All practice questions

A

3.50\text{ g}

sample of impure potassium trioxochlorate(V),

\text{KClO}_3

, was completely decomposed by heating in the presence of a manganese(IV) oxide catalyst according to the equation:

2\text{KClO}_3(\text{s}) \rightarrow 2\text{KCl}(\text{s}) + 3\text{O}_2(\text{g})

If

0.672\text{ dm}^3

of oxygen gas was collected at STP, what is the percentage purity of the

\text{KClO}_3

sample?

(\text{K} = 39.0, \text{Cl} = 35.5, \text{O} = 16.0, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1})

70.0\%

105.0\%

65.3\%

35.0\%

Show answer & explanation

Answer: $70.0\%$

Answer

The percentage purity of the potassium trioxochlorate(V) sample is 70.0%.

The option stating 70.0% is correct. Converting

0.672\text{ dm}^3

of

\text{O}_2

gas at STP yields

0.030\text{ mol}

of

\text{O}_2

. Using the balanced stoichiometric mole ratio (

2\text{ KClO}_3 : 3\text{ O}_2

), this corresponds to

0.020\text{ mol}

of pure

\text{KClO}_3

. Multiplying by its molar mass (

122.5\text{ g/mol}

) gives

2.45\text{ g}

of pure

\text{KClO}_3

. Dividing

2.45\text{ g}

by the total sample mass of

3.50\text{ g}

and multiplying by

100\%

gives

70.0\%

.

Step-by-Step Solution

1

Calculate the moles of oxygen gas collected at STP

\text{Moles of O}_2 = \frac{0.672\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.030\text{ mol}

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

2

Determine the moles of pure potassium trioxochlorate(V) reacted using stoichiometric coefficients

\text{Moles of KClO}_3 = 0.030\text{ mol O}_2 \times \frac{2\text{ mol KClO}_3}{3\text{ mol O}_2} = 0.020\text{ mol}

From the balanced chemical equation, 2 moles of KClO3 decompose to yield 3 moles of O2.

3

Calculate the mass of pure potassium trioxochlorate(V)

\text{Molar mass of KClO}_3 = 39.0 + 35.5 + 3(16.0) = 122.5\text{ g/mol}

\text{Mass of pure KClO}_3 = 0.020\text{ mol} \times 122.5\text{ g/mol} = 2.45\text{ g}

Mass is obtained by multiplying moles by molar mass.

4

Calculate the percentage purity of the original sample

\text{Percentage purity} = \left(\frac{2.45\text{ g}}{3.50\text{ g}}\right) \times 100\% = 70.0\%

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

Key Concept

Percentage purity calculation via gas stoichiometry at STP

Question 8382Question →

An unknown gaseous hydrocarbon $X$ belongs to the alkane homologous series. Under identical conditions of temperature and pressure, a given volume of oxygen gas ( $O_2$ ) diffuses through a micro-porous barrier in $40\text{ s}$ , whereas the exact same volume of gas $X$ requires $60\text{ s}$ to diffuse through the same barrier. What is the molecular formula of hydrocarbon $X$ ? $[H = 1,\ C = 12,\ O = 16]$

C_5H_{12}

C_4H_{10}

C_3H_8

C_2H_6

Show answer & explanation

Answer: $C_5H_{12}$

Answer

The molecular formula of hydrocarbon X is

C_5H_{12}

.

According to Graham's Law of Diffusion, the time taken for equal volumes of two gases to diffuse under identical conditions is directly proportional to the square root of their respective molar masses:

t_X / t_{O_2} = \sqrt{M_X / M_{O_2}}

. Given

t_X = 60\text{ s}

,

t_{O_2} = 40\text{ s}

, and

M_{O_2} = 32\text{ g/mol}

, we have

60/40 = \sqrt{M_X / 32}

. Squaring both sides yields

(3/2)^2 = 9/4 = M_X / 32

, which solves to

M_X = 72\text{ g/mol}

. Matching this to the alkane formula

C_n H_{2n+2} = 14n + 2 = 72

gives

n = 5

, identifying the gas as pentane,

C_5H_{12}

.

Step-by-Step Solution

1

Calculate the molar mass of oxygen gas (

O_2

)

M_{O_2} = 2 \times 16 = 32\text{ g/mol}

Oxygen exists as a diatomic molecule under standard conditions.

2

Apply Graham's Law of Diffusion relating diffusion time (

t

) and molar mass (

M

) for equal volumes

\frac{t_X}{t_{O_2}} = \sqrt{\frac{M_X}{M_{O_2}}}

Time required for diffusion of equal gas volumes is directly proportional to the square root of the molar mass.

3

Substitute known diffusion times (

t_X = 60\text{ s}

,

t_{O_2} = 40\text{ s}

) into the equation and solve for

M_X

\frac{60}{40} = \sqrt{\frac{M_X}{32}} \implies \frac{3}{2} = \sqrt{\frac{M_X}{32}} \implies \left(\frac{3}{2}\right)^2 = \frac{M_X}{32} \implies \frac{9}{4} = \frac{M_X}{32} \implies M_X = 72\text{ g/mol}

Squaring both sides removes the radical and allows direct algebraic isolation of

M_X

.

4

Determine the molecular formula using the alkane general formula

C_n H_{2n+2}

12n + (2n + 2) = 72 \implies 14n + 2 = 72 \implies 14n = 70 \implies n = 5

Each carbon atom contributes

12\text{ g/mol}

and each hydrogen atom contributes

1\text{ g/mol}

.

