and $r = -\tfrac{1}{2}$. !answer $\dfrac{1}{1 + \tfrac{1}{2}} = \dfrac{2}{3}$. 3. Write $0.\overline{5}$ as a fraction. !hint It is a GP with $a = 0.5$ and $r = \tfrac{1}{10}$. !answer $\dfrac{0.5}{0.9} = \dfrac{5}{9}$. 4. Find the geometric mean of $2$ and $18$. !hint Multiply, then take the square root. !answer $\sqrt{36} = 6$. 5. For $x > 0$, what is the least value of $x + \dfrac{25}{x}$? !hint The product of the two parts is $25$. !answer $2\sqrt{25} = 10$, at $x = 5$.2d:T112e,A brief note on **Mitochondria, Plastids and the Nucleus**. Read it before a test. Try each question at the end before you open its answer. ## Mitochondria and plastids - **Mitochondrion**: two membranes; the inner one folds into **cristae**; outer compartment and **matrix**. It makes ATP by aerobic respiration. - **Plastids**: **chloroplasts** (chlorophyll), **chromoplasts** (carotene, xanthophylls), **leucoplasts** (stores). - **Leucoplasts**: **amyloplasts** (starch), **elaioplasts** (oils and fats), **aleuroplasts** (proteins). - **Chloroplast**: two membranes, **stroma**, **thylakoids** stacked in **grana**, joined by **stroma lamellae**. Chlorophyll is in the thylakoids. - Both have their own **circular DNA** and **70S ribosomes**, and both divide. ## Ribosomes and movement - **Ribosomes**: RNA and protein, no membrane. **80S = 60S + 40S** in the eukaryotic cytoplasm, **70S = 50S + 30S** elsewhere. - **Cytoskeleton**: microtubules, microfilaments, intermediate filaments. - **Cilia and flagella**: a **9 + 2** axoneme, growing from a **basal body**. - **Centriole**: a cartwheel of **nine triplets** round a **hub**. Centrioles form basal bodies and spindle fibres. ## The nucleus - **Nuclear envelope**: two membranes, a perinuclear space and **pores** for two-way traffic. - **Nucleolus**: no membrane, the site of **rRNA** synthesis. - **Chromatin**: DNA, histones, non-histone proteins and RNA. It forms **chromosomes** in division. - **Centromere**: middle (**metacentric**), a little off (**sub-metacentric**), near one end (**acrocentric**), at the end (**telocentric**). - **Microbodies**: small membrane-bound vesicles of enzymes. ## A worked example Name the organelle: it has two membranes, its own DNA, and stacks of flat sacs inside. 1. **Two membranes and its own DNA**: a mitochondrion, a chloroplast or the nucleus. 2. **Stacks of flat sacs**: these are grana of thylakoids. 3. **Organelle:** a chloroplast. ## Practice with new numbers ```practice { "title": "Microtubules in dividing cells", "vars": { "k": [ 2, 4, 6, 8 ] }, "values": { "t": "27*k", "wrong": "20*k" }, "question": "Some cells about to divide hold {{k}} centrioles in all. How many microtubules make up their peripheral triplets?", "answer": { "value": "t", "unit": "microtubules", "digits": 0 }, "hint": "A centriole has nine triplets.", "mistakes": [ { "value": "wrong", "say": "That treats each centriole like an axoneme. A centriole has 9 × 3 = 27 microtubules." } ], "solution": "$27 \\times {{k}} = {{t}}$ microtubules." } ``` ## A program to try The program lists six organelles with the membranes round them and whether they have their own DNA. ```python # Organelles with their own DNA and 70S ribosomes, and the rest. organelles = { "mitochondrion": (2, True), # (membranes around it, own DNA?) "chloroplast": (2, True), "nucleus": (2, True), "ribosome": (0, False), "centriole": (0, False), "microbody": (1, False), } for name, (membranes, dna) in organelles.items(): print(f"{name:14} membranes: {membranes} DNA: {'yes' if dna else 'no'}") no_membrane = [n for n, (m, d) in organelles.items() if m == 0] print("no membrane:", no_membrane) ``` !try Run it. Which two organelles have no membrane? 1. Add `"lysosome": (1, False)` to `organelles`. How many organelles now have exactly one membrane? !hint Count the microbody and the lysosome. !answer Two: the microbody and the lysosome. The ribosome and the centriole still have none. ## Mistakes to avoid - **Putting cristae on the outer membrane.** They are folds of the inner one. - **Giving mitochondria 80S ribosomes.** They have 70S ribosomes. - **Giving a centriole a central pair.** It has nine triplets and no central pair. - **Saying an acrocentric chromosome has no short arm.** That is a telocentric one. ## Check yourself 1. Which plastids are colourless, and what do they do? !hint Leuco means white. !answer Leucoplasts. They store food: starch, oils and fats, or proteins. 2. What are the subunits of a 70S ribosome? !hint One large and one small. !answer A 50S subunit and a 30S subunit. 3. What does the nucleolus make? !hint It is linked to ribosomes. !answer Ribosomal RNA. 4. Where is the centromere of a metacentric chromosome? !hint Meta means middle here. !answer In the middle, so the two arms are equal.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:Tf3c,A brief note on **Superposition and Standing Waves**. Read it before a test. Try each question at the end before you open its answer. ## Adding waves - **Superposition:** where waves meet, the displacements add, $y = y_1 + y_2$. - Two equal waves with a phase difference $\phi$: $A = 2a\cos(\phi/2)$. - Unequal waves: $A^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi$. - **In phase** ($\phi = 0$): they help each other. **Out of phase** ($\phi = \pi$): they cancel. - Path difference to phase difference: $\phi = \dfrac{2\pi}{\lambda}\Delta x$. Intensity $I \propto A^2$. ## Reflection and standing waves - **Fixed end:** the wave comes back inverted. **Free end:** it comes back upright. - Two equal waves going opposite ways add to $y = 2a\sin kx\cos\omega t$. - Nodes never move. Next nodes are $\lambda/2$ apart, and a node is $\lambda/4$ from the nearest antinode. ## Strings and pipes - **String tied at both ends:** $f_n = \dfrac{nv}{2L}$, every harmonic, with $v = \sqrt{T/\mu}$. - **Pipe open at both ends:** $f_n = \dfrac{nv}{2L}$, every harmonic. - **Pipe closed at one end:** $f = \dfrac{(2n - 1)v}{4L}$, odd harmonics only. - A closed end is a displacement node; an open end is an antinode. ## A worked example A string $0.65\ \text{m}$ long has a tension of $90\ \text{N}$ and $2.5\ \text{g}$ of mass in each metre. Find its fundamental. 1. **Speed.** $v = \sqrt{\dfrac{90\ \text{N}}{0.0025\ \text{kg/m}}} = 189.7\ \text{m/s}$. 2. **Wavelength.** In the fundamental, $\lambda = 2L = 1.3\ \text{m}$. 3. **Frequency.