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1526 questions

Question 1021Question

In an industrial chlor-alkali membrane cell, concentrated sodium chloride solution (brine) is electrolyzed using a constant current of 19.3 A19.3\text{ A} for 50 minutes50\text{ minutes}. What is the mass, in grams, of sodium hydroxide (NaOH\text{NaOH}) produced in the solution? [Molar mass of NaOH=40.0 g mol1\text{NaOH} = 40.0\text{ g mol}^{-1}, Faraday constant F=96,500 C mol1F = 96,500\text{ C mol}^{-1}]

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Answer: 24

Answer

The mass of sodium hydroxide produced is 24.0 g24.0\text{ g}.
Converting the electrolysis time to seconds (3000 s3000\text{ s}) yields a total charge of Q=19.3 A×3000 s=57,900 CQ = 19.3\text{ A} \times 3000\text{ s} = 57,900\text{ C}. Dividing by Faraday's constant (96,500 C mol196,500\text{ C mol}^{-1}) gives 0.6 mol0.6\text{ mol} of electrons. In the chlor-alkali process, reduction of water produces 1 mol1\text{ mol} of OH\text{OH}^- ions per mole of electrons, producing 0.6 mol0.6\text{ mol} of NaOH\text{NaOH}. Multiplying by the molar mass (40.0 g mol140.0\text{ g mol}^{-1}) gives a mass of 24.0 g24.0\text{ g}.

Step-by-Step Solution

1
Convert electrolysis time into seconds and calculate total charge.
Q=19.3 A×(50×60 s)=57,900 CQ = 19.3\text{ A} \times (50 \times 60\text{ s}) = 57,900\text{ C}.
Faraday's equations require time in seconds (Q=I×tQ = I \times t).
2
Calculate the moles of electrons transferred using the Faraday constant.
n(e)=57,900 C96,500 C mol1=0.6 moln(e^-) = \frac{57,900\text{ C}}{96,500\text{ C mol}^{-1}} = 0.6\text{ mol}.
One Faraday (96,500 C96,500\text{ C}) corresponds to one mole of electrons.
3
Determine the stoichiometry of the cathode reaction.
Cathode reaction: 2H2O(l)+2eH2(g)+2OH(aq)2\text{H}_2\text{O}_{(l)} + 2e^- \rightarrow \text{H}_{2(g)} + 2\text{OH}^-_{(aq)}. Thus, 1 mol e1\text{ mol } e^- forms 1 mol OH1\text{ mol } \text{OH}^-, giving 0.6 mol0.6\text{ mol} of NaOH\text{NaOH}.
During brine electrolysis, water is preferentially reduced at the cathode, generating hydroxide ions that pair with sodium ions.
4
Calculate the mass of sodium hydroxide produced.
Mass=0.6 mol×40.0 g mol1=24.0 g\text{Mass} = 0.6\text{ mol} \times 40.0\text{ g mol}^{-1} = 24.0\text{ g}.
Mass is obtained by multiplying the number of moles by the molar mass.

Key Concept

Industrial Electrolysis Stoichiometry (Chlor-Alkali Process) and Faraday's First Law
Estimated Time:1m 30s
Question 1022Question
Excess solid carbon, C(s)\text{C}(s), and 4.0 mol4.0\text{ mol} of carbon dioxide gas, CO2(g)\text{CO}_2(g), are introduced into an evacuated 2.0 dm32.0\text{ dm}^3 rigid reaction vessel at a constant temperature. The system reaches equilibrium according to the reaction equation:
C(s)+CO2(g)2CO(g)\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)
If the equilibrium concentration of carbon monoxide gas, CO(g)\text{CO}(g), is determined to be 0.80 mol dm30.80\text{ mol dm}^{-3}, calculate the numerical value of the equilibrium constant, KcK_c, in mol dm3\text{mol dm}^{-3}.
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Answer: 0.4

Answer

The numerical value of the equilibrium constant, KcK_c, is 0.40.4 (or 0.400.40).
To find KcK_c, first convert moles of CO2\text{CO}_2 to initial concentration: [CO2]0=4.0 mol2.0 dm3=2.00 mol dm3[\text{CO}_2]_0 = \frac{4.0\text{ mol}}{2.0\text{ dm}^3} = 2.00\text{ mol dm}^{-3}. From stoichiometry (C(s)+CO2(g)2CO(g)\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)), creating 0.80 mol dm30.80\text{ mol dm}^{-3} of CO\text{CO} consumes 0.802=0.40 mol dm3\frac{0.80}{2} = 0.40\text{ mol dm}^{-3} of CO2\text{CO}_2. At equilibrium, [CO2]eq=2.000.40=1.60 mol dm3[\text{CO}_2]_{eq} = 2.00 - 0.40 = 1.60\text{ mol dm}^{-3}. Omitting the solid carbon C(s)\text{C}(s) from the expression gives Kc=[CO]2[CO2]=(0.80)21.60=0.40 mol dm3K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]} = \frac{(0.80)^2}{1.60} = 0.40\text{ mol dm}^{-3}.

Step-by-Step Solution

1
Calculate initial concentration of reactant gas
[CO2]0=2.00 mol dm3[\text{CO}_2]_0 = 2.00\text{ mol dm}^{-3}
Concentration must be calculated in mol dm3\text{mol dm}^{-3} by dividing moles by vessel volume (2.0 dm32.0\text{ dm}^3).
2
Determine equilibrium concentrations using stoichiometric ratios
[CO2]eq=1.60 mol dm3[\text{CO}_2]_{eq} = 1.60\text{ mol dm}^{-3} and [CO]eq=0.80 mol dm3[\text{CO}]_{eq} = 0.80\text{ mol dm}^{-3}
The stoichiometric mole ratio of CO2\text{CO}_2 to CO\text{CO} is 1:21:2. Thus, consuming 0.40 mol dm30.40\text{ mol dm}^{-3} of CO2\text{CO}_2 yields 0.80 mol dm30.80\text{ mol dm}^{-3} of CO\text{CO}.
3
Formulate the equilibrium constant expression for the heterogeneous reaction
Kc=[CO]2[CO2]K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]}
Pure solids such as C(s)\text{C}(s) have constant concentration and are omitted from the equilibrium constant expression.
4
Calculate the value of KcK_c
Kc=(0.80)21.60=0.40K_c = \frac{(0.80)^2}{1.60} = 0.40
Substitute equilibrium concentrations into the KcK_c expression and solve.

