kld unit 4 2014 student notes - home: future students :...

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2014 KLD Chemistry VCE Revision Lecture Unit 4

Kwong Lee Dow Young Scholars Program Chemistry Revision Unit 4 2014

VCE Chemistry Exam 150 minutes (+15 mins reading time) Tues 11 November 9.00 - 11.45 am

Presenter: Phil Ponder (Penleigh and Essendon Grammar)


Useful Websites

• The VCAA website is www.vcaa.vic.edu.au • The Chemistry Education Association website is

www.cea.asn.au • A lot of bright and conscientious students seem to use

www.atarnotes.com to exchange ideas and materials.

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VCAA Exams 2008 – 2013 (VCAA Website)


VCAA Exams 2000 – 2007 (CEA Website)

!

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Analyzing Practice Exams

USE THE EXAMINERS’ REPORTS These are available from the VCAA and/or CEA websites for all VCAA exams from 2000 to 2013 that have been sat and marked.

– take careful note of questions that were done poorly and the examiners’ thoughts on why this was the case (Poorly done questions are often revisited on subsequent exams) – look carefully at the mark allocation – look at ‘popular’ but incorrect multiple-choice answers and try to determine why they were selected (and why they are wrong!) – examine your own responses critically in the light of the examiners’ expectations – add noteworthy remarks to your topic summaries as reminders


Four ways to save marks (SUBS)

Calculations

•  Significant Figures: All of your numerical answers must have the correct number of significant figures

• Unit: unless a question specifically asks for a particular unit, you will be penalized if you have no unit or an incorrect unit

Equations

• Balance: Double check that any equation you write

is balanced for both elements and charge • Symbols of State: Include and use the terms

(aq), (s), (l) and (g) properly

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Significant Figures

ADDING / SUBTRACTING - for addition or subtraction, the result is rounded to the smallest number of DECIMAL PLACES:

eg if adding: NB: the periodic table in the DATA BOOK gives atomic masses to one decimal place. - so all molar masses should be calculated to one decimal place


Significant Figures

MULTIPLYING / DIVIDING - for multiplication and/or division, the result is rounded to the smallest number of SIGNIFICANT FIGURES in the data used.

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Significant Figures for pH

Unlikely to be tested on the exam, but best to be thorough!

[H+] = 1.45 x 10–4 M (3 sig figs) pH = –log10(1.45 x 10–4 ) = 3.839 (3 sig figs) The number before the decimal point in a pH refers to the power of 10, so the 3 sig figs we require come after the decimal point.

pH = 7.85 The 7 refers to the power of 10, so there are only 2 sig figs

(the digits after the decimal point) [H+] = 10–7.85 = 1.4 x 10–8 M (2 sig figs) Beware It is not acceptable to leave your answer as [H+] = 10–7.85 M


Significant Figures

Example from June 2012 VCAA Exam, Q.7

The original mass measurements had 4 significant figures. Mass of PbI2 precipitate = 0.4059 – 0.3120 = 0.0939 g

• This value is accurate to the fourth decimal place (subtraction rule) and so has 3 significant figures (initial zeros are not significant) • In a calculation involving multiplication or division, the number of significant figures in the answer should be the same as the number of significant figures in the piece of data that has the least number of significant figures (ie. the least accurate bit of data).

n(PbI2) = 0.0939 / 461.0 = 2.04 x 10-4 mol (accurate to 3 sig figs) = n(Pb(CH3COO)2) in 20.0 mL of hair dye m(Pb(CH3COO)2) in 20.0 mL of hair dye = 2.04 x 10-4 x 325.3 = 0.0663 g m(Pb(CH3COO)2) in 100.0 mL of hair dye = 0.0663 x (100.0/20.0) = 0.331 g

Many students gave the answer to 4 significant figures.