** $f_1 = \dfrac{189.7\ \text{m/s}}{1.3\ \text{m}} = 146\ \text{Hz}$. ## Practice with new numbers ```practice { "title": "Which pipe?", "vars": { "L": [ 0.2, 0.34, 0.5, 0.68 ] }, "values": { "fo": "340/(2*L)", "fc": "340/(4*L)" }, "question": "A pipe {{L}} m long is closed at one end. Sound travels at 340 m/s. What is its fundamental, in Hz?", "answer": { "value": "fc", "unit": "Hz", "digits": 0 }, "hint": "A closed pipe holds a quarter wavelength: $f = v/4L$.", "mistakes": [ { "value": "fo", "say": "That is the open pipe, $v/2L$. A closed pipe gives half of it." } ], "solution": "$f = \\dfrac{340\\ \\text{m/s}}{4 \\times {{L}}\\ \\text{m} } = {{fc:0}}\\ \\text{Hz}$." } ``` ## A program to try The program works out the example above, and the first three notes of a closed pipe. ```python # A string and a pipe: their lowest three notes. import math L_string, T, mu = 0.65, 90.0, 2.5e-3 # m, N, kg/m L_pipe, v_air = 0.34, 340.0 # m, m/s v = math.sqrt(T / mu) print("String, wave speed:", round(v, 1), "m/s") print("String:", [round(n * v / (2 * L_string), 1) for n in (1, 2, 3)], "Hz") print("Closed pipe:", [round(n * v_air / (4 * L_pipe)) for n in (1, 3, 5)], "Hz") ``` !try Run it. Compare the string's first number with the worked example. 1. Change `L_pipe = 0.34` to `L_pipe = 0.17`. Do the pipe's notes double? ## Mistakes to avoid - **Adding amplitudes when the waves are not in phase.** Use $2a\cos(\phi/2)$. - **Node spacing of $\lambda$.** It is $\lambda/2$. - **Even harmonics from a closed pipe.** It has odd ones only. - **$f \propto T$.** It is $f \propto \sqrt{T}$. ## Check yourself 1. Two equal waves meet with a phase difference of $180^\circ$. What is the amplitude of the sum? !hint $\cos 90^\circ = 0$. !answer Zero. The waves cancel. 2. Next nodes of a standing wave are $40\ \text{cm}$ apart. What is the wavelength? !hint Nodes are $\lambda/2$ apart. !answer $\lambda = 80\ \text{cm}$. 3. A string has a fundamental of $200\ \text{Hz}$. What is its third harmonic? !hint A string has every whole multiple of $f_1$. !answer $3 \times 200\ \text{Hz} = 600\ \text{Hz}$. 4. Which has the lower fundamental: an open pipe or a closed pipe of the same length? !hint Compare $v/2L$ with $v/4L$. !answer The closed pipe. Its fundamental is half that of the open pipe.31:Tfce,A brief note on **Acids, Bases and pH**. Read it before a test. Try each question at the end before you open its answer. ## Three ideas - Arrhenius: acids give $\text{H}^+$, bases give $\text{OH}^-$, in water. - Bronsted-Lowry: acids donate protons, bases accept them. A conjugate pair differs by one proton. - Lewis: acids accept an electron pair, bases donate one. $\text{BF}_3$ is a Lewis acid. - Strong acids and bases ionise fully. A strong acid has a very weak conjugate base. ## Water and pH - $K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $298\ \text{K}$. It grows with temperature. - $\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]$, and $\text{pH} + \text{pOH} = 14$ at $298\ \text{K}$. - One pH unit is a tenfold change. Neutral means equal ions. ## Weak acids and bases - $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$ and $\text{p}K_a = -\log_{10} K_a$. Smaller pKa, stronger acid. - $\alpha \approx \sqrt{K_a/C}$ and $[\text{H}_3\text{O}^+] \approx \sqrt{K_a C}$ when $\alpha$ is small. - Weak base: $[\text{OH}^-] \approx \sqrt{K_b C}$. - Conjugate pair: $K_a K_b = K_w$, and $\text{p}K_a + \text{p}K_b = 14$. ## A worked example Compare $0.050\ \text{mol/L}$ of a strong acid with $0.050\ \text{mol/L}$ of acetic acid. 1. **Strong acid.** $[\text{H}_3\text{O}^+] = 0.050\ \text{mol/L}$, so the pH is $1.30$. 2. **Acetic acid.** $[\text{H}_3\text{O}^+] \approx \sqrt{1.8 \times 10^{-5} \times 0.050} = 9.5 \times 10^{-4}\ \text{mol/L}$, so the pH is about $3.0$. 3. **Meaning.** Same concentration, but the weak acid has about fifty times fewer hydronium ions. ## Practice with new numbers ```practice { "title": "pKb from Ka", "vars": { "a": [ 2, 4, 5, 8 ], "n": [ 4, 5, 6, 9 ] }, "values": { "pkb": "14 - (n - log10(a))", "w1": "n - log10(a)", "w2": "14 + n - log10(a)" }, "question": "An acid has $K_a = {{a}} \\times 10^{-{{n}}}$ at $298\\ \\text{K}$. What is the pKb of its conjugate base?", "answer": { "value": "pkb", "unit": "", "digits": 2 }, "hint": "Find the pKa first. Then pKa and pKb add up to 14.", "mistakes": [ { "value": "w1", "say": "That is the pKa. Take it away from 14." }, { "value": "w2", "say": "pKa is minus the logarithm, so it is n minus the logarithm of {{a}}." } ], "solution": "$\\text{p}K_a = {{n}} - \\log_{10} {{a}}$, so $\\text{p}K_b = 14 - \\text{p}K_a = {{pkb:2}}$." } ``` ## A program to try The program finds the pH of the strong acid and of acetic acid, solving the weak one exactly. ```python # pH of a strong acid and of a weak acid of the same concentration import math C = 0.050 # mol/L Ka = 1.8e-5 # acetic acid strong = C weak = (-Ka + math.sqrt(Ka ** 2 + 4 * Ka * C)) / 2 print("strong acid: pH", round(-math.log10(strong), 2)) print("weak acid: pH", round(-math.log10(weak), 2)) ``` !try Run it. How far apart are the two pH values? 1. Change `C` to `0.0050`. Does the gap grow or shrink? ## Mistakes to avoid - **The minus sign in pH.** pH is minus the logarithm. - **The square root.** $\sqrt{K_a C}$, not $K_a C$. - **The pOH of a base.** Take it away from 14 for the pH. - **Two hydroxides.** $\text{Ba(OH)}_2$ gives two $\text{OH}^-$ per unit. - **Weak and dilute.** They mean different things. ## Check yourself 1. What is the conjugate acid of $\text{NH}_3$? !hint Add one proton. !answer $\text{NH}_4^+$. 2. What is the pH of a solution with $[\text{H}_3\text{O}^+] = 1.0 \times 10^{-6}\ \text{mol/L}$? !hint Minus the logarithm. !answer 6. 3. Which is the stronger acid: pKa 2 or pKa 5? !hint Smaller pKa means larger Ka. !answer pKa 2. 4. Why is $\text{BF}_3$ an acid though it has no hydrogen? !hint What does boron lack? !answer Boron has only six electrons, so it accepts an electron pair. That makes it a Lewis acid. 5. A weak acid is diluted. Does its pH rise or fall? !hint What happens to the hydronium concentration? !answer It rises, because the hydronium concentration falls, even though a bigger share ionises.32:T110d,A brief note on **Sequences, Series and Arithmetic Progressions**. Read it before a test. Try each question at the end before you open its answer. ## Sequences and series - A sequence is a function on $1, 2, 3, \dots$; its $n$th term is $a_n$. - A formula gives $a_n$ at once. A recurrence builds each term from earlier ones. - A series adds the terms: $S_n = a_1 + \dots + a_n = \sum_{k=1}^{n} a_k$. - $a_1 = S_1$, and $a_n = S_n - S_{n-1}$ for $n \geq 2$. ## Arithmetic progressions - Each term is the one before plus $d$: $a_n = a + (n - 1)d$. - From the end: the $n$th term from the end is $l - (n - 1)d$. - $S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right) = \dfrac{n}{2}(a + l)$. - A sequence is an AP exactly when $a_n$ is linear in $n$, or $S_n$ is a quadratic in $n$ with no constant term. - Adding, subtracting, multiplying or dividing every term by a constant keeps an AP. - Choose three numbers as $a - d, a, a + d$, and four as $a - 3d, a - d, a + d, a + 3d$. ## Arithmetic means - The AM of $a$ and $b$ is $\dfrac{a + b}{2}$; $a$, $b$, $c$ are in an AP exactly when $2b = a + c$. - $m$ means between $a$ and $b$: $d = \dfrac{b - a}{m + 1}$, and the $j$th mean is $a + jd$. - The $m$ means add up to $m \times \dfrac{a + b}{2}$. ## A worked example An AP has $a_2 = 11$, and the sum of its first $9$ terms is $207$. Find its first term and common difference. 