Key Concept

Equilibrium constant expression for heterogeneous equilibria and ICE table calculations.
Question 1023Question

A 16.1 g16.1\text{ g} sample of hydrated sodium tetraoxosulfate(VI), Na2SO4xH2O\text{Na}_2\text{SO}_4 \cdot x\text{H}_2\text{O}, is heated in a crucible until all the water of crystallization is driven off. The mass of the remaining anhydrous salt is 7.1 g7.1\text{ g}. Calculate the value of xx. [Relative atomic masses: Na=23\text{Na} = 23, S=32\text{S} = 32, O=16\text{O} = 16, H=1\text{H} = 1]

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Answer: 10

Answer

The integer value of xx is 10.
Heating 16.1 g16.1\text{ g} of hydrated sodium tetraoxosulfate(VI) yields 7.1 g7.1\text{ g} of anhydrous Na2SO4\text{Na}_2\text{SO}_4 (0.05 mol0.05\text{ mol}) and releases 9.0 g9.0\text{ g} of water (0.5 mol0.5\text{ mol}). The mole ratio of H2O\text{H}_2\text{O} to Na2SO4\text{Na}_2\text{SO}_4 is 0.5/0.05=100.5 / 0.05 = 10, giving x=10x = 10.

Step-by-Step Solution

1
Calculate the mass of water lost upon heating
Mass of H2O=16.1 g7.1 g=9.0 g\text{H}_2\text{O} = 16.1\text{ g} - 7.1\text{ g} = 9.0\text{ g}
The difference between the initial hydrated mass and final anhydrous mass represents the driven-off water of crystallization.
2
Calculate molar masses of anhydrous salt and water
Molar mass of Na2SO4=142 g/mol\text{Na}_2\text{SO}_4 = 142\text{ g/mol}, Molar mass of H2O=18 g/mol\text{H}_2\text{O} = 18\text{ g/mol}
Required to convert sample masses to mole quantities.
3
Calculate moles of anhydrous salt and water, and find the mole ratio
Moles of Na2SO4=0.05 mol\text{Na}_2\text{SO}_4 = 0.05\text{ mol}, Moles of H2O=0.5 mol\text{H}_2\text{O} = 0.5\text{ mol}, x=0.50.05=10x = \frac{0.5}{0.05} = 10
The subscript xx gives the ratio of moles of water to moles of anhydrous salt per mole of compound.

Key Concept

Stoichiometric determination of water of crystallization in hydrated salts
Question 1024Question
Concentrated tetraoxosulfate(VI) acid reacts with copper metal as an oxidizing agent according to the balanced chemical equation:
Cu(s)+2H2SO4(aq)CuSO4(aq)+2H2O(l)+SO2(g)Cu(s) + 2H_2SO_4(aq) \rightarrow CuSO_4(aq) + 2H_2O(l) + SO_2(g)
If 31.75 g31.75\text{ g} of copper turnings react completely with an excess of concentrated tetraoxosulfate(VI) acid, what volume of sulfur(IV) oxide (SO2SO_2) gas, in dm3\text{dm}^3, is evolved at standard temperature and pressure (STP)? [Molar mass of Cu=63.5 g mol1Cu = 63.5\text{ g mol}^{-1}; Molar volume of gas at STP = 22.4 dm3 mol122.4\text{ dm}^3\text{ mol}^{-1}]
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Answer: 11.2

Answer

The volume of sulfur(IV) oxide gas evolved at STP is 11.2 dm311.2\text{ dm}^3.
Concentrated tetraoxosulfate(VI) acid acts as an oxidizing agent when heated with copper metal, being reduced to sulfur(IV) oxide gas. From the balanced reaction equation, 1 mole1\text{ mole} (63.5 g63.5\text{ g}) of copper yields 1 mole1\text{ mole} (22.4 dm322.4\text{ dm}^3 at STP) of SO2SO_2 gas. Therefore, 31.75 g31.75\text{ g} (0.5 moles0.5\text{ moles}) of copper yields 0.5×22.4 dm3=11.2 dm30.5 \times 22.4\text{ dm}^3 = 11.2\text{ dm}^3 of SO2SO_2 gas.

Step-by-Step Solution

1
Calculate the moles of copper metal reacted
n(Cu)=31.75 g63.5 g mol1=0.5 moln(Cu) = \frac{31.75\text{ g}}{63.5\text{ g mol}^{-1}} = 0.5\text{ mol}
Converting the given mass of copper to moles allows stoichiometric comparison with the reaction products.
2
Determine the moles of sulfur(IV) oxide (SO2SO_2) gas produced
n(SO2)=0.5 moln(SO_2) = 0.5\text{ mol}
From the balanced chemical equation, 1 mole1\text{ mole} of CuCu reacts to produce 1 mole1\text{ mole} of SO2SO_2 gas.
3
Calculate the volume of SO2SO_2 gas evolved at STP
V(SO2)=0.5 mol×22.4 dm3 mol1=11.2 dm3V(SO_2) = 0.5\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3
At STP, 1 mole1\text{ mole} of any ideal gas occupies a standard molar volume of 22.4 dm322.4\text{ dm}^3.