5

Write the molecular formula for

n = 5

C_5H_{12}

Substituting

n = 5

into

C_n H_{2n+2}

gives

C_5H_{12}

(pentane).

Key Concept

Graham's Law of Diffusion and Effusion

Estimated Time:2m 0s

Question 8383Question →

Malachite is an important copper ore with the chemical formula

\text{CuCO}_3\cdot\text{Cu(OH)}_2

. Upon strong heating, it undergoes thermal decomposition according to the following balanced equation:

\text{CuCO}_3\cdot\text{Cu(OH)}_2(s) \rightarrow 2\text{CuO}(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)

What is the mass of copper(II) oxide (

\text{CuO}

), in grams, formed when

22.2\text{ g}

of malachite is completely decomposed? [Relative atomic masses:

\text{Cu} = 64

,

\text{C} = 12

,

\text{O} = 16

,

\text{H} = 1

]

Show answer & explanation

Answer: 16

Answer

The mass of copper(II) oxide produced is 16.0 g.

Thermal decomposition of 1 mole of malachite (

\text{CuCO}_3\cdot\text{Cu(OH)}_2

, molar mass

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

) produces 2 moles of copper(II) oxide (

\text{CuO}

, molar mass

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

). Given

22.2\text{ g}

of malachite (

0.1\text{ mol}

), exactly

0.2\text{ mol}

of

\text{CuO}

is formed, giving a mass of

0.2\text{ mol} \times 80\text{ g mol}^{-1} = 16.0\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of malachite,

\text{CuCO}_3\cdot\text{Cu(OH)}_2

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

Summing the relative atomic masses:

2(64) + 12 + 5(16) + 2(1) = 222\text{ g mol}^{-1}

.

2

Calculate the number of moles of malachite in the sample

0.1\text{ mol}

Dividing given mass by molar mass:

\frac{22.2\text{ g}}{222\text{ g mol}^{-1}} = 0.1\text{ mol}

.

3

Determine the moles of

\text{CuO}

produced using stoichiometry

0.2\text{ mol}

of

\text{CuO}

The balanced chemical equation shows a

1:2

mole ratio between malachite and

\text{CuO}

.

4

Calculate the mass of

\text{CuO}

formed

16.0\text{ g}

Multiplying moles of

\text{CuO}

by its molar mass (

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

):

0.2 \times 80 = 16.0\text{ g}

.

Key Concept

Stoichiometry of copper compounds thermal decomposition

Question 8384Question →

Given that the molar mass of methane ( $CH_4$ ) is $16\text{ g/mol}$ and that of sulfur(IV) oxide ( $SO_2$ ) is $64\text{ g/mol}$ , what is the ratio of the rate of diffusion of methane to that of sulfur(IV) oxide under identical conditions of temperature and pressure?

2 : 1

4 : 1

1 : 2

1 : 4

Show answer & explanation

Answer: $2 : 1$

Answer

2 : 1

According to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r \propto \frac{1}{\sqrt{M}}

). Comparing methane (

CH_4

,

M = 16\text{ g/mol}

) to sulfur(IV) oxide (

SO_2

,

M = 64\text{ g/mol}

), the ratio of their rates is

\sqrt{\frac{64}{16}} = \sqrt{4} = 2

. Therefore, methane diffuses twice as fast as sulfur(IV) oxide, yielding a ratio of

2 : 1

.

Step-by-Step Solution

1

State Graham's Law of Diffusion formula

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

2

Substitute the given molar masses into the formula

\frac{r_{CH_4}}{r_{SO_2}} = \sqrt{\frac{64}{16}}

Molar mass of

SO_2 = 64\text{ g/mol}

and molar mass of

CH_4 = 16\text{ g/mol}

.

3

Simplify the ratio

\frac{r_{CH_4}}{r_{SO_2}} = \sqrt{4} = 2

Taking the square root of 4 gives 2, which corresponds to a ratio of

2 : 1

.

Key Concept

Graham's Law of Diffusion

Estimated Time:45s

Question 8385Question →

Under specific laboratory conditions of temperature and pressure, a $50\text{ cm}^3$ sample of sulfur(IV) oxide ( $SO_2$ ) gas diffuses through a tiny orifice in $24\text{ seconds}$ . What is the time, in seconds, required for an equal volume of helium ( $He$ ) gas to diffuse through the same orifice under identical conditions? [Relative atomic masses: $\text{He} = 4$ , $\text{O} = 16$ , $\text{S} = 32$ ]

Show answer & explanation

Answer: 6

Answer

The time required for an equal volume of helium gas to diffuse under identical conditions is

6\text{ seconds}

.

According to Graham's Law of Diffusion, the time required for equal volumes of two gases to diffuse at constant temperature and pressure is directly proportional to the square root of their molar masses (

t_1 / t_2 = \sqrt{M_1 / M_2}

). Because helium (

4\text{ g/mol}

) has one-sixteenth the molar mass of sulfur(IV) oxide (

64\text{ g/mol}

), its diffusion time is

\sqrt{1/16} = 1/4

of the time taken by

SO_2

. Therefore,

t_{He} = 24 \text{ s} \times 0.25 = 6\text{ seconds}

.

Step-by-Step Solution

1

Calculate the molar masses of both gases from their chemical formulas and relative atomic masses.