1. **Use the sum.** $S_9 = \dfrac{9}{2}(a_1 + a_9)$, and $a_1 + a_9 = 2a_5$, because they are the same distance from the middle. So $S_9 = 9a_5$. 2. **The middle term.** $9a_5 = 207$, so $a_5 = 23$. 3. **The difference.** $a_5 - a_2 = 3d = 12$, so $d = 4$. 4. **The first term.** $a = a_2 - d = 7$. ## Practice with new numbers ```practice { "title": "A term of an AP", "vars": { "a": [ 5, 9, 16 ], "d": [ 4, 7, -3 ], "n": [ 12, 15, 25 ] }, "values": { "ans": "a + (n - 1)*d", "wrong": "a + n*d" }, "question": "An AP has first term ${{a}}$ and common difference ${{d}}$. Find its ${{n}}$th term.", "answer": { "value": "ans", "digits": 0 }, "hint": "Use $a_n = a + (n - 1)d$.", "mistakes": [ { "value": "wrong", "say": "After the first term there are $n - 1$ steps of $d$, not $n$." } ], "solution": "$a_{ {{n}} } = {{a}} + ({{n}} - 1) \\times ({{d}}) = {{ans}}$." } ``` ## A program to try The program checks the formulas of this note on one AP: the $n$th term, the sum, recovering the terms from the sums, and the middle term as a mean. ```python # Check the rules of the note on one progression. a, d, n = 11, -2, 15 terms = [a + (k - 1) * d for k in range(1, n + 1)] print("a_n by the formula:", a + (n - 1) * d, " last term listed:", terms[-1]) print("S_n by the formula:", n * (2 * a + (n - 1) * d) // 2, " by adding:", sum(terms)) # Recover the terms from the partial sums. sums = [sum(terms[:k]) for k in range(1, n + 1)] back = [sums[0]] + [sums[k] - sums[k - 1] for k in range(1, n)] print("terms recovered from the sums:", back == terms) # The arithmetic mean of the first and last terms is the middle term. print("middle term:", terms[n // 2], " (a + l)/2 =", (terms[0] + terms[-1]) / 2) ``` !try Run it. Do the formulas agree with the listed terms? 1. Change `d` to `5`. Predict the sum of the $15$ terms first. ## Mistakes to avoid - **$a + nd$ for the $n$th term.** It is $a + (n - 1)d$. - **$a_n = S_n - S_{n-1}$ at $n = 1$.** Use $a_1 = S_1$. - **Dividing by $m$ for $m$ means.** There are $m + 1$ steps. - **Rejecting a second root for $n$.** Two values can both be right when some terms add up to zero. - **$b^2 = ac$ for an AP.** That is the geometric test. An AP needs $2b = a + c$. ## Check yourself 1. Find the $15$th term of the AP $-4, 1, 6, \dots$. !hint Here $d = 5$. !answer $-4 + 14 \times 5 = 66$. 2. The partial sums of a sequence are $S_n = 5n^2$. Find $a_3$. !hint Use $a_3 = S_3 - S_2$. !answer $45 - 20 = 25$. 3. Find the sum $9 + 13 + 17 + \dots$ to $12$ terms. !hint $a = 9$ and $d = 4$. !answer $S_{12} = 6(18 + 44) = 372$. 4. Insert $2$ arithmetic means between $-5$ and $13$. !hint Two means make three steps. !answer $d = 6$, so the means are $1$ and $7$. 5. Four numbers in an AP add up to $40$. What do the two middle ones add up to? !hint Take them as $a - 3d, a - d, a + d, a + 3d$. !answer $4a = 40$, so $a = 10$, and the middle two add up to $2a = 20$.33:Tfe1,A brief note on **The Cell Membrane, Wall and Endomembrane System**. Read it before a test. Try each question at the end before you open its answer. ## The cell membrane - A **lipid bilayer**: **polar heads** outwards, **hydrophobic tails** inwards; mostly **phosphoglycerides**, with **cholesterol**. - **Proteins**: **integral** (buried) and **peripheral** (on the surface); **carbohydrates** on the outer side. - Human red blood cell membrane: about **52 per cent protein** and **40 per cent lipid**. - **Fluid mosaic model** (**Singer and Nicolson, 1972**): a quasi-fluid lipid with proteins that move **sideways**. - **Passive transport**: diffusion and **osmosis**, no energy. **Active transport**: against the gradient, using **ATP**, as in the **Na+/K+ pump**. ## The cell wall - Non-living and rigid, outside the membrane of **plants** and **fungi**. - **Algae:** cellulose, **galactans**, **mannans**, **calcium carbonate**. **Land plants:** cellulose, **hemicellulose**, **pectins**, proteins. - **Primary wall:** young, can grow. **Secondary wall:** on the inner side. - **Middle lamella:** **calcium pectate**, glues cells. **Plasmodesmata:** join the cytoplasm of neighbours. ## The endomembrane system - **ER, Golgi complex, lysosomes, vacuoles.** Not mitochondria, chloroplasts or peroxisomes. - **Rough ER:** ribosomes, proteins for secretion. **Smooth ER:** lipids, steroid hormones. - **Golgi** (Camillo Golgi, 1898): stacked cisternae; **cis** (forming) and **trans** (maturing) faces; packaging; **glycoproteins** and **glycolipids**. - **Lysosomes:** from the Golgi; **hydrolases** active at **acidic pH**. - **Vacuoles:** bound by the **tonoplast**; up to **90 per cent** of a plant cell; contractile vacuole in *Amoeba*. ## A worked example A membrane sample from red blood cells weighs $300$ µg. How much protein and how much lipid does it hold? 1. **Protein.** $\dfrac{52}{100} \times 300 = 156$ µg. 2. **Lipid.** $\dfrac{40}{100} \times 300 = 120$ µg. 3. **The rest.** $300 - 156 - 120 = 24$ µg, mostly carbohydrate. ## Practice with new numbers ```practice { "title": "Protein in a membrane sample", "vars": { "mm": [ 50, 75, 125 ] }, "values": { "pr": "52*mm/100", "wrong": "40*mm/100" }, "question": "A sample of red blood cell membrane has a mass of {{mm}} µg. About what mass of it is protein?", "answer": { "value": "pr", "unit": "µg", "digits": 1 }, "hint": "The red blood cell membrane is about 52 per cent protein by mass.", "mistakes": [ { "value": "wrong", "say": "That uses 40 per cent, the share of lipid. Use the share of protein." } ], "solution": "$\\dfrac{52}{100} \\times {{mm}} = {{pr}}$ µg of protein." } ``` ## A program to try The program checks which organelles belong to the endomembrane system. ```python # Is it part of the endomembrane system? endomembrane = {"endoplasmic reticulum", "Golgi apparatus", "lysosome", "vacuole"} for part in ["Golgi apparatus", "mitochondrion", "vacuole", "chloroplast"]: print(part, "->", part in endomembrane) ``` !try Run it. Which two print `False`? 