Key Concept

Oxidizing action of concentrated tetraoxosulfate(VI) acid on metals and gas volume stoichiometry at STP
Question 1025Question

In an industrial electroplating plant, a steel component is coated with silver in an electrolytic bath. If a constant electric current of 2.0 A2.0\text{ A} is passed through the bath for 48.25 minutes48.25\text{ minutes}, what is the mass of silver, in grams, deposited on the cathode? [Molar mass of Ag=108 g mol1\text{Ag} = 108\text{ g mol}^{-1}, 1 F=96500 C mol11\text{ F} = 96500\text{ C mol}^{-1}]

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Answer: 6.48

Answer

The mass of silver deposited on the cathode during electroplating is 6.48 g.
According to Faraday's first law of electrolysis, charge Q=2.0 A×(48.25×60 s)=5790 CQ = 2.0\text{ A} \times (48.25 \times 60\text{ s}) = 5790\text{ C}. The number of moles of electrons passed is 5790/96500=0.06 mol5790 / 96500 = 0.06\text{ mol}. Since silver reduction (Ag++eAg\text{Ag}^+ + e^- \rightarrow \text{Ag}) requires 1 mole1\text{ mole} of electrons per mole of silver, 0.06 mol0.06\text{ mol} of Ag\text{Ag} is formed. The mass of silver deposited is 0.06 mol×108 g mol1=6.48 g0.06\text{ mol} \times 108\text{ g mol}^{-1} = 6.48\text{ g}.

Step-by-Step Solution

1
Convert time to seconds
t=48.25 min×60 s/min=2895 st = 48.25 \text{ min} \times 60 \text{ s/min} = 2895 \text{ s}
Time must be expressed in seconds to calculate electric charge in coulombs.
2
Calculate total quantity of electricity QQ
Q=I×t=2.0 A×2895 s=5790 CQ = I \times t = 2.0 \text{ A} \times 2895 \text{ s} = 5790 \text{ C}
Electric charge is the product of current and duration of electrolysis.
3
Determine moles of electrons transferred
n(e)=5790 C96500 C mol1=0.06 moln(e^-) = \frac{5790 \text{ C}}{96500 \text{ C mol}^{-1}} = 0.06 \text{ mol}
Faraday's constant indicates that 96500 C96500\text{ C} corresponds to 1 mole1\text{ mole} of electrons.
4
Relate electron flow to silver discharge at cathode
Ag++eAg\text{Ag}^+ + e^- \rightarrow \text{Ag}, so 0.06 mol of e yields 0.06 mol of Ag0.06 \text{ mol of } e^- \text{ yields } 0.06 \text{ mol of Ag}
Silver ion discharge requires one electron per silver atom deposited.
5
Calculate mass of silver deposited
Mass=0.06 mol×108 g mol1=6.48 g\text{Mass} = 0.06 \text{ mol} \times 108 \text{ g mol}^{-1} = 6.48 \text{ g}
Multiplying the molar quantity by relative atomic mass yields the total mass deposited.

Key Concept

Quantitative application of Faraday's laws of electrolysis in industrial electroplating.
Question 1026Question

In sweet pea plants (*Lathyrus odoratus*), purple flower color (PP) is dominant over red flower color (pp), and long pollen grain (LL) is dominant over round pollen grain (ll). If a heterozygous dihybrid plant with the genotype PpLlPpLl is self-pollinated and yields a total of 160160 F2 seeds, how many of these seeds are expected to produce plants with the double recessive phenotype of red flowers and round pollen grains?

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Answer: 10

Answer

The expected number of seeds producing plants with red flowers and round pollen grains is 10.
In a dihybrid cross involving two heterozygous parents (PpLl×PpLlPpLl \times PpLl), independent assortment results in a 9:3:3:1 phenotypic ratio in the F2 generation. The double recessive phenotype (red flowers and round pollen grains, genotype ppllppll) accounts for 1 out of 16 total offspring. Multiplying this fraction (1/16) by the total yield of 160 seeds produces an expected value of 10 seeds.

Step-by-Step Solution

1
Determine the dihybrid F2 phenotypic ratio
The phenotypic ratio for a cross between two heterozygous dihybrid parents (PpLl×PpLlPpLl \times PpLl) is 9:3:3:1.
According to Mendel's Law of Independent Assortment, the alleles for flower color and pollen shape segregate independently during gamete formation.
2
Calculate the proportion of double recessive offspring
The fraction of offspring displaying both recessive traits (red flowers and round pollen grains, genotype ppllppll) is 1/16.
Out of 16 equal Punnett square combinations, exactly 1 combination represents the homozygous double recessive phenotype.
3
Compute the expected number of double recessive seeds
(1 / 16) * 160 = 10 seeds.
Multiplying the phenotypic probability (1/16) by the total seed population (160) yields the expected quantity.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
Question 1027Question
Potassium tetraoxomanganate(VII), KMnO4\text{KMnO}_4, reacts with excess concentrated hydrochloric acid according to the redox equation:
2KMnO4(s)+16HCl(aq)2KCl(aq)+2MnCl2(aq)+8H2O(l)+5Cl2(g)2\text{KMnO}_4(s) + 16\text{HCl}(aq) \rightarrow 2\text{KCl}(aq) + 2\text{MnCl}_2(aq) + 8\text{H}_2\text{O}(l) + 5\text{Cl}_2(g)
Calculate the volume of chlorine gas (in dm3\text{dm}^3) produced at s.t.p. when 15.8 g15.8\text{ g} of KMnO4\text{KMnO}_4 reacts completely. [Molar mass of KMnO4=158 g mol1\text{KMnO}_4 = 158\text{ g mol}^{-1}, molar volume of gas at s.t.p. = 22.4 dm3mol122.4\text{ dm}^3\text{mol}^{-1}]
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Answer: 5.6

Answer

The volume of chlorine gas liberated at s.t.p. is 5.6 dm35.6\text{ dm}^3.
Converting 15.8 g15.8\text{ g} of KMnO4\text{KMnO}_4 gives 0.10 mol0.10\text{ mol}. According to the balanced equation, 2 mol2\text{ mol} of KMnO4\text{KMnO}_4 yields 5 mol5\text{ mol} of Cl2\text{Cl}_2, giving 0.25 mol0.25\text{ mol} of Cl2\text{Cl}_2. Multiplying 0.25 mol0.25\text{ mol} by the molar volume at s.t.p. (22.4 dm3mol122.4\text{ dm}^3\text{mol}^{-1}) yields 5.6 dm35.6\text{ dm}^3.