Molar mass of

SO_2 = 64\text{ g/mol}

; Molar mass of

He = 4\text{ g/mol}

.

Graham's Law requires the molar mass of each gas to determine relative diffusion rates and times.

2

Set up the Graham's Law expression relating diffusion time to molar mass for equal volumes.

\frac{t_{He}}{t_{SO_2}} = \sqrt{\frac{M_{He}}{M_{SO_2}}}

Diffusion time for a given volume is directly proportional to the square root of the molar mass.

3

Substitute given values and simplify the square root expression.

\frac{t_{He}}{24} = \sqrt{\frac{4}{64}} = \sqrt{\frac{1}{16}} = \frac{1}{4}

Simplifying the molar mass ratio gives a simple fraction.

4

Multiply to solve for the diffusion time of helium gas.

t_{He} = 24 \times \frac{1}{4} = 6\text{ seconds}

Helium is lighter and diffuses 4 times faster than

SO_2

, requiring 1/4th of the time.

Key Concept

Graham's Law of Diffusion and Effusion

Estimated Time:1m 30s

Question 8386Question →

Which of the following structural isomers of hexane, $C_6H_{14}$ , contains exactly one tertiary carbon atom?

2-Methylpentane

2,3-Dimethylbutane

2,2-Dimethylbutane

Hexane

Show answer & explanation

Answer: 2-Methylpentane

Answer

2-Methylpentane contains exactly one tertiary carbon atom.

A tertiary carbon atom is defined as a carbon atom bonded to three other carbon atoms. In 2-methylpentane, the carbon atom at position 2 is bonded to C-1, C-3, and the branch methyl group, making it the only tertiary carbon atom in the molecule.

Step-by-Step Solution

1

Define a tertiary carbon atom

A tertiary carbon atom (

3^\circ

) is directly bonded to three other carbon atoms.

Carbon classification depends on the number of attached alkyl/carbon groups.

2

Analyze the carbon skeleton of 2-Methylpentane

In

CH_3-CH(CH_3)-CH_2-CH_2-CH_3

, C-2 is bonded to C-1, C-3, and the methyl group.

This structural arrangement yields exactly one tertiary carbon atom.

3

Evaluate the remaining options

2,3-Dimethylbutane has two tertiary carbons, 2,2-Dimethylbutane has one quaternary carbon (zero tertiary), and unbranched hexane has zero tertiary carbons.

Only 2-Methylpentane satisfies the condition of having exactly one tertiary carbon atom.

Key Concept

Classification of carbon atoms in structural isomers

Question 8387Question →

When ammonia ( $NH_3$ ) reacts with an acid to form the ammonium ion ( $NH_4^+$ ), the central nitrogen atom shares its lone pair of electrons to form a dative covalent bond with a proton ( $H^+$ ). According to Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular geometry and hybridization state of the nitrogen atom in the $NH_4^+$ ion?

Tetrahedral geometry with

sp^3

hybridization

Trigonal pyramidal geometry with

sp^3

hybridization

Trigonal planar geometry with

sp^2

hybridization

Square planar geometry with

sp^3d^2

hybridization

Show answer & explanation

Answer: Tetrahedral geometry with $sp^3$ hybridization

Answer

Tetrahedral geometry with

sp^3

hybridization

In the ammonium ion (

NH_4^+

), the central nitrogen atom is surrounded by four single covalent bonds (including one dative covalent bond formed with

H^+

) and zero lone pairs. Four electron domains arrange themselves to minimize repulsion into a symmetric tetrahedral arrangement with bond angles of

109.5^\circ

. Mixing one s orbital and three p orbitals gives

sp^3

hybridization.

Step-by-Step Solution

1

Determine the valence electron count and electron domain total around the central nitrogen atom in

NH_4^+

.

Nitrogen brings 5 valence electrons, four hydrogen atoms contribute 1 electron each, and the +1 charge indicates the loss of 1 electron. Total valence electrons =

5 + 4 - 1 = 8

electrons (4 pairs).

VSEPR theory uses the total number of electron pairs around the central atom to predict spatial geometry.

2

Identify the number of bonding pairs and lone pairs on the nitrogen atom.

All 4 electron pairs are involved in single covalent bonds (3 covalent, 1 dative covalent) with hydrogen atoms, leaving 0 lone pairs.

Molecular geometry depends specifically on the arrangement of bonding pairs when 4 electron domains are present.

3

Determine the molecular geometry and hybridization from the electron pair arrangement.

Four bonding domains with zero lone pairs yield a symmetrical regular tetrahedral shape with a bond angle of

109.5^\circ

and

sp^3

hybridization.

Four equivalent steric domains require one s orbital and three p orbitals to hybridize into four

sp^3

hybrid orbitals.

Key Concept

Molecular geometry and hybridization of polyatomic ions using VSEPR theory

Question 8388Question →

The solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water is $3.5\text{ mol dm}^{-3}$ at $70^\circ\text{C}$ and $1.2\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . What mass of $\text{KNO}_3$ will crystallize out of solution when $200\text{ cm}^3$ of a saturated solution at $70^\circ\text{C}$ is cooled to $25^\circ\text{C}$ ? $[\text{K} = 39, \text{N} = 14, \text{O} = 16]$

46.46\text{ g}

0.46\text{ g}

70.70\text{ g}

24.24\text{ g}

Show answer & explanation

Answer: $46.46\text{ g}$

Answer

The mass of potassium trioxonitrate(V) that crystallizes out is

46.46\text{ g}

.