1. Add `"lysosome"` to the list in the loop. What does it print? ## Mistakes to avoid - **The tails, not the heads, face the middle** of the bilayer. - **Osmosis is passive**; the Na+/K+ pump is active and uses ATP. - **The secondary wall forms inside** the primary wall, towards the membrane. - **Mitochondria are not in the endomembrane system.** - **ER vesicles join the cis face**; products leave from the trans face. ## Check yourself 1. Who proposed the fluid mosaic model? !hint Two scientists, in the early 1970s. !answer Singer and Nicolson. 2. Is the movement of water by osmosis active or passive? !hint Does it use ATP? !answer Passive. 3. What is the middle lamella mainly made of? !hint A calcium compound of pectin. !answer Calcium pectate. 4. Which kind of ER makes steroid hormones? !hint Steroids are lipid-like. !answer Smooth ER. 5. Name the membrane round a plant vacuole. !hint Its name starts with t. !answer The tonoplast.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:T10e4,A brief note on **Travelling Waves and Wave Speed**. Read it before a test. Try each question at the end before you open its answer. ## Kinds of wave - A wave carries energy and a pattern. Each particle of the medium only moves about its own place. - **Transverse:** the particles move across the direction of travel. Crests and troughs. - **Longitudinal:** the particles move along the direction of travel. Compressions and rarefactions. - Mechanical waves need a medium. Gases and liquids carry only longitudinal waves in their bulk. ## The equation of a wave - Towards $+x$: $y = a\sin(kx - \omega t + \phi)$. Towards $-x$: $kx + \omega t$. - $k = \dfrac{2\pi}{\lambda}$, in rad/m. $\omega = \dfrac{2\pi}{T} = 2\pi f$, in rad/s. - **Wave speed:** $v = \dfrac{\omega}{k} = f\lambda$. - **Particle velocity:** $u = -a\omega\cos(kx - \omega t + \phi)$, at most $a\omega$. - Phase difference between two points: $\Delta\phi = k\Delta x$. ## What sets the speed - **String:** $v = \sqrt{\dfrac{T}{\mu}}$, with $\mu$ in kg/m. - **Fluid:** $v = \sqrt{\dfrac{B}{\rho}}$. **Solid rod:** $v = \sqrt{\dfrac{Y}{\rho}}$. - **Gas:** $v = \sqrt{\dfrac{\gamma p}{\rho}} = \sqrt{\dfrac{\gamma R T}{M}}$ (Laplace). Newton's $\sqrt{p/\rho}$ is too low. - $v \propto \sqrt{T}$, in kelvin. Pressure alone does not change it. Damp air carries sound a little faster. ## A worked example A wave is $y = 0.02\sin(4x - 48t)$, in SI units, on a rope with $\mu = 0.25\ \text{kg/m}$. Find its wavelength, speed and the tension in the rope. 1. **Wavelength.** $\lambda = \dfrac{2\pi}{4\ \text{rad/m}} = 1.57\ \text{m}$. 2. **Speed.** $v = \dfrac{48\ \text{rad/s}}{4\ \text{rad/m}} = 12\ \text{m/s}$. 3. **Tension.** $v^2 = T/\mu$, and we want $T$, so $T = \mu v^2 = 0.25\ \text{kg/m} \times (12\ \text{m/s})^2 = 36\ \text{N}$. ## Practice with new numbers ```practice { "title": "From the equation to the speed", "vars": { "k": [ 2, 3, 4 ], "w": [ 24, 36, 60 ], "a": [ 1, 2, 3 ] }, "values": { "v": "w/k", "u": "a*w/100", "wrong": "a*w/100" }, "question": "A wave is $y = {{a}}\\sin({{k}}x - {{w}}t)$, with $y$ in cm, $x$ in m and $t$ in s. What is the speed of the wave, in m/s?", "answer": { "value": "v", "unit": "m/s", "digits": 2 }, "hint": "The wave speed is $\\omega/k$.", "mistakes": [ { "value": "wrong", "say": "That is the largest speed of a particle, $a\\omega$. The wave speed is $\\omega/k$." } ], "solution": "$v = \\dfrac{ {{w}}\\ \\text{rad/s} }{ {{k}}\\ \\text{rad/m} } = {{v:2}}\\ \\text{m/s}$." } ``` ## A program to try The program works through the example above. ```python # A wave on a rope: its numbers from y = a sin(kx - wt), and its speed from the tension. import math a, k, w = 0.02, 4.0, 48.0 # m, rad/m, rad/s print("Wavelength:", round(2 * math.pi / k, 3), "m") print("Frequency:", round(w / (2 * math.pi), 2), "Hz") print("Wave speed:", w / k, "m/s") mu = 0.25 # mass of each metre of rope, kg/m print("Tension that gives this speed:", (w / k) ** 2 * mu, "N") ``` !try Run it. Compare the numbers with the worked example. 1. Change `mu = 0.25` to `mu = 1.0`. By how many times does the tension grow? ## Mistakes to avoid - **Wave speed and particle speed.** They are different. One is $\omega/k$, the other at most $a\omega$. - **$k$ and $\lambda$.** $k = 2\pi/\lambda$. A larger $k$ is a shorter wave. - **Grams in $\mu$.** Use kg/m. - **Celsius in $\sqrt{T}$.** Use kelvin. ## Check yourself 1. Is a wave on a plucked string transverse or longitudinal? !hint Which way does the string move? !answer Transverse: the string moves sideways while the wave runs along it. 2. A wave has $k = 2\pi\ \text{rad/m}$ and $\omega = 10\pi\ \text{rad/s}$. Find its speed. !hint $v = \omega/k$. !answer $v = \dfrac{10\pi\ \text{rad/s}}{2\pi\ \text{rad/m}} = 5\ \text{m/s}$. 3. The tension in a string is made nine times as large. What happens to the wave speed? !hint $v \propto \sqrt{T}$. !answer It becomes three times as large. 4. Why does pressure alone not change the speed of sound in a gas? !hint What happens to the density when the pressure rises at a fixed temperature? !answer At a fixed temperature the density rises in step with the pressure, so $p/\rho$, and with it the speed, stays the same.37:T1059,A brief note on **The Equilibrium Constant**. Read it before a test. Try each question at the end before you open its answer. ## Equilibrium - In a **closed system**, forward and backward processes reach **equal rates**. The concentrations then stay constant. - Equilibrium is **dynamic**, and the same mixture is reached from either side. - Physical examples: liquid and vapour, a saturated solution, a gas dissolved in a liquid (Henry's law). ## The constants - $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$: $K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$, constant at a given temperature. - $K_p = K_c(RT)^{\Delta n}$, with $R = 0.0831$ L bar K$^{-1}$ mol$^{-1}$ and $\Delta n$ the change in moles of gas. - **Pure solids and liquids** are left out. - Reverse the equation: $1/K$. Multiply it by n: $K^n$. - K above about 1000: products favoured. Below about 0.001: reactants favoured. - $\Delta G^\circ = -2.303\,RT\log K$. ## Q and the direction - $Q_c < K_c$: forward. $Q_c > K_c$: backward. $Q_c = K_c$: at equilibrium. ## Le Chatelier's principle - A disturbed equilibrium shifts to undo part of the change. - Add a reactant or remove a product: forward. - Higher pressure: towards fewer moles of gas. - Heating: towards the endothermic direction. Only temperature changes K. - A catalyst: equilibrium sooner, no shift. ## A worked example For $\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$, $K_c = 9.0$. Start with 1.0 mol/L each of A and B. Find the equilibrium mixture. 1. **ICE.** At equilibrium, $[\text{A}] = [\text{B}] = 1.0 - x$ and $[\text{C}] = [\text{D}] = x$. 