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Formulae

MOLE / STOICHIOMETRY OTHER

n = m / Mm pH= –log[H+] n = V / Vm [H+]= 10–pH n = c × V calibration factor = ∆E /∆T n = No. / NA d (density) = m/V pV = nRT Q = It n(e–) = Q / F E = VIt (hence E=VQ)

E = 4.18 × m × ∆T


Electrochemical Series

Write ‘oxidant’ at the top left of the table (above F2 (g)) and ‘reductant’ at the top right (above 2F–

(aq)). F2 is the strongest oxidant - a substance that causes the oxidation

of something else, and is itself reduced.

BEWARE: a common mistake is to name the substance being oxidised as ‘the oxidant’, or calling the substance being reduced ‘the reductant’.

The Electrochemical Series can be used to predict : • relative strengths of oxidants and reductants • half reactions that occur in galvanic and electrolytic cells • whether or not a redox reaction is likely to occur between 2 reactants.

Note that many of the reductants on the right hand side of the table are metal elements.

- if a metal element is reacting, it will be oxidised, and the substance it reacts with will be reduced.

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• Half reactions for the H2/O2 fuel cell using acid or alkaline electrolyte • If an aqueous solution is involved, don’t forget H2O as a possible reactant when predicting electrolytic cell reactions or general redox reactions


H2 / O2 Fuel Cell

The overall equation is the same as for the combustion of hydrogen: 2H2 (g) + O2 (g) → 2H2O(g) ΔH = –ve

Being an exothermic redox reaction, electrical energy can be obtained directly from chemical energy if the two half reactions can be separated,

• oxidation of hydrogen occurs at the anode • reduction of oxygen at the cathode (‘red cat’ helps you to remember that!)

Acid electrolyte, you want H+ in the half equations (you should be able to find them at Eo = +1.23 v and 0.00 v)

Alkaline electrolyte, you want OH– in the half equations (at Eo = +0.40 v and –0.83 v)

Note that in both electrolytes, the predicted cell voltage under standard conditions is 1.23v, as you would expect, given that both cells would have the same overall reaction.

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Predicting Redox Reactions

A USEFUL GENERAL IDEA for Galvanic Cell and General Redox Reaction Predictions • in any situation where two half-equations are combined

- the one with the higher Eo value will go forward - the one with the lower Eo value will go backward

• hence - “top left → bottom right” - “backward-Z” - “favourable diagonal” (fav diag)

• a reaction is predicted to occur if Eo

oxidant > Eo

reductant



The following reactions occur spontaneously as written.

2Cr2+(aq) + Co2+

(aq) → 2Cr3+(aq) + Co(s)

Co(s) + Pb2+(aq) → Co2+

(aq) + Pb(s) Fe(s) + 2Cr3+

(aq) → Fe2+(aq) + 2Cr2+

(aq)

Using this information, predict which one of the following pairs of reactants will react spontaneously.

A. Co(s) + Fe2+(aq)

B. Cr2+(aq) + Fe2+

(aq) C. Cr2+

(aq) + Pb2+(aq)

D. Pb(s) + Co2+(aq)

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BEWARE: Know the limitations of Electrochemical Series predictions: • A predicted reaction may seem not to occur if it is very slow

(eg H2O2 reacting with itself) The Series cannot predict rate of reaction (which is related to EA)

• An oxide coating may make a reactive metal seem unreactive

(eg Al2O3 on Al) • Non-standard conditions

(eg Cl–(aq) can be oxidised in preference to H2O if [Cl–] is high) However, in general, predictions can be relied on,

even if conditions are not standard.



BEWARE: When predicting electrolysis half reactions, don’t forget:

• to consider H2O in your list of possible reactants if it is an aqueous electrolyte

• that reactive metals such as Li, Na, K, Mg, Ca, Al cannot be produced by electrolysing an aqueous solution of their cations.