The net solubility decrease when cooling from

70^\circ\text{C}

to

25^\circ\text{C}

is

2.3\text{ mol dm}^{-3}

. In

200\text{ cm}^3

(

0.2\text{ dm}^3

) of saturated solution,

0.46\text{ mol}

of

\text{KNO}_3

precipitates out. Multiplying

0.46\text{ mol}

by the molar mass of

\text{KNO}_3

(

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

) yields

46.46\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of potassium trioxonitrate(V),

\text{KNO}_3

.

\text{Molar mass} = 39 + 14 + (3 \times 16) = 101\text{ g mol}^{-1}

.

Molar mass is needed to convert molar concentration into mass.

2

Find the change in solubility per

\text{dm}^3

upon cooling from

70^\circ\text{C}

to

25^\circ\text{C}

.

\Delta S = 3.5\text{ mol dm}^{-3} - 1.2\text{ mol dm}^{-3} = 2.3\text{ mol dm}^{-3}

.

The crystallization amount depends on the difference between initial and final solubilities.

3

Determine the amount of solute precipitated in

200\text{ cm}^3

(

0.2\text{ dm}^3

) of solution.

\text{Moles} = 2.3\text{ mol dm}^{-3} \times 0.2\text{ dm}^3 = 0.46\text{ mol}

.

Solubility values are given per

\text{dm}^3

, so they must be scaled to the given volume of

200\text{ cm}^3

.

4

Convert the precipitated moles into mass in grams.

\text{Mass} = 0.46\text{ mol} \times 101\text{ g mol}^{-1} = 46.46\text{ g}

.

Multiplying the number of moles by molar mass gives the required mass in grams.

Key Concept

Solubility Curves and Temperature Effects

Question 8389Question →

A sample of $9.0\text{ g}$ of anhydrous ethanedioic acid, $\text{H}_2\text{C}_2\text{O}_4$ , is completely dehydrated using excess concentrated tetraoxosulfate(VI) acid. The evolved gaseous mixture is passed through an excess concentrated sodium hydroxide solution to absorb carbon(IV) oxide. What volume of pure carbon(II) oxide gas is collected at STP? [Molar mass of $\text{H}_2\text{C}_2\text{O}_4 = 90\text{ g mol}^{-1}$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{ mol}^{-1}$ ]

2.24\text{ dm}^3

2.40\text{ dm}^3

4.48\text{ dm}^3

4.80\text{ dm}^3

Show answer & explanation

Answer: $2.24\text{ dm}^3$

Answer

The volume of pure carbon(II) oxide collected at STP is

2.24\text{ dm}^3

.

Dehydration of

9.0\text{ g}

(

0.10\text{ mol}

) of ethanedioic acid produces

0.10\text{ mol}

of

\text{CO}

and

0.10\text{ mol}

of

\text{CO}_2

. Concentrated sodium hydroxide absorbs all the

\text{CO}_2

, leaving only

0.10\text{ mol}

of neutral

\text{CO}

gas. At STP (

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

),

0.10\text{ mol}

of

\text{CO}

occupies

2.24\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the number of moles of ethanedioic acid (

\text{H}_2\text{C}_2\text{O}_4

).

\text{Moles} = \frac{9.0\text{ g}}{90\text{ g mol}^{-1}} = 0.10\text{ mol}

.

Determining initial mole quantity from given mass and molar mass.

2

Write the balanced chemical equation for the dehydration of ethanedioic acid.

\text{H}_2\text{C}_2\text{O}_4(\text{s}) \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \text{CO}(\text{g}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

.

1 mole of ethanedioic acid yields 1 mole of

\text{CO}

gas and 1 mole of

\text{CO}_2

gas.

3

Determine the mole amount of carbon(II) oxide collected after passing through sodium hydroxide.

Sodium hydroxide reacts with acidic oxide

\text{CO}_2

to form

\text{Na}_2\text{CO}_3

, leaving

0.10\text{ mol}

of unreacted neutral

\text{CO}

gas to be collected.

Sodium hydroxide selectively absorbs carbon(IV) oxide, allowing pure carbon(II) oxide gas to pass through.

4

Calculate the volume of pure

\text{CO}

gas at STP.

\text{Volume} = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3$.

Converting moles of gas to volume at standard temperature and pressure.

Key Concept

Dehydration of ethanedioic acid yields an equimolar mixture of carbon(II) oxide and carbon(IV) oxide. Sodium hydroxide selectively absorbs carbon(IV) oxide, allowing pure carbon(II) oxide to be collected.

Question 8390Question →

Match each solubility phenomenon or term on the left with its correct thermodynamic or practical description on the right. Which pair correctly connects each term to its appropriate description?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fractional crystallization pairs with separation technique using solubility differences; Endothermic dissolution pairs with heat absorption increasing solubility with temperature; Exothermic dissolution pairs with heat evolution decreasing solubility with temperature; Supersaturated solution pairs with holding excess dissolved solute beyond equilibrium limit.

Each term directly aligns with its fundamental thermodynamic property or chemical application. Fractional crystallization isolates salts using temperature-dependent solubility variations; endothermic dissolution absorbs heat so solubility increases with temperature; exothermic dissolution gives off heat so solubility decreases with temperature; and a supersaturated solution holds excess dissolved solute beyond standard equilibrium capacity.

