All practice questions

Question 261Question →

A $10.0\text{ g}$ sample of impure calcium trioxocarbonate(IV), $\text{CaCO}_3$ , was strongly heated until decomposition was complete. If the volume of carbon(IV) oxide gas evolved at STP was $1.792\text{ dm}^3$ , what is the percentage purity of the $\text{CaCO}_3$ sample? [Molar volume of gas at STP = $22.4\text{ dm}^3\text{mol}^{-1}$ ; relative atomic masses: $\text{Ca}=40, \text{C}=12, \text{O}=16$ ]

Show answer & explanation

Answer: 80

Answer

The percentage purity of the calcium trioxocarbonate(IV) sample is

80\%

.

Thermal decomposition of pure calcium trioxocarbonate(IV) releases carbon(IV) oxide gas according to

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

. Dividing the gas volume (

1.792\text{ dm}^3

) by the molar gas volume at STP (

22.4\text{ dm}^3\text{mol}^{-1}

) yields

0.08\text{ mol}

of

\text{CO}_2

. Due to the 1:1 stoichiometry,

0.08\text{ mol}

of pure

\text{CaCO}_3

reacted. Multiplying by the molar mass of

\text{CaCO}_3

(

100\text{ g/mol}

) gives

8.0\text{ g}

of pure

\text{CaCO}_3

. The percentage purity is calculated as

\left(\frac{8.0\text{ g}}{10.0\text{ g}}\right) \times 100\% = 80\%

.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of calcium trioxocarbonate(IV).

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

Establishes the 1:1 stoichiometric mole ratio between

\text{CaCO}_3

and

\text{CO}_2

.

2

Calculate the number of moles of

\text{CO}_2

gas produced at STP.

\text{Moles of CO}_2 = \frac{1.792\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.08\text{ mol}

Uses the molar gas volume relationship at standard temperature and pressure (

V / V_m

).

3

Calculate the mass of pure

\text{CaCO}_3

in the original sample.

\text{Mass of CaCO}_3 = 0.08\text{ mol} \times 100\text{ g/mol} = 8.0\text{ g}

Because

1\text{ mol}

of

\text{CaCO}_3

produces

1\text{ mol}

of

\text{CO}_2

,

0.08\text{ mol}

of pure

\text{CaCO}_3

reacted.

4

Calculate the percentage purity of the sample.

\text{Percentage purity} = \left(\frac{8.0\text{ g}}{10.0\text{ g}}\right) \times 100\% = 80\%

Compares the mass of active pure reactant to the total mass of the impure sample.

Key Concept

Stoichiometry of thermal decomposition of trioxocarbonate(IV) salts and gas molar volume calculations at STP.

Question 262Question →

A

10.0\text{ g}

sample of impure calcium carbonate (

\text{CaCO}_3

) is completely decomposed by strong heating according to the chemical equation:

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

The carbon(IV) oxide gas evolved is passed into an excess solution of sodium hydroxide, causing the mass of the solution to increase by

3.52\text{ g}

. Assuming the impurities present in the sample do not react or produce any gas, what is the percentage purity of the calcium carbonate sample?

[\text{Ca} = 40, \text{C} = 12, \text{O} = 16]

Show answer & explanation

Answer: 80

Answer

The percentage purity of the calcium carbonate sample is 80%.

Sodium hydroxide absorbs carbon(IV) oxide (

\text{CO}_2

) gas released during the thermal decomposition of calcium carbonate (

\text{CaCO}_3

). The

3.52\text{ g}

mass gain of the solution equals the mass of

\text{CO}_2

evolved. Dividing this mass by the molar mass of

\text{CO}_2

(

44\text{ g/mol}

) yields

0.08\text{ mol}

of

\text{CO}_2

. According to the 1:1 stoichiometric relationship,

0.08\text{ mol}

of pure

\text{CaCO}_3

decomposed. Multiplying by the molar mass of

\text{CaCO}_3

(

100\text{ g/mol}

) gives

8.00\text{ g}

of pure

\text{CaCO}_3

. The percentage purity is

(8.00\text{ g} / 10.0\text{ g}) \times 100\% = 80\%

.

Step-by-Step Solution

1

Calculate the molar masses of carbon(IV) oxide and calcium carbonate

Molar mass of CO2 = 44 g/mol, Molar mass of CaCO3 = 100 g/mol

Molar masses are required to convert between mass and moles.

2

Determine the moles of carbon(IV) oxide gas evolved

Moles of CO2 = 3.52 g / 44 g/mol = 0.08 mol

Sodium hydroxide reacts with and absorbs acidic carbon(IV) oxide, so mass increase equals the mass of CO2.