Step-by-Step Solution

1
Calculate the moles of potassium tetraoxomanganate(VII), KMnO4\text{KMnO}_4
Moles of KMnO4=15.8 g158 g mol1=0.10 mol\text{Moles of KMnO}_4 = \frac{15.8\text{ g}}{158\text{ g mol}^{-1}} = 0.10\text{ mol}
Converting the given mass of reactant to moles allows stoichiometric comparison.
2
Use the balanced redox equation to determine the mole ratio between KMnO4\text{KMnO}_4 and Cl2\text{Cl}_2
Moles of Cl2=0.10 mol×52=0.25 mol\text{Moles of Cl}_2 = 0.10\text{ mol} \times \frac{5}{2} = 0.25\text{ mol}
The equation shows that 2 moles2\text{ moles} of KMnO4\text{KMnO}_4 produce 5 moles5\text{ moles} of Cl2\text{Cl}_2 gas.
3
Calculate the volume of Cl2\text{Cl}_2 gas produced at s.t.p.
Volume=0.25 mol×22.4 dm3mol1=5.6 dm3\text{Volume} = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 5.6\text{ dm}^3
At s.t.p., 1 mole1\text{ mole} of any gas occupies 22.4 dm322.4\text{ dm}^3.

Key Concept

Stoichiometric calculations in the laboratory preparation of chlorine from oxidation of hydrochloric acid
Question 1028Question

A steady electric current of 5.0 A5.0\text{ A} is passed through molten lead(II) bromide (PbBr2PbBr_2) for 32 minutes32\text{ minutes} and 10 seconds10\text{ seconds}. What mass of lead, in grams, is deposited at the cathode? [Pb=207\text{Pb} = 207, 1 F=96 500 C mol11\text{ F} = 96\text{ }500\text{ C mol}^{-1}]

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Answer: 10.35

Answer

The mass of lead deposited at the cathode is 10.35 g10.35\text{ g}.
Passing a steady current of 5.0 A5.0\text{ A} for 1930 s1930\text{ s} transfers 9650 C9650\text{ C} of charge, corresponding to 0.10 mol0.10\text{ mol} of electrons (0.10 F0.10\text{ F}). Since reduction of lead(II) ions (Pb2++2ePbPb^{2+} + 2e^- \rightarrow Pb) requires 2 moles of electrons per mole of lead metal, 0.05 mol0.05\text{ mol} of lead is deposited. Multiplying by the relative atomic mass of lead (207 g/mol207\text{ g/mol}) gives 10.35 g10.35\text{ g}.

Step-by-Step Solution

1
Convert time from minutes and seconds into total seconds
t=(32×60 s)+10 s=1930 st = (32 \times 60\text{ s}) + 10\text{ s} = 1930\text{ s}
Time must be expressed in seconds to calculate electric charge in coulombs.
2
Calculate total quantity of electricity (QQ) passed
Q=I×t=5.0 A×1930 s=9650 CQ = I \times t = 5.0\text{ A} \times 1930\text{ s} = 9650\text{ C}
Electric charge is the product of current in amperes and duration in seconds.
3
Calculate the moles of electrons transferred
\text{Moles of } e^- = \frac{9650\text{ C}}{96500\text{ C mol}^{-1}} = 0.10\text{ mol e}^-
One Faraday (96500 C96500\text{ C}) corresponds to one mole of electrons.
4
Relate moles of electrons to moles of lead metal using the cathode half-reaction
Pb^{2+} + 2e^- \rightarrow Pb(s) \implies \text{Moles of } Pb = \frac{0.10\text{ mol e}^-}{2} = 0.05\text{ mol}
Lead has a valency of 2 in PbBr2PbBr_2, requiring 2 moles of electrons per mole of lead deposited.
5
Calculate the mass of deposited lead
\text{Mass} = 0.05\text{ mol} \times 207\text{ g mol}^{-1} = 10.35\text{ g}
Mass is obtained by multiplying the amount of substance in moles by its relative atomic mass.

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations
Question 1029Question
A 1.0 dm31.0\text{ dm}^3 rigid reaction vessel contains an equilibrium mixture of 0.20 mol0.20\text{ mol} of sulfur dioxide (SO2\text{SO}_2), 0.10 mol0.10\text{ mol} of oxygen (O2\text{O}_2), and 0.40 mol0.40\text{ mol} of sulfur trioxide (SO3\text{SO}_3) at a constant temperature according to the equation:
2SO2(g)+O2(g)2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)
What is the numerical value of the equilibrium constant, KcK_c, for this reaction?
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Answer: 40

Answer

The numerical value of the equilibrium constant KcK_c is 40.
Substituting the given equilibrium concentrations into the stoichiometric expression Kc=[SO3]2[SO2]2[O2]K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} gives Kc=(0.40)2(0.20)2×0.10=0.160.04×0.10=40K_c = \frac{(0.40)^2}{(0.20)^2 \times 0.10} = \frac{0.16}{0.04 \times 0.10} = 40.

Step-by-Step Solution

1
Write the equilibrium constant expression (KcK_c) for the reaction.
Kc=[SO3]2[SO2]2[O2]K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}
Products are in the numerator and reactants in the denominator, each raised to the power of their respective stoichiometric coefficients.
2
Convert equilibrium moles into molar concentrations (mol dm⁻³).
[SO₂] = 0.20 mol dm⁻³, [O₂] = 0.10 mol dm⁻³, [SO₃] = 0.40 mol dm⁻³
Concentration equals moles divided by volume (1.0 dm31.0\text{ dm}^3).
3
Substitute the values into the KcK_c expression and solve.
Kc=(0.40)2(0.20)2×0.10=0.160.004=40K_c = \frac{(0.40)^2}{(0.20)^2 \times 0.10} = \frac{0.16}{0.004} = 40
Evaluating the square terms yields 0.16 in the numerator and 0.004 in the denominator, which simplifies to 40.