Step-by-Step Solution

1

Identify the separation technique based on solubility curve variance.

Fractional crystallization separates solutes based on their distinct solubility curves across temperatures.

Differences in solubility curves allow one component to crystallize out of solution before another when cooled.

2

Analyze how enthalpy changes govern temperature dependence on solubility.

Endothermic processes absorb heat leading to increased solubility with temperature rise, whereas exothermic processes release heat leading to decreased solubility with temperature rise.

According to Le Chatelier's principle, adding heat favors the endothermic direction of a solution equilibrium.

3

Determine the saturation state definition for excess solute concentration.

A supersaturated solution contains a higher concentration of dissolved solute than a saturated solution at the specified temperature.

It represents a metastable condition created by careful cooling without crystallization.

Key Concept

Temperature effects on solubility, enthalpy of solution, and fractional crystallization

Question 8391Question →

A $17.2\text{ g}$ sample of hydrated calcium tetraoxosulfate(VI), $\text{CaSO}_4 \cdot x\text{H}_2\text{O}$ , is heated at $120^\circ\text{C}$ until it partially dehydrates, losing $2.70\text{ g}$ of water vapor to form plaster of Paris, $\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}$ . What is the value of $x$ , the number of molecules of water of crystallization per formula unit in the original hydrated salt? $[\text{Ca} = 40, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 2

Answer

The value of

x

in the hydrated salt formula is 2.

Applying mass conservation, the mass of plaster of Paris (

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

) remaining is

17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

, corresponding to

0.10\text{ mol}

. The mass of evolved water is

2.70\text{ g}

, which equals

0.15\text{ mol}

. The mole ratio of evolved water to salt formula units is

\frac{0.15}{0.10} = 1.5

. Since the reaction is

\text{CaSO}_4 \cdot x\text{H}_2\text{O} \rightarrow \text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} + (x - 0.5)\text{H}_2\text{O}

,

x - 0.5 = 1.5 \implies x = 2

.

Step-by-Step Solution

1

Calculate the molar masses of

\text{CaSO}_4

,

\text{H}_2\text{O}

, and the residue

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

.

Molar mass of

\text{CaSO}_4 = 40 + 32 + (4 \times 16) = 136\text{ g/mol}

;

\text{H}_2\text{O} = 18\text{ g/mol}

;

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = 136 + (0.5 \times 18) = 145\text{ g/mol}

.

Molar masses are necessary to perform mole calculations from mass measurements.

2

Determine the mass and amount in moles of plaster of Paris formed.

Mass of residue

= 17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

. Moles of

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = \frac{14.50\text{ g}}{145\text{ g/mol}} = 0.10\text{ mol}

.

Subtracting the mass of lost water gives the mass of solid product remaining.

3

Calculate the moles of water vapor driven off.

Moles of

\text{H}_2\text{O} = \frac{2.70\text{ g}}{18\text{ g/mol}} = 0.15\text{ mol}

.

Determining the quantity of lost water allows finding the mole ratio of lost water to salt units.

4

Relate the moles of lost water to the stoichiometry of partial dehydration to solve for

x

.

Moles of water lost per mole of salt

= \frac{0.15\text{ mol}}{0.10\text{ mol}} = 1.5\text{ mol}

. Since partial dehydration yields

(x - 0.5)

moles of lost water,

x - 0.5 = 1.5 \implies x = 2

.

Connecting empirical mole ratios to chemical formula coefficients yields the integer hydration number

x

.

Key Concept

Stoichiometric determination of water of crystallization from mass loss during partial dehydration.

Question 8392Question →

The standard reduction potentials for four half-cell reactions at $25^\circ\text{C}$ are given below:

1. $\text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s) \quad E^\circ = -0.40\text{ V}$
2. $\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \quad E^\circ = -0.13\text{ V}$
3. $\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\circ = +0.77\text{ V}$
4. $\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ = +0.80\text{ V}$

Which of the following reaction processes is non-spontaneous under standard conditions?

Oxidation of lead metal by aqueous cadmium ions

Oxidation of cadmium metal by aqueous lead ions

Oxidation of iron(II) ions by aqueous silver ions

Reduction of iron(III) ions by cadmium metal

Show answer & explanation

Answer: Oxidation of lead metal by aqueous cadmium ions

Answer

Oxidation of lead metal by aqueous cadmium ions is non-spontaneous because its standard cell potential is negative (

E^\circ_{\text{cell}} = -0.27\text{ V}

).

The reaction process involving the oxidation of lead metal by aqueous cadmium ions requires cadmium ions to act as the oxidizing agent (reduced at the cathode) and lead metal to act as the reducing agent (oxidized at the anode). Calculating the standard cell potential gives

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.40\text{ V} - (-0.13\text{ V}) = -0.27\text{ V}

. Because

E^\circ_{\text{cell}}

is negative, the reaction is non-spontaneous under standard conditions.

Step-by-Step Solution

1

Identify the reduction and oxidation half-reactions for the proposed reaction.

For the oxidation of lead metal by cadmium ions: Cathode (reduction):

\text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s)

, Anode (oxidation):

\text{Pb}(s) \rightarrow \text{Pb}^{2+}(aq) + 2e^-

.

Reaction spontaneity depends on the overall standard cell potential, calculated as

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

.