3

Determine the mass of pure calcium carbonate in the sample

Mass of pure CaCO3 = 0.08 mol * 100 g/mol = 8.00 g

The mole ratio of CaCO3 to CO2 in the thermal decomposition reaction is 1:1.

4

Calculate the percentage purity of the sample

(8.00 g / 10.0 g) * 100 = 80%

Percentage purity is the ratio of pure reactive substance mass to total sample mass expressed as a percentage.

Key Concept

Thermal decomposition of trioxocarbonates and percentage purity stoichiometry

Question 263Question →

Calculate the quantity of electricity, in Coulombs, required to deposit $0.108\text{ g}$ of silver at the cathode during the electrolysis of silver trioxonitrate(V) solution. ( $\text{Ag} = 108\text{ g/mol}$ , $1\text{ F} = 96,500\text{ C/mol}$ )

Show answer & explanation

Answer: 96.5

Answer

The quantity of electricity required is

96.5\text{ C}

.

Depositing

0.108\text{ g}

of silver (molar mass

108\text{ g/mol}

) requires

0.001\text{ moles}

of silver atoms. According to the cathodic reaction

\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}

,

1\text{ mole}

of electrons (

96,500\text{ C}

) is required to deposit

1\text{ mole}

of

\text{Ag}

. Therefore, the total charge required is

0.001 \times 96,500\text{ C} = 96.5\text{ C}

.

Step-by-Step Solution

1

Calculate the number of moles of silver deposited

\text{Moles of Ag} = \frac{0.108\text{ g}}{108\text{ g/mol}} = 0.001\text{ mol}

Number of moles is calculated by dividing mass by molar mass.

2

Determine the quantity of electricity (charge) required

Q = 0.001\text{ mol} \times 96,500\text{ C/mol} = 96.5\text{ C}

The reduction reaction

\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}

shows

1\text{ mole}

of electrons (

1\text{ F} = 96,500\text{ C}

) deposits

1\text{ mole}

of silver.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Mass-Charge Relationship

Question 264Question →

During the electrolytic refining of copper, a steady current of $5.0\text{ A}$ is passed through an aqueous copper(II) tetraoxosulfate(VI) ( $CuSO_4$ ) solution for $965\text{ seconds}$ . What mass of copper, in grams, is deposited at the cathode? (Faraday's constant $F = 96,500\text{ C mol}^{-1}$ ; Molar mass of $Cu = 64\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 1.6

Answer

1.6 g

Passing a

5.0\text{ A}

current for

965\text{ s}

transfers

4825\text{ C}

of electric charge, corresponding to

0.05\text{ mol}

of electrons. Because the deposition of copper from

CuSO_4

follows

Cu^{2+} + 2e^- \rightarrow Cu

, every

2\text{ moles}

of electrons deposit

1\text{ mole}

of copper metal. Thus,

0.025\text{ mol}

of copper is deposited, which corresponds to

0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}

.

Step-by-Step Solution

1

Calculate the total quantity of electricity (

Q

) passed through the electrolyte.

Q = I \times t = 5.0\text{ A} \times 965\text{ s} = 4825\text{ C}

Electric charge is the product of current in amperes and time in seconds.

2

Calculate the moles of electrons transferred.

n(e^-) = \frac{Q}{F} = \frac{4825\text{ C}}{96500\text{ C mol}^{-1}} = 0.05\text{ mol}

of electrons

One mole of electrons carries a charge equivalent to 1 Faraday (

96,500\text{ C}

).

3

Use the cathode half-equation to find the moles of deposited copper.

Cathode reaction:

Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}

. Moles of

Cu = \frac{0.05\text{ mol}}{2} = 0.025\text{ mol}

Reduction of one mole of copper(II) ions requires two moles of electrons.

4

Convert the moles of deposited copper into mass.

Mass of

Cu = n \times M = 0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}

Multiplying the chemical amount of copper by its molar mass yields the mass in grams.

Key Concept

Quantitative electrolysis of copper(II) ions using Faraday's laws of electrolysis

Question 265Question →

A sample of

0.13\text{ g}

of zinc granules reacts completely with an excess of dilute hydrochloric acid according to the reaction equation:

\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)

If the reaction takes exactly

40\text{ seconds}

to reach completion, what is the average rate of consumption of hydrochloric acid in

\text{mol s}^{-1}

? (Molar mass of

\text{Zn} = 65\text{ g mol}^{-1}

)

Show answer & explanation

Answer: 0.0001

Answer

The average rate of consumption of hydrochloric acid is

0.0001\text{ mol s}^{-1}

(or

1.0 \times 10^{-4}\text{ mol s}^{-1}

).