2. **The equation.** $\dfrac{x^2}{(1.0 - x)^2} = 9.0$, so $\dfrac{x}{1.0 - x} = 3.0$. 3. **Solve.** $x = 3.0 - 3.0x$, so $x = 0.75$ mol/L. 4. **Answer.** $[\text{A}] = [\text{B}] = 0.25$ mol/L and $[\text{C}] = [\text{D}] = 0.75$ mol/L. ## Practice with new numbers ```practice { "title": "Kp from Kc", "vars": { "kc": [ 0.15, 0.85, 2.6 ], "T": [ 350, 450, 600 ], "dn": [ 1, 2 ] }, "values": { "kp": "kc*pow(0.0831*T, dn)", "wrong": "kc*pow(8.314*T, dn)" }, "question": "A gas reaction has $K_c = {{kc}}$ at {{T}} K, and $\\Delta n = {{dn}}$. What is $K_p$? Use $R = 0.0831$ L bar K$^{-1}$ mol$^{-1}$.", "answer": { "value": "kp", "digits": 2 }, "hint": "$K_p = K_c(RT)^{\\Delta n}$.", "mistakes": [ { "value": "wrong", "say": "With pressures in bar, use R = 0.0831, not 8.314." } ], "solution": "$RT = 0.0831 \\times {{T}}$. $K_p = {{kc}} \\times (0.0831 \\times {{T}})^{ {{dn}} } = {{kp:2}}$." } ``` ## A program to try The program works through the example above. ```python # A + B ⇌ C + D, Kc = 9.0, starting with 1.0 mol/L each of A and B. import math K, a0 = 9.0, 1.0 # x^2 / (a0 - x)^2 = K, so x / (a0 - x) = sqrt(K) r = math.sqrt(K) x = a0 * r / (1 + r) print(f"x = {x:.2f} mol/L") print(f"[A] = [B] = {a0 - x:.2f} mol/L, [C] = [D] = {x:.2f} mol/L") print("Check: Qc =", round(x * x / (a0 - x) ** 2, 2)) ``` !try Run it. Compare each line with the worked example. 1. Change `K` to `4.0`. What fraction of A reacts now? ## Mistakes to avoid - **Thinking the reactions stop** at equilibrium. - **Putting solids or pure liquids** into K. - **Using starting amounts** in K instead of equilibrium amounts. - **Saying a catalyst gives more product.** - **Saying a pressure change alters K.** Only temperature does. ## Check yourself 1. Write $K_p$ for $\text{C(s)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO(g)} + \text{H}_2\text{(g)}$. !hint Leave out the solid. !answer $K_p = \dfrac{p_{\text{CO}}\,p_{\text{H}_2}}{p_{\text{H}_2\text{O}}}$ 2. A reaction has $K_c = 0.20$, and a mixture has $Q_c = 0.050$. Which way does it go? !hint Compare $Q_c$ with $K_c$. !answer $Q_c < K_c$, so it goes forward. 3. Does halving the volume shift $\text{H}_2 + \text{Br}_2 \rightleftharpoons 2\text{HBr}$ (all gases)? !hint Count the moles of gas on each side. !answer No. Both sides have two moles of gas. 4. An endothermic reaction is cooled. What happens to K? !hint Cooling favours the exothermic direction. !answer K falls, and the mixture shifts backward.38:Te5e,A brief note on **Probability with Coins, Dice and Cards**. Read it before a test. Try each question at the end before you open its answer. ## Coins and dice - Two coins: HH, HT, TH, TT. HT and TH are different outcomes. - Three coins: $8$ outcomes. Each extra coin doubles the count. - Two dice: $36$ pairs (first, second). Sums are not equally likely: $7$ comes in $6$ ways, $2$ and $12$ in one way each. - "At least one" is usually quickest by the complement. ## A pack of cards - $52$ cards, $4$ suits of $13$. Hearts and diamonds are red, spades and clubs black. - $12$ face cards (jack, queen, king). $4$ cards of each kind. The ace is not a face card. - In an "or" question, count a card in both groups once. - When cards are removed, divide by the cards that are left. ## Bags, discs and spinners - The total is every ball, disc or sector that could come up. - From a probability, write an equation in the unknown number and solve it. - Unequal sectors on a spinner are not equally likely. ## A worked example Two dice are rolled. Find the probability that the sum is $5$ or less. 1. **Count each sum.** Sum $2$: $1$ way. Sum $3$: $2$. Sum $4$: $3$. Sum $5$: $4$. 2. **Add.** $1 + 2 + 3 + 4 = 10$ outcomes. 3. **Divide.** $\dfrac{10}{36} = \dfrac{5}{18}$. Always simplify the fraction at the end. ## Practice with new numbers ```practice { "title": "Three coins", "vars": { "h": [ 0, 1, 2, 3 ] }, "values": { "w": "1 + h*(3 - h)", "wrong": "h" }, "question": "Three coins are tossed. In how many of the $8$ outcomes are there exactly ${{h}}$ heads?", "answer": { "value": "w", "digits": 0 }, "hint": "List the outcomes from HHH to TTT and count the heads in each.", "mistakes": [ { "value": "wrong", "say": "Count outcomes, not heads. For example, one head comes from HTT, THT and TTH." } ], "solution": "${{w}}$ outcomes have exactly ${{h}}$ heads, so the probability is $\\dfrac{ {{w}} }{8}$." } ``` ## A program to try The program counts outcomes for three coins and for two dice by listing them. ```python # Count favourable outcomes by listing them from itertools import product from fractions import Fraction coins = list(product("HT", repeat=3)) at_least_two_heads = [c for c in coins if c.count("H") >= 2] print("three coins:", len(coins), "outcomes,", len(at_least_two_heads), "with at least two heads:", Fraction(len(at_least_two_heads), len(coins))) dice = list(product(range(1, 7), repeat=2)) no_six = [d for d in dice if 6 not in d] print("two dice:", len(no_six), "with no six, so at least one six:", Fraction(36 - len(no_six), 36)) ``` !try Run it. Do the answers match the ones in the lessons? 1. Change it to count the outcomes of two dice where the sum is odd. ## Mistakes to avoid - **Three outcomes for two coins** in place of four. - **Treating every sum of two dice as equally likely.** - **Counting a card twice** in an "or" question. - **Dividing by $52$** after cards have been removed. ## Check yourself 1. Two dice are rolled. What is the probability of a double? !hint A double has the same number on both dice. !answer $6$ doubles, so $\dfrac{6}{36} = \dfrac{1}{6}$. 2. One card is drawn from a full pack. What is the probability that it is a black ace? !hint The black suits are spades and clubs. !answer $\dfrac{2}{52} = \dfrac{1}{26}$. 3. A bag holds $4$ red and $6$ green balls. What is the probability of a green ball? !hint How many balls are there in all? !answer $\dfrac{6}{10} = \dfrac{3}{5}$. 