2

Substitute the standard reduction potentials into the cell potential formula.

E^\circ_{\text{cell}} = E^\circ(\text{Cd}^{2+}/\text{Cd}) - E^\circ(\text{Pb}^{2+}/\text{Pb}) = -0.40\text{ V} - (-0.13\text{ V}) = -0.27\text{ V}

.

The cathode potential is the reduction potential of the species being reduced, and the anode potential is the reduction potential of the species being oxidized.

3

Evaluate reaction spontaneity based on the sign of

E^\circ_{\text{cell}}

.

Since

E^\circ_{\text{cell}} = -0.27\text{ V} < 0

, this reaction cannot occur spontaneously under standard conditions.

A chemical reaction is spontaneous under standard conditions if and only if

E^\circ_{\text{cell}} > 0

.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Question 8393Question →

Match each of the following chemical species on the left with its corresponding central atom hybridization state and molecular geometry on the right, as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. Which set of pairs accurately connects each species to its geometry and hybridization?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Iodine trifluoride (

IF_3

) matches with

sp^3d

hybridization and T-shaped geometry; Pentafluoroxenate ion (

XeF_5^-

) matches with

sp^3d^3

hybridization and pentagonal planar geometry; Sulfite ion (

SO_3^{2-}

) matches with

sp^3

hybridization and trigonal pyramidal geometry; Nitronium ion (

NO_2^+

) matches with

sp

hybridization and linear geometry.

Each chemical species is correctly paired according to its total steric number (sum of bonding electron domains and non-bonding lone pairs).

IF_3

has steric number 5 (

sp^3d

, T-shaped),

XeF_5^-

has steric number 7 (

sp^3d^3

, pentagonal planar),

SO_3^{2-}

has steric number 4 (

sp^3

, trigonal pyramidal), and

NO_2^+

has steric number 2 (

sp

, linear).

Step-by-Step Solution

1

Determine valence electron count and steric number for

IF_3

Iodine has 7 valence electrons + 3 from fluorine = 10 electrons (5 pairs). Steric number = 5 (3 bonding pairs, 2 lone pairs).

Steric number 5 corresponds to

sp^3d

hybridization. Two equatorial lone pairs force the 3 terminal fluorines into a T-shaped arrangement.

2

Determine valence electron count and steric number for

XeF_5^-

Xenon has 8 valence electrons + 5 from fluorine + 1 from overall negative charge = 14 electrons (7 pairs). Steric number = 7 (5 bonding pairs, 2 lone pairs).

Steric number 7 corresponds to

sp^3d^3

hybridization. The two lone pairs occupy axial positions above and below the equatorial plane, creating a pentagonal planar molecular shape.

3

Determine valence electron count and steric number for

SO_3^{2-}

Sulfur has 6 valence electrons + 2 from charge = 8 valence shell electrons. It forms 3 sigma bonds with oxygen and retains 1 lone pair. Steric number = 4.

Steric number 4 corresponds to

sp^3

hybridization. Three bonding domains and 1 lone pair produce a trigonal pyramidal molecular shape.

4

Determine valence electron count and steric number for

NO_2^+

Nitrogen has 5 valence electrons - 1 from positive charge = 4 electrons. It forms two double bonds with oxygen atoms and has 0 lone pairs. Steric number = 2.

Steric number 2 corresponds to

sp

hybridization, resulting in a linear geometry with a

180^\circ

bond angle.

Key Concept

VSEPR Theory, Steric Numbers, and Central Atom Hybridization States

Estimated Time:2m 30s

Question 8394Question →

A weather balloon is launched containing $3.20\text{ dm}^3$ of helium gas at an initial temperature of $47^\circ\text{C}$ and a pressure of $1.50\text{ atm}$ . As the balloon rises into the atmosphere, the temperature drops to $-23^\circ\text{C}$ and the pressure decreases to $0.60\text{ atm}$ . Calculate the final volume of the helium gas in $\text{dm}^3$ .

Show answer & explanation

Answer: 6.25

Answer

The final volume of the helium gas is

6.25\text{ dm}^3

.

Converting the given temperatures to Kelvin yields

T_1 = 47 + 273 = 320\text{ K}

and

T_2 = -23 + 273 = 250\text{ K}

. Applying the General Gas Law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

and rearranging for

V_2

gives

V_2 = \frac{1.50 \times 3.20 \times 250}{0.60 \times 320} = 6.25\text{ dm}^3

.

Step-by-Step Solution

1

Convert both initial and final temperatures from Celsius to Kelvin.

T_1 = 47^\circ\text{C} + 273 = 320\text{ K}

and

T_2 = -23^\circ\text{C} + 273 = 250\text{ K}

.

Gas laws require absolute temperature units (Kelvin).

2

State the General Gas Law equation.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Relates initial and final values of pressure, volume, and temperature for a fixed mass of gas.

3

Rearrange the equation to solve for the final volume (

V_2

).

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}

Isolates the target variable on one side.

4

Substitute the values into the equation and compute the result.

V_2 = \frac{1.50\text{ atm} \times 3.20\text{ dm}^3 \times 250\text{ K}}{0.60\text{ atm} \times 320\text{ K}} = 6.25\text{ dm}^3

Yields the exact final volume of the gas.