To determine the average rate of consumption of hydrochloric acid, first convert the mass of zinc to moles (

0.13\text{ g} / 65\text{ g mol}^{-1} = 0.002\text{ mol}

). According to the stoichiometric coefficients in the balanced equation

\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2

,

2\text{ moles}

of

\text{HCl}

react for every

1\text{ mole}

of

\text{Zn}

. Therefore,

0.004\text{ mol}

of

\text{HCl}

is consumed. Dividing this quantity by the reaction time (

40\text{ seconds}

) gives an average rate of

0.0001\text{ mol s}^{-1}

.

Step-by-Step Solution

1

Calculate the amount in moles of zinc reacted

\text{Moles of Zn} = \frac{0.13\text{ g}}{65\text{ g mol}^{-1}} = 0.002\text{ mol}

Mass divided by molar mass yields the quantity in moles.

2

Determine the moles of hydrochloric acid consumed using the mole ratio

\text{Moles of HCl} = 2 \times 0.002\text{ mol} = 0.004\text{ mol}

The balanced chemical equation shows a

1:2

stoichiometric ratio between

\text{Zn}

and

\text{HCl}

.

3

Calculate the average rate of consumption of HCl per unit time

\text{Rate of HCl consumption} = \frac{0.004\text{ mol}}{40\text{ s}} = 0.0001\text{ mol s}^{-1}

Rate of reaction is defined as the change in moles of reactant divided by elapsed time.

Key Concept

Stoichiometric determination of reaction rate from reactant consumption

Question 266Question →

What volume of carbon(IV) oxide gas, in $\text{dm}^3$ , measured at STP, is produced by the complete thermal decomposition of $20.0\text{ g}$ of pure calcium trioxocarbonate(IV), $\text{CaCO}_3$ ? [ $\text{Ca} = 40$ , $\text{C} = 12$ , $\text{O} = 16$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ ]

Show answer & explanation

Answer: 4.48

Answer

4.48 dm³

Complete thermal decomposition of 20.0 g of CaCO₃ (molar mass 100 g/mol) generates 0.20 mol of CO₂ gas according to the equation CaCO₃(s) -> CaO(s) + CO₂(g). Since 1 mol of gas at STP occupies 22.4 dm³, 0.20 mol occupies 4.48 dm³.

Step-by-Step Solution

1

Determine the molar mass and number of moles of calcium trioxocarbonate(IV).

Molar mass of CaCO₃ = 100 g/mol; Moles of CaCO₃ = 20.0 g / 100 g/mol = 0.20 mol

Converting given mass to moles is required to apply stoichiometric ratios.

2

Apply the balanced reaction mole ratio to find moles of carbon(IV) oxide produced.

Moles of CO₂ = 0.20 mol

The equation CaCO₃(s) -> CaO(s) + CO₂(g) shows a 1:1 molar ratio between CaCO₃ and CO₂.

3

Multiply moles of CO₂ by molar gas volume at STP.

Volume of CO₂ = 0.20 mol × 22.4 dm³/mol = 4.48 dm³

At STP, 1 mole of any ideal gas occupies 22.4 dm³.

Key Concept

Thermal decomposition of trioxocarbonate(IV) salts and gas volume calculations at STP

Estimated Time:45s

Question 267Question →

The hydration of ethene to produce liquid ethanol is represented by the chemical equation:

\text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)

Given the following thermochemical equations:
1. $\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \quad \Delta H^\circ = -1367\text{ kJ mol}^{-1}$
2. $\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H^\circ = -1411\text{ kJ mol}^{-1}$
3. $\text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -44\text{ kJ mol}^{-1}$

Calculate the standard enthalpy change, $\Delta H^\circ$ , for the hydration reaction in $\text{kJ mol}^{-1}$ .

Show answer & explanation

Answer: -88

Answer

The standard enthalpy change for the hydration reaction is -88 kJ/mol.

According to Hess's law, the standard enthalpy change of a net reaction can be determined by algebraically combining component reactions and their enthalpy values. Adding Equation 2 as written, Equation 3 as written, and the reverse of Equation 1 cancels out intermediate species (carbon dioxide, oxygen, and liquid water), leaving the net hydration reaction. Summing their respective enthalpy values gives -1411 kJ/mol + (-44 kJ/mol) + 1367 kJ/mol = -88 kJ/mol.

Step-by-Step Solution

1

Identify the required target thermochemical equation

Target:

\text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)

This establishes the stoichiometry and physical states required for reactants and products.