Key Concept

Calculation of equilibrium constant (Kc) from equilibrium concentrations
Question 1030Question
During the first stage of the Contact Process for the industrial manufacture of tetraoxosulfate(VI) acid, pure sulfur is burned in dry air to produce sulfur(IV) oxide gas according to the equation:
S(s)+O2(g)SO2(g)S(s) + O_2(g) \rightarrow SO_2(g)
What volume of sulfur(IV) oxide gas, in dm3\text{dm}^3 measured at standard temperature and pressure (STP), is produced by the complete combustion of 16.0 g16.0\text{ g} of sulfur?
[Relative atomic mass: S=32S = 32; Molar volume of gas at STP = 22.4 dm3 mol122.4\text{ dm}^3\text{ mol}^{-1}]
Show answer & explanation

Answer: 11.2

Answer

The volume of sulfur(IV) oxide gas produced at STP is 11.2 dm311.2\text{ dm}^3.
According to the balanced equation S(s)+O2(g)SO2(g)S(s) + O_2(g) \rightarrow SO_2(g), 1 mol1\text{ mol} (32 g32\text{ g}) of sulfur yields 1 mol1\text{ mol} (22.4 dm322.4\text{ dm}^3 at STP) of sulfur(IV) oxide gas. Therefore, 16.0 g16.0\text{ g} of sulfur corresponds to 16.032=0.50 mol\frac{16.0}{32} = 0.50\text{ mol}, which produces 0.50×22.4 dm3=11.2 dm30.50 \times 22.4\text{ dm}^3 = 11.2\text{ dm}^3 of SO2SO_2 gas at STP.

Step-by-Step Solution

1
Calculate the amount in moles of sulfur reacted.
0.50 mol0.50\text{ mol} of sulfur.
Using the formula moles=massmolar mass=16.0 g32.0 g mol1=0.50 mol\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{16.0\text{ g}}{32.0\text{ g mol}^{-1}} = 0.50\text{ mol}.
2
Use the mole ratio from the balanced chemical equation to find moles of sulfur(IV) oxide gas formed.
0.50 mol0.50\text{ mol} of SO2(g)SO_2(g).
The equation shows a 1:1 stoichiometric ratio between S(s)S(s) and SO2(g)SO_2(g).
3
Calculate the gas volume at standard temperature and pressure (STP).
11.2 dm311.2\text{ dm}^3.
Multiply the moles of gas by the molar volume at STP: V=0.50 mol×22.4 dm3 mol1=11.2 dm3V = 0.50\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3.

Key Concept

Stoichiometric Volume Calculations for Gas Generation in the Contact Process
Question 1031Question

In humans, Duchenne muscular dystrophy is inherited as an X-linked recessive disorder (XdX^d), while the normal allele is dominant (XDX^D). A phenotypically normal woman seeks genetic counseling. Her maternal grandfather had Duchenne muscular dystrophy, whereas her maternal grandmother was homozygous normal. Her father is phenotypically normal. If this woman marries a phenotypically normal man, what is the probability (expressed as a percentage) that their first male child will be affected by the disorder?

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Answer: 25

Answer

The probability that their first male child will be affected by Duchenne muscular dystrophy is 25%.
The maternal grandfather (XdYX^d Y) passes his XdX^d chromosome to his daughter (the woman's mother), making her an obligate carrier (XDXdX^D X^d). When this carrier mother has a daughter with a normal male (XDYX^D Y), the daughter has a 50%50\% (0.50.5) chance of being a carrier (XDXdX^D X^d). If the woman is a carrier, any male child she has has a 50%50\% (0.50.5) chance of receiving the XdX^d allele and being affected. Multiplying these independent probabilities (0.5×0.50.5 \times 0.5) yields 0.250.25, or 25%25\%.

Step-by-Step Solution

1
Determine the genotype of the woman's mother from her maternal grandparents.
The woman's mother inherited XdX^d from her father (XdYX^d Y) and XDX^D from her mother (XDXDX^D X^D), making her an obligate carrier (XDXdX^D X^d).
Fathers always pass their single X chromosome to their daughters.
2
Calculate the probability that the woman inherited the recessive allele from her mother.
Probability that the woman is a carrier (XDXdX^D X^d) is 0.50.5 (or 50%50\%).
A carrier mother (XDXdX^D X^d) and normal father (XDYX^D Y) have a 50%50\% chance of producing a carrier daughter.
3
Calculate the probability that a male child of a carrier woman receives the recessive X-linked allele.
If the woman is a carrier, the probability of an affected son (XdYX^d Y) is 0.50.5 (or 50%50\%).
A male child receives his only X chromosome from his mother.
4
Multiply the independent probabilities to find the overall risk for the first male child.
P(Affected male child)=0.5 (mother is carrier)×0.5 (son inherits Xd)=0.25=25%P(\text{Affected male child}) = 0.5 \text{ (mother is carrier)} \times 0.5 \text{ (son inherits } X^d) = 0.25 = 25\%.
Both independent events (mother being a carrier and son inheriting the mutated allele) must occur.

Key Concept

Sex-Linked Recessive Inheritance and Pedigree Carrier Probability
Question 1032Question

In humans, red-green color blindness is an X-linked recessive trait (XcX^c), whereas normal vision is controlled by the dominant allele (XCX^C). Albinism is an autosomal recessive disorder (aa), whereas normal skin pigmentation is controlled by the dominant allele (AA). A woman with normal vision and normal skin pigmentation, whose father was both color-blind and albino, marries a man with normal vision who is a carrier for albinism. If this couple produces a male child (son), what is the percentage probability that the son will be both color-blind and albino?