Key Concept

General Gas Law

Question 8395Question →

During the chromatographic analysis of an ink sample, a component spot moves a distance of $3.0\text{ cm}$ from the origin line while the solvent front advances $10.0\text{ cm}$ . What is the retardation factor ( $R_f$ ) for this component?

0.30

3.33

0.70

3.00

Show answer & explanation

Answer: $0.30$

Answer

The retardation factor (

R_f

) for the component is

0.30

.

The retardation factor (

R_f

) is calculated by dividing the distance moved by the solute spot (

3.0\text{ cm}

) by the distance moved by the solvent front (

10.0\text{ cm}

), giving

0.30

.

Step-by-Step Solution

1

Identify the distance traveled by the component spot and the distance traveled by the solvent front.

Distance moved by solute spot =

3.0\text{ cm}

; Distance moved by solvent front =

10.0\text{ cm}

.

These values are required to calculate the retardation factor.

2

Apply the retardation factor formula:

R_f = \frac{\text{Distance moved by solute spot}}{\text{Distance moved by solvent front}}

.

R_f = \frac{3.0\text{ cm}}{10.0\text{ cm}} = 0.30

.

The

R_f

value represents the relative movement of the solute with respect to the mobile phase.

Key Concept

Calculation of Retardation Factor (Rf) in paper chromatography

Estimated Time:45s

Question 8396Question →

A

10.0\text{ g}

sample of impure limestone containing calcium trioxocarbonate(IV),

\text{CaCO}_3

, was strongly heated until decomposition was complete according to the equation:

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

If

1.792\text{ dm}^3

of carbon(IV) oxide gas was evolved at s.t.p., what is the percentage purity of the limestone sample?

[\text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

44.8\%

74.7\%

80.0\%

89.6\%

Show answer & explanation

Answer: $80.0\%$

Answer

The percentage purity of the limestone sample is 80.0%.

From the balanced chemical equation, 1 mole of calcium trioxocarbonate(IV) produces 1 mole of carbon(IV) oxide gas. At s.t.p.,

1.792\text{ dm}^3

of

\text{CO}_2

equals

\frac{1.792}{22.4} = 0.08\text{ mol}

. Consequently,

0.08\text{ mol}

of pure

\text{CaCO}_3

was present. With a molar mass of

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

, the mass of pure

\text{CaCO}_3

is

8.00\text{ g}

. Dividing this pure mass by the

10.0\text{ g}

total sample mass yields a purity of

80.0\%

.

Step-by-Step Solution

1

Calculate the number of moles of carbon(IV) oxide gas evolved at s.t.p.

\text{Moles of CO}_2 = \frac{\text{Volume at s.t.p.}}{\text{Molar volume at s.t.p.}} = \frac{1.792\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.08\text{ mol}

Gas volume at standard temperature and pressure directly determines mole quantity using standard molar gas volume.

2

Determine the molar mass of pure calcium trioxocarbonate(IV),

\text{CaCO}_3

.

\text{Molar mass of CaCO}_3 = 40 + 12 + (3 \times 16) = 100\text{ g mol}^{-1}

Required to convert moles of pure reactant into mass.

3

Use the mole ratio from the balanced chemical equation to find the mass of pure

\text{CaCO}_3

.

Mole ratio of

\text{CaCO}_3 : \text{CO}_2 = 1 : 1

. Moles of pure

\text{CaCO}_3 = 0.08\text{ mol}

. Mass of pure

\text{CaCO}_3 = 0.08\text{ mol} \times 100\text{ g mol}^{-1} = 8.00\text{ g}

.

Stoichiometry dictates that 1 mole of

\text{CaCO}_3

produces 1 mole of

\text{CO}_2

upon complete decomposition.

4

Calculate the percentage purity of the limestone sample.

\text{Percentage purity} = \left( \frac{\text{Mass of pure CaCO}_3}{\text{Mass of impure sample}} \right) \times 100 = \left( \frac{8.00\text{ g}}{10.0\text{ g}} \right) \times 100 = 80.0\%

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

Key Concept

Percentage Purity and Stoichiometry from Gas Volumes

Estimated Time:1m 30s

Question 8397Question →

In a gas counter-diffusion experiment using a horizontal glass tube of length $100\text{ cm}$ , gas $A$ with a molar mass of $36\text{ g/mol}$ is released from end $P$ , while an unknown gas $B$ is released simultaneously from end $Q$ under identical conditions of temperature and pressure. The two gases meet and form a visible reaction ring at a distance of $40\text{ cm}$ from end $P$ . What is the molar mass of gas $B$ in $\text{g/mol}$ ?

Show answer & explanation

Answer: 16

Answer

16 g/mol

According to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r_A / r_B = \sqrt{M_B / M_A}

). Because both gases diffuse over the same duration, the distance ratio equals the rate ratio (

d_A / d_B = 40 / 60 = 2/3

). Substituting

M_A = 36\text{ g/mol}

into

2/3 = \sqrt{M_B / 36}

and squaring both sides gives

4/9 = M_B / 36

, which yields

M_B = 16\text{ g/mol}

.

Step-by-Step Solution

1

Calculate the distance traveled by gas B

d_B = 100 cm - 40 cm = 60 cm

The total length of the diffusion tube is 100 cm, and gas A traveled 40 cm from end P before meeting gas B.

2

Determine the ratio of the rates of diffusion of gas A and gas B

r_A / r_B = 40 / 60 = 2/3

The rate of diffusion is directly proportional to the distance traveled in a given time period.