2

Apply Hess's law to reverse and combine the given thermochemical equations

Keep Equation 2:

\Delta H_2 = -1411\text{ kJ mol}^{-1}

Keep Equation 3:

\Delta H_3 = -44\text{ kJ mol}^{-1}

Reverse Equation 1:

\Delta H_1' = +1367\text{ kJ mol}^{-1}

Reversing Equation 1 places liquid ethanol on the product side, requiring the sign of its enthalpy change to be inverted.

3

Sum the enthalpy changes of the modified reaction steps

\Delta H^\circ = (-1411) + (-44) + (+1367) = -88\text{ kJ mol}^{-1}

According to Hess's law, the total enthalpy change of an overall reaction equals the sum of the enthalpy changes for individual component steps.

Key Concept

Hess's Law of Constant Heat Summation

Question 268Question →

During the industrial extraction of aluminium using the Hall-Héroult process, a steady current of $96.5\text{ A}$ is passed through an electrolytic cell containing molten alumina ( $Al_2O_3$ ) dissolved in molten cryolite for $5.0\text{ hours}$ . What is the mass of pure aluminium, in grams, deposited at the cathode?

(Take $1\text{ Faraday} = 96,500\text{ C mol}^{-1}$ , Relative atomic mass: $\text{Al} = 27$ )

Show answer & explanation

Answer: 162

Answer

The mass of pure aluminium deposited at the cathode is

162\text{ g}

.

Using

Q = I \times t

, a current of

96.5\text{ A}

for

5.0\text{ hours}

(

18,000\text{ s}

) yields a total charge of

1,737,000\text{ C}

, which equals

18\text{ Faradays}

(

18\text{ moles of electrons}

). Since the reduction of

\text{Al}^{3+}

to

\text{Al}

requires 3 electrons per atom (

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

),

18\text{ moles}

of electrons liberate

6\text{ moles}

of aluminium metal. Multiplying by the molar mass of aluminium (

27\text{ g mol}^{-1}

) gives

162\text{ g}

.

Step-by-Step Solution

1

Convert time to seconds and calculate total electric charge transferred

Q = 96.5\text{ A} \times (5.0 \times 3600\text{ s}) = 1,737,000\text{ C}

Electric charge is defined as current multiplied by time in seconds (

Q = I \times t

).

2

Calculate the moles of electrons transferred using Faraday's constant

Moles of

e^- = \frac{1,737,000\text{ C}}{96,500\text{ C mol}^{-1}} = 18\text{ moles}

One Faraday (

96,500\text{ C}

) corresponds to the charge carried by one mole of electrons.

3

Relate moles of electrons to moles of aluminium deposited using the half-equation

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}_{(s)}

, so

3\text{ mol } e^-

produces

1\text{ mol Al}

. Moles of

\text{Al} = \frac{18}{3} = 6\text{ moles}

Aluminium ion

\text{Al}^{3+}

requires three electrons for reduction to metallic aluminium.

4

Multiply moles of aluminium by its relative atomic mass

\text{Mass} = 6\text{ mol} \times 27\text{ g mol}^{-1} = 162\text{ g}

Mass equals molar amount multiplied by molar mass.

Key Concept

Quantitative application of Faraday's laws of electrolysis in the extraction of metals

Question 269Question →

During a chemical reaction between dilute hydrochloric acid and calcium carbonate, $60\text{ cm}^3$ of carbon dioxide gas is collected over a period of $30\text{ seconds}$ . What is the average rate of evolution of the gas in $\text{cm}^3\text{ s}^{-1}$ ?

Show answer & explanation

Answer: 2

Answer

The average rate of gas evolution is

2.0\text{ cm}^3\text{ s}^{-1}

.

The average rate of a reaction yielding a gas is given by dividing the total volume of gas produced by the total time taken. Dividing

60\text{ cm}^3

by

30\text{ seconds}

gives an average rate of

2.0\text{ cm}^3\text{ s}^{-1}

.

Step-by-Step Solution

1

Identify the volume of gas evolved and the time duration.

Volume =

60\text{ cm}^3

, Time =

30\text{ seconds}

.

These are the measured parameters required to calculate the average rate of reaction.

2

Calculate the average rate of reaction.

\text{Rate} = \frac{60\text{ cm}^3}{30\text{ s}} = 2.0\text{ cm}^3\text{ s}^{-1}

.

The reaction rate measures the change in concentration or amount of a reactant or product per unit time.

Key Concept

Rate of Reaction Calculation

Question 270Question →

Adamu and Zainab are partners in a firm. They agree to value the firm's goodwill on the basis of $3$ years' purchase of the average super profit of the past $4$ years. The net profits of the firm for the last $4$ years were $\mathcal{N}45,000$ , $\mathcal{N}55,000$ , $\mathcal{N}60,000$ , and $\mathcal{N}80,000$ . The capital employed in the business is $\mathcal{N}400,000$ , and the normal rate of return expected on capital employed in a similar business is $10\%$ . What is the value of the firm's goodwill in Naira?