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Answer: 12.5

Answer

The percentage probability that a son born to this couple will be both color-blind and albino is 12.5%.
The mother's father was albino (aaaa) and color-blind (XcYX^c Y), meaning she inherited aa and XcX^c from him. Given her normal phenotype, her genotype is AaXCXcAa X^C X^c. The father is AaXCYAa X^C Y. When determining traits for a son, the son receives the YY chromosome from the father, so his vision phenotype depends entirely on which XX chromosome he receives from his mother (50% chance of XcX^c). The probability of being albino from two carrier parents (Aa×AaAa \times Aa) is 25% (14\frac{1}{4}). Multiplying these independent probabilities yields 12×14=18\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}, which equals 12.5%.

Step-by-Step Solution

1
Determine parental genotypes from the pedigree information provided.
Mother's genotype: AaXCXcAa X^C X^c; Father's genotype: AaXCYAa X^C Y.
The mother received recessive alleles aa and XcX^c from her affected father (aaXcYaa X^c Y). The father is stated to have normal vision (XCYX^C Y) and to be a carrier for albinism (AaAa).
2
Calculate the probability of the male child inheriting the X-linked color blindness trait.
Probability of color-blind son = 12\frac{1}{2} (50%).
For male offspring, sex is fixed by inheriting the YY chromosome from the father. The mother has a 50% chance of passing her XcX^c allele.
3
Calculate the probability of the child inheriting autosomal albinism.
Probability of albino phenotype (aaaa) = 14\frac{1}{4} (25%).
Crossing two heterozygous carriers (Aa×AaAa \times Aa) yields a 1 in 4 chance of an autosomal recessive aaaa offspring.
4
Apply the product rule for independent genetic events.
Combined probability = 12×14=18=12.5%\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} = 12.5\%.
Autosomal inheritance and X-linked inheritance are independent genetic events, so their probabilities are multiplied.

Key Concept

Independent assortment of an autosomal recessive trait and an X-linked recessive trait in human pedigree analysis.
Question 1033Question

During an ecological study of an agricultural fish pond in Ibadan, Oyo State, a student employed the mark-release-recapture technique to estimate the population size of tilapia (*Oreochromis niloticus*). In the initial sampling, 120120 fish were captured, marked with harmless plastic tags, and released back into the pond. Two days later, a second sample of 150150 fish was netted, out of which 3030 individuals were found to be marked. What is the estimated total population size of tilapia fish in the pond?

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Answer: 600

Answer

The estimated total population size of tilapia fish in the pond is 600.
The correct answer is derived using the Lincoln index formula for population estimation: N=M×CRN = \frac{M \times C}{R}, where M=120M = 120 (initial marked sample), C=150C = 150 (total second sample), and R=30R = 30 (recaptured marked sample). Substituting these values yields N=120×15030=600N = \frac{120 \times 150}{30} = 600 fish.

Step-by-Step Solution

1
Identify the values for the Lincoln Index (Lincoln-Petersen estimator) parameters from the problem statement.
Number marked in first capture (MM) = 120120, total caught in second capture (CC) = 150150, recaptured marked individuals (RR) = 3030.
The capture-recapture method relies on the proportion of marked individuals in the second sample being equal to the proportion of marked individuals in the total population.
2
Apply the Lincoln Index formula N=M×CRN = \frac{M \times C}{R}.
N=120×15030N = \frac{120 \times 150}{30}.
Multiplying the initial sample size by the second sample size and dividing by the recaptured marked count yields the total estimated population.
3
Perform the division and multiplication to solve for NN.
N=600N = 600.
Simplifying 15030=5\frac{150}{30} = 5, then 120×5=600120 \times 5 = 600 fish.

Key Concept

Lincoln Index (Mark-Release-Recapture Method)
Question 1034Question

In guinea pigs (*Cavia porcellus*), black coat color (BB) is dominant over white coat color (bb), and short hair (SS) is dominant over long hair (ss). If a heterozygous black, short-haired guinea pig (BbSsBbSs) is mated with a white, long-haired guinea pig (bbssbbss) and they produce a total of 640 offspring, how many of the offspring are expected to display a black coat and long hair?

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Answer: 160

Answer

160 offspring are expected to have a black coat and long hair.
In a dihybrid testcross involving a double heterozygote (BbSsBbSs) and a homozygous recessive individual (bbssbbss), the offspring phenotypes appear in equal ratios of 1:1:1:1 (25% for each phenotypic class). The black coat, long hair phenotype (BbssBbss) corresponds to 1/4 of the total offspring. Multiplying 1/4 by 640 yields exactly 160 expected offspring.

Step-by-Step Solution

1
Determine the type of genetic cross and parental genotypes.
The cross is a dihybrid testcross between BbSsBbSs and bbssbbss.
One parent is heterozygous for both independently assorting traits (BbSsBbSs), and the other parent is homozygous recessive (bbssbbss).
2
Determine the proportion of offspring expected to have the phenotype black coat and long hair (BbssBbss).
The proportion of BbssBbss offspring is 14\frac{1}{4} (or 25%25\%).
The BbSsBbSs parent produces four types of gametes (BSBS, BsBs, bSbS, bsbs) in equal proportions (14\frac{1}{4} each). Combining BsBs with bsbs yields BbssBbss.
3
Calculate the expected count out of 640 total offspring.
14×640=160\frac{1}{4} \times 640 = 160.
Multiplying the expected phenotypic fraction by the total offspring count yields the absolute expected count.

Key Concept

Dihybrid testcross ratio and probability calculation
Question 1035Question

What is the molar concentration (in mol dm3\text{mol dm}^{-3}) of a tetraoxosulfate(VI) acid solution if 20.0 cm320.0\text{ cm}^3 of the acid is required to completely neutralize 25.0 cm325.0\text{ cm}^3 of a 0.08 mol dm30.08\text{ mol dm}^{-3} sodium hydroxide solution?

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Answer: 0.05

Answer

The correct molar concentration of the tetraoxosulfate(VI) acid solution is 0.05 mol dm30.05\text{ mol dm}^{-3}.
From the balanced chemical equation H2SO4+2NaOHNa2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}, 1 mole1\text{ mole} of tetraoxosulfate(VI) acid reacts with 2 moles2\text{ moles} of sodium hydroxide (na=1,nb=2n_a = 1, n_b = 2). Substituting the given values into the titration equation CaVaCbVb=nanb\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} gives Ca=0.08×25.0×120.0×2=0.05 mol dm3C_a = \frac{0.08 \times 25.0 \times 1}{20.0 \times 2} = 0.05\text{ mol dm}^{-3}.