3

Apply Graham's Law of Diffusion relating diffusion rates to molar masses

2/3 = sqrt(M_B / 36)

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

4

Solve for the unknown molar mass M_B

4/9 = M_B / 36 => M_B = 16 g/mol

Squaring both sides of the equation eliminates the square root, allowing straightforward algebraic solution.

Key Concept

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r_1 / r_2 = \sqrt{M_2 / M_1}

).

Question 8398Question →

Complete the statement regarding the transition metal complex ion $[Fe(CN)_6]^{3-}$ by filling in the missing values.

Fill in the blanks below

In the hexacyanoferrate(III) complex ion,

[Fe(CN)_6]^{3-}

, the coordination number of the central iron ion is and the oxidation state of iron is .

Show answer & explanation

Answer

The coordination number of the central iron ion in

[Fe(CN)_6]^{3-}

is 6, and its oxidation state is +3.

In

[Fe(CN)_6]^{3-}

, iron is bonded to six cyano (

CN^-

) ligands, establishing a coordination number of 6. Balancing the total charge gives

x + 6(-1) = -3

, yielding an oxidation state of +3 for the iron central metal ion.

Step-by-Step Solution

1

Determine the coordination number from the formula of the complex ion.

The central iron ion is bonded to 6 monodentate cyano (

CN^-

) ligands, so the coordination number is 6.

The coordination number represents the total number of ligand donor atoms attached directly to the central transition metal ion.

2

Calculate the oxidation state of the central iron ion.

Let

x

be the oxidation state of

Fe

. Each cyano ligand has a charge of

-1

, and the total complex ion charge is

-3

. Setting up the equation:

x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3

.

The sum of the oxidation state of the central metal and the charges of all ligands equals the net charge of the complex ion.

Key Concept

Coordination Number and Oxidation State Determination in Complex Ions

Question 8399Question →

Complete the statement below regarding the characteristic laboratory test observation and chemical role of acidified potassium tetraoxomanganate(VII) when reacted with iron(II) tetraoxosulfate(VI).

Fill in the blanks below

When acidified potassium tetraoxomanganate(VII) solution is added to a solution containing iron(II) tetraoxosulfate(VI), the purple color of the permanganate solution turns , because the permanganate ion acts as an agent.

Show answer & explanation

Answer

The purple solution turns colorless because potassium tetraoxomanganate(VII) acts as an oxidizing agent.

Acidified potassium tetraoxomanganate(VII) contains the intensely purple

MnO_4^-

ion. Upon reacting with a reducing agent like

Fe^{2+}

, the

MnO_4^-

ion accepts electrons and is reduced to the colorless

Mn^{2+}

ion. Because it removes electrons from iron(II) ions and oxidizes them to iron(III), potassium tetraoxomanganate(VII) functions as an oxidizing agent.

Step-by-Step Solution

1

Determine the color change of acidified potassium tetraoxomanganate(VII) when reacted with a reducing agent such as iron(II) ions.

The manganese in

MnO_4^-

(purple) is reduced to

Mn^{2+}

, which is colorless in dilute aqueous solution.

Permanganate ions undergo reduction from an oxidation state of +7 to +2 in acidic media.

2

Identify the chemical role of potassium tetraoxomanganate(VII) in this redox reaction.

Potassium tetraoxomanganate(VII) accepts electrons from iron(II) ions, converting

Fe^{2+}

to

Fe^{3+}

, making it an oxidizing agent.

A chemical species that causes another substance to be oxidized while undergoing reduction itself is defined as an oxidizing agent.

Key Concept

Laboratory test for reducing agents using acidified potassium tetraoxomanganate(VII)

Question 8400Question →

Under the influence of ultraviolet radiation in the stratosphere, chlorofluorocarbons (CFCs) undergo photolysis to yield reactive chlorine free radicals ( $\text{Cl}^\bullet$ ). Which of the following chemical equations correctly represents the propagation step in which the chlorine radical directly reacts with ozone to cause its depletion?

\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2

\text{CF}_2\text{Cl}_2 + h\nu \rightarrow \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet

\text{CO}_2 + \text{O}_3 \rightarrow \text{CO}_3 + \text{O}_2

\text{CO}_2 + h\nu \rightarrow \text{CO} + \text{O}^\bullet

Show answer & explanation

Answer: $\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2$

Answer

\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2

In the catalytic cycle of ozone destruction, the reactive chlorine free radical (

\text{Cl}^\bullet

) collides with an ozone molecule (

\text{O}_3

), abstracts an oxygen atom to form a chlorine monoxide radical (

\text{ClO}^\bullet

), and releases diatomic oxygen (

\text{O}_2

).

Step-by-Step Solution

1

Identify the initiation reaction of CFCs in the stratosphere

UV radiation breaks a C-Cl bond in CFCs:

\text{CF}_2\text{Cl}_2 + h\nu \rightarrow \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet

.

High-energy UV photons supply sufficient energy to homolytically cleave the carbon-chlorine bond.

2

Determine the propagation step involving ozone

The produced chlorine radical acts as a homogeneous catalyst, reacting with ozone:

\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2

.

This specific step directly consumes atmospheric ozone while maintaining the catalytic cycle.

Key Concept

Catalytic mechanism of stratospheric ozone depletion by chlorine free radicals

Estimated Time:1m 0s

Answer

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