Show answer & explanation

Answer: 60000

Answer

The value of the firm's goodwill is 60,000 Naira.

Goodwill under the super profit method is obtained by taking the excess of average annual profits over normal expected profits and multiplying by the number of years' purchase. The average profit is 60,000 Naira and normal profit is 40,000 Naira (10% of 400,000 Naira). The super profit is 20,000 Naira, which when multiplied by 3 years' purchase gives 60,000 Naira.

Step-by-Step Solution

1

Calculate the average annual profit of the firm over the 4-year period

Average profit = 60,000 Naira

Sum the total profits of the four years (240,000 Naira) and divide by 4.

2

Calculate the normal profit expected from the capital employed

Normal profit = 40,000 Naira

Multiply the capital employed (400,000 Naira) by the normal rate of return (10%).

3

Determine the super profit of the firm

Super profit = 20,000 Naira

Subtract normal profit (40,000 Naira) from average annual profit (60,000 Naira).

4

Compute goodwill using the 3 years' purchase multiplier

Goodwill = 60,000 Naira

Multiply the super profit (20,000 Naira) by the agreed 3 years' purchase.

Key Concept

Valuation of Goodwill using the Super Profit Method

Estimated Time:1m 30s

Question 271Question →

In an industrial Contact Process plant, sulfur(IV) oxide (

SO_2

) gas is catalytically oxidized to sulfur(VI) oxide (

SO_3

) according to the equation:

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

If

67.2\text{ dm}^3

of

SO_2

measured at STP is reacted with excess oxygen gas, and the reaction achieves a

90\%

conversion yield of

SO_3

, what is the mass in grams of tetraoxosulfate(VI) acid (

H_2SO_4

) produced when all the formed

SO_3

is absorbed in concentrated

H_2SO_4

and subsequently diluted with water? (Molar mass of

H_2SO_4 = 98\text{ g/mol}

, molar volume of gas at STP

= 22.4\text{ dm}^3\text{/mol}

)

Show answer & explanation

Answer: 264.6

Answer

264.6 g

The molar volume at STP (

22.4\text{ dm}^3\text{/mol}

) converts

67.2\text{ dm}^3

of

SO_2

into

3.0\text{ moles}

. Accounting for the

90\%

conversion efficiency yields

2.7\text{ moles}

of

SO_3

. Absorption of

SO_3

into concentrated

H_2SO_4

to form oleum (

H_2S_2O_7

) followed by dilution with water yields a

1:1

molar ratio of

H_2SO_4

relative to

SO_3

. Multiplying

2.7\text{ moles}

by the molar mass of

H_2SO_4

(

98\text{ g/mol}

) gives the correct mass of

264.6\text{ g}

.

Step-by-Step Solution

1

Calculate the moles of

SO_2

gas at STP.

n(SO_2) = \frac{67.2\text{ dm}^3}{22.4\text{ dm}^3\text{/mol}} = 3.0\text{ moles}

Molar volume of any ideal gas at STP is

22.4\text{ dm}^3\text{/mol}

.

2

Apply the

90\%

conversion efficiency to find the moles of

SO_3

produced.

n(SO_3) = 3.0\text{ mol} \times 0.90 = 2.7\text{ moles}

Only

90\%

of the reacted

SO_2

is converted to

SO_3

under operating conditions.

3

Relate the moles of

SO_3

to the moles of

H_2SO_4

produced.

n(H_2SO_4) = n(SO_3) = 2.7\text{ moles}

The absorption of

SO_3

into concentrated

H_2SO_4

forms oleum (

H_2S_2O_7

), which upon dilution with water yields

H_2SO_4

with an overall

1:1

stoichiometric molar equivalence to

SO_3

.

4

Calculate the total mass of

H_2SO_4

formed.

\text{Mass} = 2.7\text{ mol} \times 98\text{ g/mol} = 264.6\text{ g}

Mass equals number of moles multiplied by molar mass.

Key Concept

Stoichiometry of the Contact Process including STP gas conversion and oleum dilution stoichiometry

Question 272Question →

Calculate the standard enthalpy of combustion of liquid carbon disulfide (

\text{CS}_2(l)

) in

\text{kJ mol}^{-1}

, given the following standard enthalpies of formation:

\Delta H_f^\circ[\text{CS}_2(l)] = +88\text{ kJ mol}^{-1}

\Delta H_f^\circ[\text{CO}_2(g)] = -394\text{ kJ mol}^{-1}

\Delta H_f^\circ[\text{SO}_2(g)] = -297\text{ kJ mol}^{-1}

The balanced chemical equation for the combustion process is:

\text{CS}_2(l) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g)

What is the standard enthalpy change of combustion in $\text{kJ mol}^{-1}$ ?