4. Three coins are tossed. What is the probability of at least one tail? !hint Use the complement: no tail at all is HHH. !answer $1 - \dfrac{1}{8} = \dfrac{7}{8}$.39:T11a7,A brief note on **Cell Theory, Prokaryotes and Eukaryotes**. Read it before a test. Try each question at the end before you open its answer. ## The cell theory - **Hooke** named cells in cork. **Leeuwenhoek** first saw live cells. **Brown** found the nucleus. - **Schleiden** (plants) and **Schwann** (animals) gave the cell theory. Schwann noted the cell wall is unique to plants. - **Virchow**: *Omnis cellula-e cellula*, every cell from a cell. - Today: all organisms are made of cells and their products, and all cells arise from pre-existing cells. ## Two kinds of cell - **Eukaryotic**: a membrane-bound nucleus and membrane-bound organelles (protists, fungi, plants, animals). - **Prokaryotic**: no nuclear membrane and no membrane-bound organelles (bacteria, blue-green algae, mycoplasma, PPLO). - Every cell has a plasma membrane, cytoplasm and ribosomes. ## Sizes and shapes - Mycoplasma about $0.3\ \mu\text{m}$; bacteria $3$ to $5\ \mu\text{m}$; a red blood cell about $7.0\ \mu\text{m}$. - Nerve cells are among the longest; the ostrich egg is the largest single cell. - Shape suits the job. A cube of side $s$ has surface area to volume $\dfrac{6}{s}$, which falls as the cell grows. ## The prokaryotic cell - **Envelope**: glycocalyx (slime layer or capsule), cell wall, plasma membrane. Gram positive cells take up the stain; Gram negative ones do not. - **Nucleoid** of naked DNA; **plasmids** carry traits such as antibiotic resistance. - **Mesosomes**: infoldings of the membrane for wall formation, DNA copying, respiration and secretion. **Chromatophores** hold pigments. - **Flagellum**: filament, hook, basal body. **Pili** and **fimbriae** do not move the cell; fimbriae help it stick. - **70S ribosomes** (50S and 30S) form polysomes. **Inclusion bodies** store phosphate, cyanophycean and glycogen granules. ## A worked example A cell has a cell wall, naked DNA, 70S ribosomes and no mitochondria. What kind of cell is it? 1. **Naked DNA.** There is no nuclear membrane. 2. **No mitochondria.** There are no membrane-bound organelles. 3. **70S ribosomes.** These are the ribosomes of prokaryotes. 4. **So** it is a prokaryotic cell, most likely a bacterium. ## Practice with new numbers ```practice { "title": "Bacteria that do not take the stain", "vars": { "n": [ 20, 40, 50 ], "g": [ 6, 10, 14 ] }, "values": { "pct": "100*g/n", "wrong": "100*(n - g)/n" }, "question": "A laboratory grows {{n}} kinds of bacteria. {{g}} of them do not take up the Gram stain. What per cent of them are Gram negative?", "answer": { "value": "pct", "unit": "%", "digits": 1 }, "hint": "Gram negative bacteria are the ones that do not take up the stain.", "mistakes": [ { "value": "wrong", "say": "That is the share that takes up the stain, the Gram positive ones." } ], "solution": "$\\dfrac{ {{g}} }{ {{n}} } \\times 100 = {{pct}}\\%$." } ``` ## A program to try The program prints the main differences between the two kinds of cell as a table. ```python # Prokaryotic and eukaryotic cells side by side. rows = [ ("nuclear membrane", "absent", "present"), ("genetic material", "naked, in a nucleoid", "in a nucleus"), ("membrane-bound organelles", "absent", "present"), ("ribosomes", "70S", "80S in the cytoplasm"), ("examples", "bacteria, blue-green algae", "protists, fungi, plants, animals"), ] for feature, pro, eu in rows: print(f"{feature:<26} {pro:<26} {eu}") ``` !try Run it. Which row tells you where the DNA lies? 1. Add a row for the cell wall. What would you write for each kind of cell? ## Mistakes to avoid - **Thinking Hooke saw living cells.** He saw the walls of dead cork cells. - **Putting the cell wall outermost in a bacterium.** The glycocalyx lies outside it. - **Adding S units.** 50S and 30S make a 70S ribosome. - **Giving mycoplasma a cell wall.** They have none. ## Check yourself 1. What did Virchow add to the cell theory? !hint Think of where new cells come from. !answer That all cells arise from pre-existing cells. 2. Name two organisms with prokaryotic cells. !hint Think of the bacteria and their relatives. !answer Any two of: bacteria, blue-green algae, mycoplasma, PPLO. 3. What is a polysome? !hint Several of one structure on one mRNA. !answer A chain of ribosomes attached to a single mRNA, all making protein from it. 4. Which prokaryotic structure helps a bacterium stick to the tissues of a host? !hint They are small and bristle-like. !answer Fimbriae.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:Tfca,A brief note on **The Simple Pendulum, Damping and Resonance**. Read it before a test. Try each question at the end before you open its answer. ## The simple pendulum - Pull back $mg\sin\theta \approx mg\theta$ for a small angle, in radians. - $T = 2\pi\sqrt{L/g}$. No mass in it, and no small amplitude. - **Seconds pendulum:** $T = 2\ \text{s}$, $L \approx 1\ \text{m}$ on Earth. - **Changed gravity:** use $g_{\text{eff}}$. Lift speeding up upwards, $g + a$. Downwards, $g - a$. Free fall, $0$, so no swing. - **Wide swings** take a little longer than the formula says. ## Other oscillators - **Rule:** pull back in proportion to displacement gives SHM. - **U-tube:** $T = 2\pi\sqrt{L/2g}$, with $L$ the whole column. Density does not matter. - **Floating body:** $T = 2\pi\sqrt{h/g}$, with $h$ the depth under the surface. - **Physical pendulum:** $T = 2\pi\sqrt{I/mgd}$. A rod about one end: $T = 2\pi\sqrt{2L/3g}$. ## Damping and resonance - **Damping force:** $-bv$. - **Amplitude:** $A_0e^{-bt/2m}$, the same factor in each equal time. - **Energy:** $\propto e^{-bt/m}$. Half the amplitude means a quarter of the energy. - **Time to halve the amplitude:** $\dfrac{2m\ln 2}{b}$. - **Forced oscillation:** steady motion at the driving frequency. - **Resonance:** largest amplitude when the driving frequency equals the natural frequency. Less damping, a sharper peak. ## A worked example A swing on chains $1.8\ \text{m}$ long is taken to the Moon, where $g$ is a sixth of its value on Earth. 1. **On Earth.** $T = 2\pi\sqrt{\dfrac{1.8\ \text{m}}{9.8\ \text{m/s}^2}} = 2.69\ \text{s}$. 2. **On the Moon.** $g$ is six times smaller, so $T$ is $\sqrt6$ times larger: $6.60\ \text{s}$. 3. **To keep it going.** Push once in each $6.60\ \text{s}$. That is resonance. ## Practice with new numbers ```practice { "title": "Quick period check", "vars": { "L": [ 0.45, 0.85, 1.25, 1.95 ] }, "values": { "T": "2*pi*sqrt(L/9.8)", "Tu": "2*pi*sqrt(L/19.6)" }, "question": "A pendulum and a U-tube liquid column are both {{L}} m long. Find the period of the pendulum, in s. ($g = 9.8$ m/s²)", "answer": { "value": "T", "unit": "s", "digits": 2 }, "hint": "For the pendulum, $T = 2\\pi\\sqrt{L/g}$.", "mistakes": [ { "value": "Tu", "say": "That is the U-tube, with $2g$. The pendulum uses $g$." } ], "solution": "$T = 2\\pi\\sqrt{\\dfrac{ {{L}}\\ \\text{m} }{9.8\\ \\text{m/s}^2}} = {{T:2}}\\ \\text{s}$. The U-tube is $\\sqrt2$ times faster: ${{Tu:2}}\\ \\text{s}$." } ``` ## A program to try The program finds the length of a seconds pendulum in two places. ```python # A pendulum clock beats seconds: T = 2.00 s. How long is its pendulum, and what if it is moved? import math T = 2.00 for place, g in [("Earth", 9.8), ("a hill town", 9.79)]: L = g * T**2 / (4 * math.pi**2) print(f"{place}: g = {g} m/s^2, L = {L:.4f} m") ``` !try Run it. Which place needs the shorter pendulum? 