(ie cations with Eo < – 0.83 v cannot be reduced in aqueous solution, because H2O will reduce in preference)

Likewise, anions with Eo > 1.23 v (eg F–) cannot (theoretically) be oxidised in aqueous solution. (note, however, an important limitation of electrochemical series predictions; if the [Cl–(aq)] is high enough, Cl2 (g) will be produced at the anode in an electrolytic cell)

• that the subscript for an ion in a molten electrolyte will not be (aq). Use (l) eg Na+(aq) + e– → Na(s)

Na+ would not be reduced if it were in aqueous solution!

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BEWARE: Don’t blindly use ‘non-standard conditions’ as a reason for failure to observe a predicted reaction.

Refer more specifically to • T < 25˚C or • concentration < 1M or • pressure < 1 atm.

Any (or all) of these could be predicted to slow the reaction rate.


Reaction Rate and Particle Collision Theory

• the rate of a reaction is determined by the frequency of successful (‘fruitful’) collisions

increased by: increased by: • higher reactant concentrations • higher collision energies • larger solid surface area – higher temperature • higher temperature • lower activation energy • higher gas pressure – catalyst

• note the ‘double-effect’ of a temperature increase Beware of using “proportion of successful collisions”

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Chemical Equilibrium

What is chemical equilibrium? • a dynamic, steady state resulting from the equal rates of forward and reverse reactions

What is an ‘equilibrium shift’? • a change in the composition of a mixture resulting from the rates of forward and reverse reactions being temporarily unequal

When will an equilibrium shift occur? • when any change causes the rates of forward and reverse reactions to become unequal


Le Chatelier’s Principle

Remember: at equilibrium : • rates of forward and back reactions are equal • amounts and concentrations are constant (but not usually equal)

• the ‘naughty boy’ principle (LCP) states that if a change is imposed on a system that is at equilibrium, the system will try to partially oppose the change. • LCP (Le Chatelier’s Principle) is merely a useful guide as to what may happen when a change is imposed on a system that is at equilibrium. - it does not cause the response!

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A system at equilibrium will only respond to a change if the • ratef ≠ rateb (Rate of forward reaction no longer equals the rate of the back reaction) • Cf ≠ Kc (Concentration Fraction no longer equals the Equilibrium Constant)


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colourless reddish brown

N2O4 (g) ⇌ 2 NO2 (g) ; ∆H = +ve (ie forward reaction is endothermic)

Kc = 5.5 x 10-3 M (at 25˚C), and increases as T increases = Kc if the system is at equilibrium ie


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Beware of situations where there is a pressure change at constant volume

When the pressure change is not a result of a volume change, LCP will probably not correctly predict the effect.

Examples - adding inert gas at constant volume

- heating a gas mixture at constant volume


LCP Problem ?

Argon, a colourless, inert gas is added to a vessel of fixed volume in order to increase the total pressure. What effect does this have on the colour of the mixture? colourless reddish brown

N2O4 (g) ⇌ 2 NO2 (g) After addition of argon, the concentration fraction has not changed, since neither of the reactant or product concentrations has changed.

If the concentration fraction is still equal to K, then the system is still at equilibrium. ⇒ no change in the colour of the mixture

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LCP Problem ?

Alternatively,

• the addition of argon does not change the concentration of reactants or products ⇒ the rate of collisions between reactant particles (and between product particles) is not changed

so that either • the rates of forward and reverse reactions are unaffected by the argon, or • the argon has a catalytic effect that increases the rates of forward and reverse reactions equally

Either way, the rates remain equal and hence: The addition of argon affects the forward and reverse reactions equally (or not at all) so that no shift occurs

⇒ no change in the colour of the mixture


Physical Constants

The unit for the Faraday constant (C mol–1) allows you to deduce the relevant formula:

F = Q (in coulomb) / n(e–) hence n(e–) = Q / F

You also need Q = It (which looks like advice to smokers from the anti-cancer council!)