Show answer & explanation

Answer: -1076

Answer

The standard enthalpy of combustion of liquid carbon disulfide is -1076 kJ mol^-1.

Applying Hess's law using standard enthalpies of formation gives

\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

. For the combustion of carbon disulfide, this evaluates to

[(-394) + 2(-297)] - (+88) = -988 - 88 = -1076\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Formulate the standard enthalpy of reaction equation using standard enthalpies of formation.

\Delta H^\circ = \sum n\Delta H_f^\circ(\text{products}) - \sum m\Delta H_f^\circ(\text{reactants})

According to Hess's Law, the net standard enthalpy change for a chemical process equals the sum of standard formation enthalpies of products minus reactants.

2

Substitute the provided standard enthalpy of formation values into the expression, taking into account stoichiometry.

\Delta H^\circ = [-394 + 2(-297)] - [88] = -1076\text{ kJ mol}^{-1}

Sulfur dioxide is formed with a mole ratio of 2, so its formation enthalpy must be doubled; oxygen gas has a formation enthalpy of zero.

Key Concept

Standard Enthalpy Changes and Hess's Law

Question 273Question →

Consider the standard reduction potentials for the following two half-cell reactions:

\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\circ = -0.76\text{ V}

\text{Fe}^{3+}(aq) + 3e^- \rightarrow \text{Fe}(s) \quad E^\circ = -0.04\text{ V}

What is the standard cell potential (

E^\circ_{\text{cell}}

), in volts, for the spontaneous redox reaction represented by the balanced chemical equation:

3\text{Zn}(s) + 2\text{Fe}^{3+}(aq) \rightarrow 3\text{Zn}^{2+}(aq) + 2\text{Fe}(s)

Show answer & explanation

Answer: 0.72

Answer

The standard cell potential for the spontaneous reaction is +0.72 V.

To calculate the standard cell potential (

E^\circ_{\text{cell}}

), identify the cathode (reduction) and anode (oxidation) processes from the balanced chemical equation. Iron(III) ions are reduced to iron metal at the cathode (

E^\circ = -0.04\text{ V}

), while zinc metal is oxidized to zinc ions at the anode (

E^\circ = -0.76\text{ V}

). Using

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

, we calculate

E^\circ_{\text{cell}} = -0.04\text{ V} - (-0.76\text{ V}) = +0.72\text{ V}

. Because standard electrode potential is an intensive property, the stoichiometric coefficients (3 for Zn and 2 for Fe³⁺) do not alter the half-cell potentials.

Step-by-Step Solution

1

Determine the oxidation and reduction species from the overall equation

Zinc is oxidized at the anode (

\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-

), and

\text{Fe}^{3+}

is reduced at the cathode (

\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}

).

The equation shows elemental Zn losing electrons to form

\text{Zn}^{2+}

and

\text{Fe}^{3+}

gaining electrons to form Fe.

2

Recall that standard electrode potential is an intensive property

The values

E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}

and

E^\circ(\text{Fe}^{3+}/\text{Fe}) = -0.04\text{ V}

remain unchanged regardless of stoichiometric coefficients.

Potential measures electrical potential energy per unit charge, which does not depend on the total amount of substance reacting.

3

Calculate the standard cell potential using

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

E^\circ_{\text{cell}} = -0.04\text{ V} - (-0.76\text{ V}) = +0.72\text{ V}

.

Subtracting the anode reduction potential from the cathode reduction potential yields the net electromotive force of the spontaneous cell.

Key Concept

Standard Cell Potential Calculation and Independence of E° from Stoichiometric Coefficients

Question 274Question →

In an experiment to measure the rate of a chemical reaction, $0.50\text{ g}$ of calcium carbonate reacts completely with excess dilute hydrochloric acid in $25\text{ seconds}$ . What is the average rate of reaction with respect to the loss of mass of calcium carbonate in $\text{g s}^{-1}$ ?

Show answer & explanation

Answer: 0.02

Answer

The average rate of reaction with respect to the mass of calcium carbonate consumed is

0.02\text{ g s}^{-1}

.

The average rate of a reaction is calculated as the ratio of the change in amount of reactant or product to the time taken. Substituting the given values gives

\text{Rate} = \frac{0.50\text{ g}}{25\text{ s}} = 0.02\text{ g s}^{-1}

.

Step-by-Step Solution

1

Extract the given values from the problem statement.

Mass of

\text{CaCO}_3 = 0.50\text{ g}

, Time

= 25\text{ s}

.