1. Change `T = 2.00` to `T = 1.00`. By what factor does each length change? ## Mistakes to avoid - **Putting the mass into $T$.** It cancels. - **Using $g$ for the U-tube.** It is $2g$. - **A straight-line fall of amplitude.** It falls by the same factor in equal times. - **Saying a forced body swings at its own frequency.** It swings at the driving frequency. ## Check yourself 1. How long is a seconds pendulum on Earth? !hint Put $T = 2\ \text{s}$ into $L = \dfrac{gT^2}{4\pi^2}$. !answer About $0.99\ \text{m}$, close to one metre. 2. What is the period of the liquid in a U-tube with a column $L$ long? !hint The pull back comes from a level difference of $2y$. !answer $T = 2\pi\sqrt{L/2g}$. 3. The amplitude of an oscillator falls to a third. What fraction of the energy is left? !hint Energy goes as the square of the amplitude. !answer $\left(\tfrac13\right)^2 = \tfrac19$. 4. Why do soldiers break step on a bridge? !hint Think of a push at the natural frequency. !answer Steps in time could match a natural frequency of the bridge. Resonance would make it sway more and more.3d: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.3e:Te52,A brief note on **Probability: Equally Likely Outcomes**. Read it before a test. Try each question at the end before you open its answer. ## Outcomes and events - An **experiment** has results you cannot know in advance. Each result is an **outcome**. - An **event** is a set of outcomes. Its **favourable outcomes** make it happen. - Outcomes are **equally likely** when none has a better chance: a fair die, equal sectors, identical tickets. - Two or three possible results need **not** be equally likely, as with win, lose or draw. ## The probability of an event - With equally likely outcomes, $P(E) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$. - $0 \le P(E) \le 1$. An **impossible** event has $P = 0$; a **sure** event has $P = 1$. - The probabilities of all the **elementary events** add up to $1$. - For unequal sectors, a sector's chance is its angle out of $360^\circ$. ## The complement - $\text{not } E$ is every outcome that does not favour $E$. - $P(E) + P(\text{not } E) = 1$, so $P(\text{not } E) = 1 - P(E)$. - When an event is hard to count, count its complement. ## A worked example One letter is picked at random from **SUCCESS**. Find $P(\text{S})$ and $P(\text{not S})$. 1. **Outcomes.** S, U, C, C, E, S, S: $7$ letters. 2. **The event.** Three S's, so $P(\text{S}) = \dfrac{3}{7}$. 3. **The complement.** $P(\text{not S}) = 1 - \dfrac{3}{7} = \dfrac{4}{7}$. ## Practice with new numbers ```practice { "title": "One die", "vars": { "m": [ 1, 2, 3, 4 ] }, "values": { "p": "(6 - m)/6", "wrong": "(7 - m)/6" }, "question": "One fair die is rolled. What is the probability of a number greater than ${{m}}$? Give it as a decimal.", "answer": { "value": "p", "digits": 3, "absolute": 0.006 }, "hint": "Count the faces from ${{m}} + 1$ to $6$, then divide by $6$.", "mistakes": [ { "value": "wrong", "say": "Greater than ${{m}}$ leaves out ${{m}}$ itself." } ], "solution": "There are $6 - {{m}}$ such faces, so the probability is $\\dfrac{6 - {{m}}}{6} \\approx {{p}}$." } ``` ## A program to try The program finds three probabilities for one die with exact fractions. ```python from fractions import Fraction # One fair die: six equally likely outcomes. outcomes = [1, 2, 3, 4, 5, 6] def prob(event): favourable = [k for k in outcomes if event(k)] return Fraction(len(favourable), len(outcomes)) print("P(a factor of 6) =", prob(lambda k: 6 % k == 0)) print("P(not a factor of 6) =", 1 - prob(lambda k: 6 % k == 0)) print("P(a 7) =", prob(lambda k: k == 7)) ``` !try Run it. Which of the three events is impossible? 1. Add a line for the probability of a number more than $4$. ## Mistakes to avoid - **Using the formula** when the outcomes are not equally likely. - **Dividing by the number of different letters** instead of all the letters. - **A probability above $1$ or below $0$:** recheck the counts. - **Treating "does not win" as "loses"** when a draw is possible. ## Check yourself 1. A spinner has nine equal sectors numbered $1$ to $9$. What is the probability of an odd number? !hint The odd numbers are $1, 3, 5, 7, 9$. !answer $\dfrac{5}{9}$. 2. What is the probability of a $0$ on one fair die? !hint Does any face show $0$? !answer $0$: it is impossible. 3. $P(E) = 0.35$. What is $P(\text{not } E)$? !hint Take it away from $1$. !answer $1 - 0.35 = 0.65$. 4. Why is the chance of red on a spinner with a red half and two blue quarters not $\dfrac{1}{3}$? !hint Are the three sectors the same size? !answer The sectors are not equal, so the outcomes are not equally likely. Red covers half the circle, so its chance is $\dfrac{1}{2}$.3f:Tf2e,A brief note on **Our Impact on the Environment**. Read it before a test. Try each question at the end before you open its answer. ## The ozone layer - **Ozone** is $\mathrm{O_3}$: three oxygen atoms. Near the ground it is a **poison**. - High in the atmosphere it forms the **ozone layer**, which absorbs most of the Sun's **UV radiation**. - UV can cause **skin cancer** and harm the eyes and the body's defences. - It forms in two steps: $\mathrm{O_2} \to \mathrm{O} + \mathrm{O}$ under UV, then $\mathrm{O} + \mathrm{O_2} \to \mathrm{O_3}$. - **CFCs** from refrigerants and fire extinguishers thinned the layer. - In **1987** a **UNEP** agreement froze CFC production at 1986 levels. ## Two kinds of waste - **Biodegradable:** broken down by decomposers. Peels, food, paper, cotton, leaves. - **Non-biodegradable:** not broken down. Plastic, glass, metal cans, some pesticides. - Decomposers' **enzymes** act only on certain substances, so plastics last for centuries. - Plastic blocks drains, harms animals that eat it, and fills landfills. ## Managing garbage - **Reduce, reuse, recycle**, in that order. - **Green bin:** wet waste, for compost. **Blue bin:** dry waste, for recycling. - Burning garbage releases **harmful gases**. - Every choice has a cost: **kulhads** use up fertile topsoil. ## A worked example A family's weekly waste is $5$ kg of wet waste and $3$ kg of dry waste. About $40$ % of the wet waste remains as compost. What is the wet share, and how much compost is made? 1. **All the waste.** $5 + 3 = 8$ kg. 2. **Wet share.** $\dfrac{5}{8} \times 100 = 62.5$ %. 3. **Compost.