Faraday Calculations n(substance) ⇔ n(e–) ⇔ Q ⇔ I and/or t

use half equation use Q = F × n(e–) use Q = It

Also, since E = VIt , you can derive a useful relationship for the energy produced by a cell:

E = VQ = V × F × n(e–).

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Faraday Questions

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Calorimetry

• Any calculation that refers to the heating of water may require use of the Specific heat capacity (c) of water.

You might prefer to memorise the relationship: E = 4.18 × m × ΔT

BEWARE: the m in this formula is the mass of water, not the mass of fuel being burned.

• It may be preferable to look at the unit (J g–1 K–1) and deduce that 4.18 J of energy will be required to heat 1g of water by 1K (1˚C). The problem can then be solved using ratios.

• Note (i) that a change in temperature measured in oC will have the same numerical value as when measured in K eg ΔT = 298 K – 273 K = 25 K ⇒ 25 ˚C – 0 ˚C = 25 ˚C (ii) 1 g of water is equivalent to 1 mL, using the density of water

(iii) E is in Joule (not kJ)


Calorimetry

• A rough value for the Calibration Factor of a calorimeter can easily be estimated if it is assumed that the water in the calorimeter absorbs most of the energy.

eg 100 mL of water in the calorimeter will absorb 418 J/˚C.

• The true value of the Calibration Factor will be higher than this (since other parts of the calorimeter will also absorb heat)

• The Calibration Factor can be more accurately found (i) using an input of electrical energy, E = VIt or (ii) by performing a reaction with a known ΔH

Calibration Factor = ∆E /∆T

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Calorimetry Question

A bomb calorimeter may be calibrated using a substance with a well-known heat of combustion. A commonly used calibrating agent is benzoic acid (C7H6O2), which has a heat of combustion of 3227 kJ mol-1

2.50 g of pure solid benzoic acid (Molar mass = 122.0 g mol–1)

is placed in a calorimeter and completely reacted with oxygen. The temperature rise of the calorimeter is observed to be 8.90˚C. Calculate the calibration factor of the calorimeter in kJ ˚C–1.


Ionic product for water

Kw = [H+] × [OH–] = 1.00 × 10–14 M2 at 25 ˚C Kw will be required for pH calculations in alkaline (basic) solutions.

pH = – log10 [H+] [H+] = 10–pH (Remember that H3O+ is interchangeable with H+)

Q.1 Calculate the pH of 0.0050 M Ca(OH)2 solution at 25°C.

The dissociation of Ca(OH)2 is often overlooked: Ca(OH)2 (s) → Ca2+

(aq) + 2OH–(aq).

This indicates that the [OH–] is 2 × [Ca(OH)2]. • The self-ionisation of water equilibrium law applies to aqueous solutions as well as pure water.

[OH–] = 2 × 0.0050 = 0.010 = 1.0 × 10–2 M [H+] = 10–14 / 10–2 = 10–12 M pH = 12

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Q.2 A 10,000 L swimming pool has a pH of 9.0. What volume of 1.0 M HCl is required to bring the pH down to 8.0?

You will need to assume that (i) The HCl added makes an insignificant contribution to the total volume (a reasonable assumption) (ii) OH– is the only base in the pool (unlikely)

(Unsurprisingly, the answer is not 0.090 mL !)



BEWARE : A neutral solution does not usually have a pH of 7. • Because Kw is an equilibrium constant, it will vary with temperature • Self ionisation of water is an endothermic reaction: H2O(l) ⇌ H+

(aq) + OH–(aq) ; ΔH = +ve

BEWARE : Kw will only equal 1.00 × 10–14 at 25 ˚C. At this temperature, [H+] in a neutral solution = [OH–] = 10–7 M, and hence pH = 7. Kw will, however increase at T > 25 ˚C, so [H+] > 10–7 M and pH of a neutral solution will be less than 7. Likewise, at T < 25 ˚C, the pH of a neutral solution will be more than 7.