These parameters define the total change in quantity and the time interval for the reaction.

2

Calculate the average rate of reaction by dividing the change in mass by the time elapsed.

\text{Rate} = \frac{0.50\text{ g}}{25\text{ s}} = 0.02\text{ g s}^{-1}

.

The rate of a chemical reaction measures how rapidly a reactant is consumed per unit time.

Key Concept

Rate of Reaction Calculation

Estimated Time:1m 0s

Question 275Question →

Calculate the quantity of electricity, in Coulombs, transferred when a steady electric current of $5.0\text{ A}$ is passed through an electrolytic cell for $20\text{ minutes}$ .

Show answer & explanation

Answer: 6000

Answer

The quantity of electricity transferred is

6000\text{ C}

.

According to Faraday's laws of electrolysis, the total charge

Q

passed through an electrolyte is calculated by

Q = I \times t

. Converting

20\text{ minutes}

to seconds gives

20 \times 60 = 1200\text{ s}

. Multiplying by the current

5.0\text{ A}

gives

Q = 5.0 \times 1200 = 6000\text{ C}

.

Step-by-Step Solution

1

Convert time from minutes to seconds

t = 1200\text{ s}

Electric current in Amperes measures charge per second, so time must be converted to seconds.

2

Calculate electric charge using

Q = I \times t

Q = 6000\text{ C}

The quantity of electricity (

Q

) in Coulombs equals current (

I

) in Amperes multiplied by time (

t

) in seconds.

Key Concept

Calculation of Quantity of Electricity (

Q = I \times t

)

Question 276Question →

An economy records a Gross Domestic Product (GDP) of 850 million Naira. The factor income earned by citizens from abroad is 30 million Naira, while factor income paid to foreigners within the domestic economy is 70 million Naira. If the capital consumption allowance (depreciation) is 65 million Naira, what is the Net National Product (NNP) of the country in million Naira?

Show answer & explanation

Answer: 745

Answer

The Net National Product (NNP) of the country is 745 million Naira.

To determine the Net National Product (NNP), first calculate Net Factor Income from Abroad (NFIA) as factor income from abroad (

30

million Naira) minus factor income paid abroad (

70

million Naira), yielding

-40

million Naira. Gross National Product (GNP) is then calculated as

\text{GDP} + \text{NFIA} = 850 + (-40) = 810

million Naira. Finally, subtract capital consumption allowance (

65

million Naira) from GNP to get

\text{NNP} = 810 - 65 = 745

million Naira.

Step-by-Step Solution

1

Calculate Net Factor Income from Abroad (NFIA)

NFIA =

30 - 70 = -40

million Naira

Net Factor Income from Abroad is the difference between income received from abroad by residents and income paid to non-residents domestically.

2

Calculate Gross National Product (GNP)

GNP =

850 + (-40) = 810

million Naira

GNP is obtained by adjusting GDP for Net Factor Income from Abroad.

3

Calculate Net National Product (NNP)

NNP =

810 - 65 = 745

million Naira

NNP is obtained by subtracting capital consumption allowance (depreciation) from GNP.

Key Concept

Calculation of Net National Product (NNP) from GDP, Net Factor Income from Abroad, and Depreciation

Question 277Question →

In an energy profile diagram for a reversible chemical reaction, the potential energy of the reactants is $85\text{ kJ mol}^{-1}$ , the potential energy of the products is $150\text{ kJ mol}^{-1}$ , and the peak potential energy of the activated complex is $235\text{ kJ mol}^{-1}$ . What is the activation energy of the reverse reaction in $\text{kJ mol}^{-1}$ ?

Show answer & explanation

Answer: 85

Answer

The activation energy of the reverse reaction is

85\text{ kJ mol}^{-1}

.

The activation energy for the reverse reaction is the difference in potential energy between the activated complex at the peak and the products. Subtracting the potential energy of the products (

150\text{ kJ mol}^{-1}

) from the peak energy (

235\text{ kJ mol}^{-1}

) gives

85\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Identify key energy values from the reaction energy profile

Energy of activated complex =

235\text{ kJ mol}^{-1}

, Energy of products =

150\text{ kJ mol}^{-1}

.

The activation energy of the reverse reaction is defined as the energy required for products to reach the transition state (activated complex).

2

Calculate the reverse activation energy (

E_{a,\text{reverse}}

)

E_{a,\text{reverse}} = 235\text{ kJ mol}^{-1} - 150\text{ kJ mol}^{-1} = 85\text{ kJ mol}^{-1}

.

Subtracting the potential energy of the products from the peak potential energy yields the minimum energy barrier for the backward process.