Step-by-Step Solution

1
Write the balanced chemical equation for the reaction to find the mole ratio of acid to base.
H2SO4+2NaOHNa2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}, giving na=1n_a = 1 and nb=2n_b = 2.
Tetraoxosulfate(VI) acid is a dibasic acid and requires two moles of sodium hydroxide for complete neutralization.
2
Apply the volumetric analysis formula relating concentration, volume, and stoichiometry.
CaVaCbVb=nanb    Ca×20.00.08×25.0=12\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} \implies \frac{C_a \times 20.0}{0.08 \times 25.0} = \frac{1}{2}
The standard titration formula relates acid concentration (CaC_a), acid volume (VaV_a), base concentration (CbC_b), base volume (VbV_b), and their mole coefficients.
3
Rearrange the equation to solve for CaC_a.
Ca=Cb×Vb×naVa×nb=0.08×25.0×120.0×2=2.040.0=0.05 mol dm3C_a = \frac{C_b \times V_b \times n_a}{V_a \times n_b} = \frac{0.08 \times 25.0 \times 1}{20.0 \times 2} = \frac{2.0}{40.0} = 0.05\text{ mol dm}^{-3}.
Simplifying the arithmetic gives the precise molar concentration of the acid.

Key Concept

Volumetric analysis stoichiometry and stoichiometric concentration calculations for acid-base neutralization.
Question 1036Question

An ecology student sampled the population of guinea grass (*Panicum maximum*) in a 600 m2600\text{ m}^2 pasture plot in Kaiama, Kwara State, using a 0.5 m×0.5 m0.5\text{ m} \times 0.5\text{ m} quadrat frame. The counts of grass clumps recorded from 10 randomly placed quadrats were 4, 6, 3, 7, 5, 2, 8, 4, 6, and 5. What is the estimated total population of guinea grass clumps in the entire pasture plot?

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Answer: 12000

Answer

The estimated total population of guinea grass clumps in the pasture plot is 12,000 clumps.
To estimate total population size from quadrat samples, first determine the total area sampled (10 quadrats×0.25 m2=2.5 m210 \text{ quadrats} \times 0.25\text{ m}^2 = 2.5\text{ m}^2). Dividing the total count of organisms (50 clumps50\text{ clumps}) by this sampled area yields a population density of 20 clumps/m220\text{ clumps/m}^2. Multiplying the density by the total area of the plot (600 m2600\text{ m}^2) gives the estimated total population of 12,000 clumps12,000\text{ clumps}.

Step-by-Step Solution

1
Determine the surface area of a single quadrat frame
Area of one quadrat = 0.5 m×0.5 m=0.25 m20.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2
Knowing the individual quadrat dimensions is necessary to calculate the sample area.
2
Calculate the total area sampled across all 10 quadrats
Total area sampled = 10×0.25 m2=2.5 m210 \times 0.25\text{ m}^2 = 2.5\text{ m}^2
Ten quadrats were placed, so the total sample area equals ten times the single quadrat area.
3
Find the total count of organisms recorded in all samples
Total count = 4+6+3+7+5+2+8+4+6+5=50 clumps4 + 6 + 3 + 7 + 5 + 2 + 8 + 4 + 6 + 5 = 50\text{ clumps}
Summing individual counts across all sampling frames gives the total sample size.
4
Compute the average population density per unit area
Population density = 50 clumps2.5 m2=20 clumps/m2\frac{50\text{ clumps}}{2.5\text{ m}^2} = 20\text{ clumps/m}^2
Density is defined as total count divided by total sampled area.
5
Extrapolate population density to the entire study area
Total estimated population = 20 clumps/m2×600 m2=12,000 clumps20\text{ clumps/m}^2 \times 600\text{ m}^2 = 12,000\text{ clumps}
Multiplying population density per square metre by total plot area gives the estimated overall population size.

Key Concept

Extrapolation of population size from sample quadrat density
Question 1037Question

An ecological survey was conducted in the Borgu sector of Kainji Lake National Park to estimate the population size of grasscutters (*Thryonomys swinderianus*). In the initial phase, 8080 grasscutters were captured, marked with ear tags, and released back into the habitat. One week later, a second sample of 100100 grasscutters was captured, out of which 2020 individuals retained their mark. If post-marking field monitoring established that 10%10\% of all originally marked animals lost their tags during the interval, what is the estimated total population size of grasscutters in the study area?

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Answer: 360

Answer

The estimated total population size of grasscutters in the study area is 360.
Accounting for a 10% tag loss reduces the active marked population from 80 to 72 individuals. Applying the Lincoln-Petersen index formula (N=M×CRN = \frac{M \times C}{R}) with M=72M = 72, C=100C = 100, and R=20R = 20 gives an estimated total population size of 360 grasscutters.

Step-by-Step Solution

1
Calculate the effective number of marked individuals (MeffM_{eff}) in the population considering the 10% tag loss.
Meff=80×(10.10)=72M_{eff} = 80 \times (1 - 0.10) = 72 marked individuals.
Animals that lost their tags no longer register as marked upon recapture, effectively reducing the active marked proportion in the population.
2
Substitute Meff=72M_{eff} = 72, total recaptured C=100C = 100, and marked recaptured R=20R = 20 into the Lincoln-Petersen index formula N=Meff×CRN = \frac{M_{eff} \times C}{R}.
N=72×10020=360N = \frac{72 \times 100}{20} = 360.
The proportion of marked individuals in the recaptured sample equals the proportion of effective marked individuals in the total population.