BEWARE : Because pH is a measure of acidity, it is easy for your brain to think that if a solution becomes more acidic (higher [H+]), the pH must be increasing. • There is an inverse relationship between [H+] and pH.

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Acidity Constants

• The higher the Ka, the stronger the acid • All acids in the Data Book tables are weak acids. (The Ka’s for strong acids are very high) • There is a relationship between the Ka for the conjugate acid in an acid-base indicator and the pH of the endpoint of that indicator. • Any indicator is a mixture of a weak acid (HIn) and its conjugate base (In–), where these two substances have different colours (see Data Book). • We can assume that the ‘official’ endpoint for the indicator (when it changes colour) is roughly when [HIn] ≈ [In–]. For the hydrolysis equation: HIn(aq) ⇌ H+

(aq) + In–(aq)

When [HIn] = [In–] (ie at the endpoint) ,

• At the endpoint, [H+] = Ka and pH = –log10Ka

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Acidity Constants

Q. Bromophenol blue is a weak acid (represented as BH) that acts as an acid-base indicator.

In solution the following equilibrium is established. BH(aq) ⇌ B–

(aq) + H+(aq); Ka = 6.3 × 10–5 M

yellow blue At low pH bromophenol blue exists mainly as the acid, BH, which is yellow in colour, while at high pH it exists mainly as its conjugate base, B–, which is blue. An intermediate colour is observed when the concentration of the acid and the concentration of the conjugate base are similar. When [BH] = [B–], the mixture appears green.

(i) Calculate the pH at which [BH] = [B–]. (ii) Calculate the ratio [B–] / [BH] when the pH of a solution of bromophenol blue is 7. (iii) What colour will the indicator solution appear at pH 7? Justify your answer.


Acidity Constants

** If a calculation involves a weak acid, you probably need to look up its Ka ** Change in pH when a weak acid is diluted

• Equilibrium and Le Chatelier’s Principle will be relevant. • Compare dilution by a factor of ten of HCl from 1.0 M to 0.10 M,

with dilution of CH3COOH from 1.0 M to 0.10 M. Because HCl can be regarded as essentially fully ionised, [H+] goes from 100 M to 10–1 M, so the pH changes from 0.0 to 1.0

For the weak ethanoic acid, Ka = 1.7 × 10–5 M (at 25˚C). This applies to the equation: CH3COOH(aq) + H2O(l) ⇌ H3O+

(aq) + CH3COO–(aq)

Making the usual 2 assumptions: (i) [CH3COOH]at eq’m ≈ [CH3COOH]initially (ii) [H+] from self-ionisation of water is negligible

leads to: For 1.0 M CH3COOH For 0.10 M CH3COOH Ka = [H+]2 / 1.0 = 1.7 × 10–5 [H+]2 / 0.10 = 1.7 × 10–5

∴ [H+] = 0.0041 M and pH = 2.4 (2.38) ∴ [H+] = 0.0013 M and pH = 2.9 (2.88)

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Acidity Constants Why is the pH change smaller for dilution of the weak acid?

• Dilution has increased the % hydrolysis of the ethanoic acid • Addition of one of the reactants in the equilibrium has resulted in a net forward reaction, leading to an increase in the n(H+) in the solution. However, the greater volume of solution means that the overall [H+] will decrease, but not by as much as it did for the strong acid, where there were effectively no equilibrium considerations. • When a system is at equilibrium, the rate of the forward reaction equals the rate of the back reaction. • Addition of water to the ethanoic acid equilibrium will hardly affect the frequency of collisions between the reactant particles in the forward reaction, but will definitely initially decrease the rate of the back reaction

– lower concentration of products will lead to less frequent collisions between product particles.

• With the two rates no longer equal, the system is not at equilibrium • Forward reaction will occur to a greater extent (thus slowing down the forward reaction) • [CH3COOH] will decrease and the [H3O+] and [CH3COO–] will increase • Eventually the two rates will be equal again.