Key Concept

Calculating reverse activation energy from potential energy levels

Question 278Question →

The daily agricultural output (in tonnes) of a cassava farm over five consecutive days was recorded as 12, 15, 18, 15, and 20. What is the mean daily output of the farm in tonnes?

Show answer & explanation

Answer: 16

Answer

The mean daily output of the farm is 16 tonnes.

The arithmetic mean is computed by adding all data values together (

12 + 15 + 18 + 15 + 20 = 80

) and dividing by the total number of data points (

5

), which gives

80 \div 5 = 16

tonnes.

Step-by-Step Solution

1

Calculate the total output by summing all daily figures.

12 + 15 + 18 + 15 + 20 = 80 tonnes

Finding the arithmetic mean requires calculating the aggregate total of all observations.

2

Divide the total output by the number of observations (days).

80 / 5 = 16 tonnes

The formula for the arithmetic mean of a sample is the sum of all values divided by the total number of values.

Key Concept

Arithmetic Mean of Ungrouped Data

Estimated Time:45s

Question 279Question →

A commercial farm in Kaduna operates with fixed total resources of $10$ hectares of arable land and $120$ worker-hours of labor per day. Producing $1$ ton of maize requires $1$ hectare of land and $12$ worker-hours of labor. Producing $1$ ton of yams requires $0.5$ hectares of land and $20$ worker-hours of labor. Initially, the farm devotes all its resources to maximize maize production, harvesting $10$ tons of maize per day. If the farm owner decides to reallocate resources to produce $3$ tons of yams per day, what is the opportunity cost of this decision expressed in tons of maize foregone?

Show answer & explanation

Answer: 5

Answer

The opportunity cost of producing 3 tons of yams is 5 tons of maize.

Producing 3 tons of yams absorbs 60 worker-hours out of the total 120 worker-hours available. The remaining 60 worker-hours can only produce 5 tons of maize. Since initial maize production was 10 tons, the farm gives up 5 tons of maize (10 - 5 = 5 tons).

Step-by-Step Solution

1

Calculate resources consumed by the production of 3 tons of yams.

Land consumed =

3 \times 0.5 = 1.5

hectares. Labor consumed =

3 \times 20 = 60

worker-hours.

Opportunity cost depends on the amount of productive inputs diverted away from maize production.

2

Determine the remaining inputs available for maize production.

Remaining land =

10 - 1.5 = 8.5

hectares. Remaining labor =

120 - 60 = 60

worker-hours.

Maize can only be produced using the unallocated land and labor inputs.

3

Identify the binding constraint for maize output.

Labor restricts maize output to

\frac{60}{12} = 5

tons (since land would permit

8.5

tons). Thus, maximum feasible maize output is

5

tons.

Production is restricted by the scarcest resource (labor).

4

Subtract the new maximum maize output from the initial maize output to find the opportunity cost.

10 \text{ tons} - 5 \text{ tons} = 5 \text{ tons of maize foregone}

.

Opportunity cost is defined as the quantity of the alternative good given up.

Key Concept

Opportunity cost with constrained multi-resource allocation

Question 280Question →

A commercial farming enterprise in Oyo State uses its fixed land and labor resources to produce Cassava and Yam. The table below presents its monthly production possibility schedule:

Combination	Cassava (bags)	Yam (bags)
P	120	0
Q	90	45
R	50	80
S	0	100

What is the opportunity cost of increasing Yam production from $45$ bags to $80$ bags, expressed in bags of Cassava foregone?

Show answer & explanation

Answer: 40

Answer

The opportunity cost of increasing Yam production from 45 bags to 80 bags is 40 bags of Cassava.

Increasing Yam production from

45

bags to

80

bags requires moving from Combination Q to Combination R. At Combination Q, Cassava production is

90

bags, whereas at Combination R, Cassava production decreases to

50

bags. The opportunity cost is the foregone Cassava production, calculated as

90 - 50 = 40

bags of Cassava.

Step-by-Step Solution

1

Locate the initial production combination

At Combination Q, the firm produces 45 bags of Yam and 90 bags of Cassava.

Opportunity cost measures what must be sacrificed when shifting resources from one alternative to another.

2

Locate the target production combination

At Combination R, the firm produces 80 bags of Yam and 50 bags of Cassava.

Increasing Yam production from 45 to 80 bags requires moving along the schedule from Q to R.

3

Calculate the quantity of the sacrificed commodity (Cassava)

Opportunity Cost = 90 - 50 = 40 bags of Cassava.

The opportunity cost is the explicit quantity of Cassava sacrificed to gain the additional 35 bags of Yam.

Key Concept

Opportunity Cost from Production Schedule

Answer

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