Key Concept

Lincoln-Petersen Mark-Recapture Index with Sampling Bias Adjustment
Question 1038Question

A subatomic particle of mass 6.63×1031 kg6.63 \times 10^{-31}\text{ kg} moves with a velocity of 1.0×106 m/s1.0 \times 10^6\text{ m/s}. Calculate its de Broglie wavelength in nanometers (nm\text{nm}). (Take Planck's constant h=6.63×1034 Jsh = 6.63 \times 10^{-34}\text{ J}\cdot\text{s} and 1 nm=109 m1\text{ nm} = 10^{-9}\text{ m})

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Answer: 1

Answer

The de Broglie wavelength of the particle is 1.0 nm1.0\text{ nm}.
Using de Broglie's wave-particle duality relation λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{m v}, substituting the given values yields λ=6.63×1034 Js(6.63×1031 kg)×(1.0×106 m/s)=1.0×109 m\lambda = \frac{6.63 \times 10^{-34}\text{ J}\cdot\text{s}}{(6.63 \times 10^{-31}\text{ kg}) \times (1.0 \times 10^6\text{ m/s})} = 1.0 \times 10^{-9}\text{ m}. In nanometers, this is equal to 1.0 nm1.0\text{ nm}.

Step-by-Step Solution

1
Calculate the linear momentum of the particle
p=6.63×1025 kgm/sp = 6.63 \times 10^{-25}\text{ kg}\cdot\text{m/s}
Momentum is the product of mass and velocity (p=mvp = m v).
2
Calculate the de Broglie wavelength
λ=1.0×109 m\lambda = 1.0 \times 10^{-9}\text{ m}
According to de Broglie's hypothesis, wavelength is given by λ=hp\lambda = \frac{h}{p}.
3
Convert the calculated wavelength to nanometers
λ=1.0 nm\lambda = 1.0\text{ nm}
Since 1 nm=109 m1\text{ nm} = 10^{-9}\text{ m}, dividing 1.0×109 m1.0 \times 10^{-9}\text{ m} by 10910^{-9} yields 1.0 nm1.0\text{ nm}.

Key Concept

de Broglie Wavelength and Wave-Particle Duality
Question 1039Question

The market demand and supply functions for a commodity are given as Qd=1202PQ_d = 120 - 2P and Qs=20+3PQ_s = 20 + 3P respectively, where PP is the price in Naira (\text{₦}) and QQ is the quantity in units. If the government levies a specific sales tax of 10\text{₦}10 per unit on the producers, what is the per-unit tax burden borne by the consumer?

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Answer: 6

Answer

The per-unit tax burden borne by the consumer is \text{₦}6.
The initial market clearing price is found by setting 1202P=20+3P120 - 2P = 20 + 3P, which gives P1=20P_1 = \text{₦}20. With a specific tax of 10\text{₦}10 per unit levied on producers, the price received by sellers becomes Ps=Pc10P_s = P_c - 10. Substituting into the supply equation gives Qs=20+3(Pc10)=3Pc10Q_s' = 20 + 3(P_c - 10) = 3P_c - 10. Equating demand and post-tax supply gives 1202Pc=3Pc10    5Pc=130    Pc=26120 - 2P_c = 3P_c - 10 \implies 5P_c = 130 \implies P_c = \text{₦}26. The consumer tax burden per unit is the price increase, 2620=626 - 20 = \text{₦}6.

Step-by-Step Solution

1
Calculate the pre-tax equilibrium price
Initial price P1=20P_1 = \text{₦}20
Equating quantity demanded Qd=1202PQ_d = 120 - 2P and quantity supplied Qs=20+3PQ_s = 20 + 3P gives 1202P=20+3P120 - 2P = 20 + 3P, which solves to P1=20P_1 = 20.
2
Adjust the supply equation to account for the specific tax of \text{₦}10 per unit
New supply function Qs=3Pc10Q_s' = 3P_c - 10
Because the tax is paid by producers, the net price received by sellers is Ps=Pc10P_s = P_c - 10. Substituting PsP_s into Qs=20+3PsQ_s = 20 + 3P_s yields Qs=20+3(Pc10)=3Pc10Q_s' = 20 + 3(P_c - 10) = 3P_c - 10.
3
Calculate the post-tax equilibrium price paid by consumers (PcP_c)
Post-tax consumer price Pc=26P_c = \text{₦}26
Equating QdQ_d and QsQ_s' gives 1202Pc=3Pc10120 - 2P_c = 3P_c - 10, which simplifies to 5Pc=130    Pc=265P_c = 130 \implies P_c = 26.
4
Calculate the consumer's share of the per-unit tax incidence
Consumer tax incidence = \text{₦}6
The per-unit tax incidence on the consumer equals the net increase in market price paid, PcP1=2620=6P_c - P_1 = 26 - 20 = 6.

Key Concept

Tax Incidence and Price Elasticity of Demand and Supply
Question 1040Question

An biology student placed a 1 m21\text{ m}^2 quadrat 10 times randomly in a grassland plot within Yankari Game Reserve to estimate the population density of wild marigold (*Tithonia diversifolia*). A total of 150 wild marigold plants were counted across all 10 quadrat samples. What is the population density of the wild marigold plants in organisms per square metre?

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Answer: 15

Answer

The population density of wild marigold plants is 15 plants/m215\text{ plants/m}^2.
Population density is calculated using the formula: Population Density=Total number of individuals countedTotal area sampled\text{Population Density} = \frac{\text{Total number of individuals counted}}{\text{Total area sampled}}. Since 10 quadrats of 1 m21\text{ m}^2 each were thrown, the total area sampled is 10 m210\text{ m}^2. Dividing 150 plants by 10 m210\text{ m}^2 yields 15 plants/m215\text{ plants/m}^2.

Step-by-Step Solution

1
Calculate total area sampled
Total area = 10 m210\text{ m}^2
The area of a single quadrat is 1 m21\text{ m}^2 and 10 quadrats were sampled in total.
2
Calculate population density
Density = 15 plants/m215\text{ plants/m}^2
Population density is determined by dividing the total count of organisms by the total sampled area.

Key Concept

Calculating population density using quadrat sampling data.
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