** $5 \times \dfrac{40}{100} = 2$ kg. ## Practice with new numbers ```practice { "title": "The wet share of a bin", "vars": { "w": [ 6, 9, 12 ], "d": [ 4, 6, 3 ] }, "values": { "pc": "100*w/(w + d)", "wrong": "100*w/d" }, "question": "A household's waste for a week is {{w}} kg of wet waste and {{d}} kg of dry waste. What percentage of it is wet?", "answer": { "value": "pc", "unit": "%", "digits": 0 }, "hint": "Divide the wet waste by all the waste, then multiply by 100.", "mistakes": [ { "value": "wrong", "say": "Divide by all the waste, wet and dry together, not by the dry waste alone." } ], "solution": "$\\dfrac{ {{w}} }{ {{w}} + {{d}} } \\times 100 = {{pc}}$ %." } ``` ## A program to try The program finds a family's garbage in a month. ```python # A family's garbage in a month, from what each person throws away a day. people = 4 grams_each = 450 # grams a person a day days = 30 kg = people * grams_each * days / 1000 print("Garbage in a month:", kg, "kg") ``` !try Run it. How many kilograms does the family throw away in a month? 1. Change `grams_each` to `300`. How much now? ## Mistakes to avoid - **Not all ozone is good.** Near the ground it is a poison. - **CFCs, not carbon dioxide,** thinned the ozone layer. - **Paper is biodegradable,** though it goes in the dry bin for recycling. - **Reduce comes before recycle.** - **Kulhads are not free of cost:** they use up topsoil. ## Check yourself 1. What is the chemical formula of ozone? !hint Count the oxygen atoms. !answer $\mathrm{O_3}$. 2. Which chemicals were the main cause of the thinning of the ozone layer? !hint They were used in refrigerators. !answer Chlorofluorocarbons, CFCs. 3. Name one biodegradable and one non-biodegradable item from a kitchen. !hint Which can bacteria break down? !answer Biodegradable: vegetable peels. Non-biodegradable: a plastic milk packet. 4. A family makes $7$ kg of wet waste and $3$ kg of dry waste in a week. What percentage is wet? !hint Divide the wet waste by all the waste. !answer $\dfrac{7}{10} \times 100 = 70$ %. 5. Why is reducing better than recycling? !hint What does recycling still need? !answer Waste that is never made needs no collecting or energy to recycle.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:T10db,A brief note on **Force Law, Springs and Energy in SHM**. Read it before a test. Try each question at the end before you open its answer. ## The force law - **Simple harmonic motion** needs a restoring force in proportion to the displacement: $F = -kx$. - **Angular frequency:** $\omega = \sqrt{k/m}$. - **Period:** $T = 2\pi\sqrt{m/k}$, and $f = 1/T$. - The period does not depend on the amplitude, or on $g$. ## Springs - **Hanging spring:** at rest $kd = mg$, and $T = 2\pi\sqrt{d/g}$. - **Side by side, in parallel:** $k = k_1 + k_2$. - **End to end, in series:** $\dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$. - **A block between two springs and two walls:** parallel. - **A piece of a spring** is stiffer: $k \times$ length stays the same. - **Two bodies on one spring:** use $\mu = \dfrac{m_1m_2}{m_1 + m_2}$. ## Energy - **Potential energy:** $U = \tfrac12 kx^2$. - **Total energy:** $E = \tfrac12 kA^2$. - **Speed:** $v = \omega\sqrt{A^2 - x^2}$, largest $v_{max} = \omega A$ at the middle. - $K = U$ at $x = A/\sqrt2$. - Each energy repeats every $T/2$ and averages to $E/2$. ## A worked example A block of $0.30\ \text{kg}$ on a spring of $120\ \text{N/m}$ swings with an amplitude of $4.0\ \text{cm}$. 1. **Angular frequency.** $\omega = \sqrt{\dfrac{120\ \text{N/m}}{0.30\ \text{kg}}} = 20\ \text{rad/s}$. 2. **Period.** $T = \dfrac{2\pi}{20\ \text{rad/s}} = 0.314\ \text{s}$. 3. **Energy.** $E = \tfrac12 \times 120\ \text{N/m} \times (0.040\ \text{m})^2 = 0.096\ \text{J}$. 4. **Largest speed.** $v_{max} = 20\ \text{rad/s} \times 0.040\ \text{m} = 0.80\ \text{m/s}$. ## Practice with new numbers ```practice { "title": "Spring, mass and energy", "vars": { "k": [ 60, 150, 240 ], "m": [ 0.3, 0.6 ], "A": [ 3, 5, 7 ] }, "values": { "E": "0.5*k*(A/100)^2", "wrong": "0.5*k*A*A" }, "question": "A {{m}} kg block on a spring of {{k}} N/m oscillates with an amplitude of {{A}} cm. What is the total energy, in J?", "answer": { "value": "E", "unit": "J", "digits": 4 }, "hint": "$E = \\tfrac12 kA^2$, with $A$ in metres. The mass is not needed.", "mistakes": [ { "value": "wrong", "say": "Put the amplitude in metres before you square it." } ], "solution": "$E = \\tfrac12 \\times {{k}}\\ \\text{N/m} \\times ({{A}} \\times 10^{-2}\\ \\text{m})^2 = {{E:4}}\\ \\text{J}$." } ``` ## A program to try The program works through the example above. ```python # The worked example of the note: a block on a spring. import math k = 120.0 # spring constant, N/m m = 0.30 # mass, kg A = 0.040 # amplitude, m w = math.sqrt(k / m) print("angular frequency:", round(w, 2), "rad/s") print("period:", round(2 * math.pi / w, 3), "s") print("total energy:", round(0.5 * k * A**2, 4), "J") print("largest speed:", round(w * A, 3), "m/s") ``` !try Run it. Compare each line with the worked example. 1. Change `m = 0.30` to `m = 1.20`. Which numbers change, and which stays the same? ## Mistakes to avoid - **Writing $T = 2\pi\sqrt{k/m}$.** The mass goes on top. - **Adding spring constants in series.** In series, add the inverses. - **Forgetting to square the amplitude** in $\tfrac12 kA^2$. - **Putting equal energies at $A/2$.** They are equal at $A/\sqrt2$. ## Check yourself 1. A block on a spring has a period of $0.5\ \text{s}$. The spring is swapped for one four times as stiff. What is the new period? !hint $T$ goes as $1/\sqrt{k}$. !answer $\sqrt4 = 2$, so the period halves to $0.25\ \text{s}$. 2. Two springs of $30\ \text{N/m}$ and $60\ \text{N/m}$ are joined end to end. What is the spring constant of the pair? !hint End to end means series: add the inverses. !answer $\dfrac{1}{k} = \dfrac{1}{30\ \text{N/m}} + \dfrac{1}{60\ \text{N/m}}$, so $k = 20\ \text{N/m}$. 3. The amplitude of an oscillator is halved. What happens to its largest speed and to its energy? !hint $v_{max} = \omega A$ and $E = \tfrac12 kA^2$. !answer The largest speed halves. The energy becomes a quarter. 4. Why does a hanging spring have the same period as the same spring on a smooth floor? !hint Measure the displacement from the new rest point. What is the net force there? !answer Gravity only moves the rest point down by $d = mg/k$. About that point the net force is still $-ky$, so the period is still $2\pi\sqrt{m/k}$.7:["$","div",null,{"className":"max-w-7xl mx-auto px-6 py-8 sm:py-16","children":[["$","div",null,{"className":"flex items-start justify-between flex-wrap gap-4","children":[["$","div",null,{"children":[["$","h1",null,{"className":"text-4xl font-bold dark:text-white","children":"Notes"}],["$","p",null,{"className":"text-slate-500 dark:text-slate-400 mt-2 max-w-3xl","children":"Reading material from teachers and learners: theory, worked examples and questions that make you think. 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