Acidity Constants

of dilution of dilution

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Molar Enthalpy of Combustion

BEWARE: Recognize the difference between the Heat of Combustion of a fuel, and the ΔH for the reaction in which the fuel reacts with oxygen.

• Heat of Combustion is merely the amount of energy produced when a given amount of a fuel is completely reacted with oxygen. • Look very carefully at the unit of the quantity involved in measuring the amount of fuel.

Heat of Combustion of ethane is 1557 kJ/mol • 1557 kJ of energy will be released if 1 mole of ethane is completely reacted with oxygen. • Heat of Combustion of ethane could also be expressed as 51.9 kJ/g or 63.6 kJ/L at SLC (25˚C and 1 atm).

Are you able to use data from the Data Book to confirm these values? Note that they are only given to 3 significant figures; do you know why?



• In the Data Book, Heats of Combustion are given in the form of ΔHc, with the unit in kJ mol-1.

- In these cases, the ‘per mol’ bit refers specifically to the fuel. • ΔH is normally used as part of a specific thermochemical equation,

- In such cases the ‘per mol’ bit does not refer to any one reactant or product in particular (ΔH doesn’t play favourites!). eg The data book can be used to write the thermochemical equation for the combustion of ethane as: C2H6 (g) + 7/2 O2 (g) → 2 CO2 (g) + 3 H2O(l) ; ΔH = – 1557 kJ mol-1

If you choose to eliminate the fraction from the co-efficients, the ΔH value must change, since the equation will no longer be referring to 1 mole of ethane reacting with 7/2 mole of oxygen, but 2 mole of ethane reacting with 7 mole of oxygen (so twice as much energy will be released).

2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O(l) ; ΔH = – 3114 kJ mol-1 Note that the unit is still kJ mol-1

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BEWARE : the Heat of Combustion of any fuel (including mixtures) can be expressed as kJ/g, but the unit kJ/mol can only be used for a pure substance with a known molar mass. Make sure you can convert between these units.

BEWARE the O in CHO when writing the equation for the combustion

of an alcohol (or biofuel) • don’t overlook the oxygen in the alcohol when balancing the equation • don’t use (aq) as the symbol of state for the alcohol.



Why can kJ/mol not be used on a biscuit packet to express the energy content of a biscuit? Why is carrot cake considered to be more fattening than an equal mass of carrots? (Both are mostly carbohydrate and would produce similar amounts of energy per gram if burned in a bomb calorimeter)

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Fun with Ethanol?


Answer to Calorimetry Question

• n(C7H6O2) = 2.50/122.0 = 0.0205 mol • E = 0.0205 x 3227 = 66.1 kJ • calibration factor = 66.1 kJ

8.90!°C= 7.43 kJ °C–1 (NB: 3 sig figs)

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Answer to swimming pool question

At pH = 9.0, [H+] = 10–9 M so [OH–] = 10–5 M n(OH–) = cV = 10–5 × 10000 = 0.10 mol

At pH = 8.0, [H+] = 10–8 M so [OH–] = 10–6 M n(OH–) = cV = 10–6 × 10000 = 0.010 mol

Hence 0.10 – 0.010 = 0.090 mol of OH–

needs to be neutralized Hence 0.090 mol of H+ will be required (0.090 mol of HCl) V(HCl) = n/c = 0.090 / 1.0 = 0.090 L = 90 mL


Answer to Bromophenol question

Answer to Bromophenol question

(i) [B! ][BH]

= 1 Ka = [H+ ][B! ][BH]

= 6.3 ! 10–5 ! [H+] = 6.3 ! 10–5 M

! pH = – log10(6.3 ! 10–5) = 4.2 At this pH the solution will be green.

(ii) [B! ][BH]

= [Ka ][H+]

= 6.3 ! 10–5 / 10–7 = 630

(iii) [B–] = 630 " [BH] So [blue] = 630 " [yellow] Hence solution will be blue.

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