All practice questions

Question 6841Question →

Sequence the evolutionary stages of exchange media in Commerce, ordering them from the earliest historical form of money to the most modern development.

Drag items to arrange them in the correct order

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Answer

The correct historical sequence from earliest to most modern form of money is: Commodity Money → Metallic Money → Paper Currency → Electronic Money.

Money evolved chronologically in response to commercial needs: first came Commodity Money (useful physical goods), followed by Metallic Money (durable minted coins), then Paper Currency (convenient representation of stored value), and finally Electronic Money (cashless digital balances).

Step-by-Step Solution

1

Identify the earliest exchange medium

Commodity money (cowries, cattle) was used immediately following the collapse/limitations of trade by barter.

Intrinsic value and direct utility made raw commodities the initial universal medium.

2

Determine the second historical phase

Metallic money (precious metal coins) replaced bulky and perishable commodities.

Metals like gold and silver offered durability, uniformity, and easy divisibility.

3

Determine the third historical phase

Paper currency developed as goldsmith receipts gained widespread acceptance in commerce.

Carrying heavy coins was risky, leading traders to use paper receipts which eventually became state fiat money.

4

Identify the latest monetary innovation

Electronic money emerged with modern computing infrastructure.

Digital balances and card-based transactions allow cashless commerce without physical banknotes.

Key Concept

Historical Evolution of Forms of Money

Question 6842Question →

A sample of gas collected over water at $27^\circ\text{C}$ occupies a volume of $250\text{ cm}^3$ at a total pressure of $750\text{ mmHg}$ . If the saturated vapor pressure of water at $27^\circ\text{C}$ is $30\text{ mmHg}$ , what is the volume of the dry gas at standard temperature and pressure (STP)?

215.5\text{ cm}^3

224.5\text{ cm}^3

236.8\text{ cm}^3

233.5\text{ cm}^3

Show answer & explanation

Answer: $215.5\text{ cm}^3$

Answer

The volume of the dry gas at STP is

215.5\text{ cm}^3

.

Subtracting the aqueous tension (

30\text{ mmHg}

) from the total pressure (

750\text{ mmHg}

) yields the true dry gas pressure (

720\text{ mmHg}

). Substituting this along with initial temperature (

300\text{ K}

) and STP conditions (

760\text{ mmHg}

and

273\text{ K}

) into the combined gas equation

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

gives

V_2 = 215.5\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the partial pressure of the dry gas

P_{\text{dry}} = P_{\text{total}} - P_{\text{water}} = 750\text{ mmHg} - 30\text{ mmHg} = 720\text{ mmHg}

According to Dalton's Law of Partial Pressures, the total pressure of a gas collected over water includes the pressure of the dry gas and the saturated vapor pressure of water.

2

Convert temperatures to Kelvin

T_1 = 27 + 273 = 300\text{ K}

,

T_2 = 273\text{ K}

(STP)

Gas law equations require absolute temperature units in Kelvin.

3

Apply the general gas law equation to solve for the volume at STP (

V_2

)

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{720 \times 250 \times 273}{760 \times 300} = \frac{49,140,000}{228,000} = 215.5\text{ cm}^3

Standard conditions (STP) are

P_2 = 760\text{ mmHg}

and

T_2 = 273\text{ K}

.

Key Concept

Dalton's Law of Partial Pressures and Collection of Gas over Water

Estimated Time:1m 30s

Question 6843Question →

A timber logging company in Ondo State intends to export a massive volume of heavy, unprocessed logs to a buyer in Liverpool, England. Given that the cargo is heavy, non-perishable, and has a low value per unit weight, which mode of transportation is most economical for this commercial transaction?

Sea transport

Air transport

Road transport

Pipeline transport

Show answer & explanation

Answer: Sea transport

Answer

Sea transport is the most economical mode for transporting heavy, low-value bulk timber internationally.

Sea transport offers huge carrying capacity and low freight charges, making it the most cost-effective mode for long-distance international shipment of heavy, low-value, non-perishable goods like timber.

Step-by-Step Solution

1

Analyze the cargo characteristics

The cargo consists of unprocessed timber logs, which are heavy, bulky, non-perishable, and low in value per unit weight.

Cargo weight, volume, urgency, and unit value dictate the choice of transport mode.

2

Evaluate transport mode economics and physical suitability

Sea transport provides massive carry capacity and the lowest cost per tonne-kilometer for international long-distance bulk trade.

Ships excel at long-haul heavy freight where speed is not the critical factor.

Key Concept

Selection of Transport Modes Based on Cargo Characteristics and Economic Functions

Estimated Time:1m 0s

Question 6844Question →

In the laboratory preparation of zinc tetraoxosulfate(VI) crystals ( $ZnSO_4 \cdot 7H_2O$ ) from dilute tetraoxosulfate(VI) acid ( $H_2SO_4$ ) and zinc metal, why is excess zinc metal added to the acid?

To ensure all the acid is completely reacted so that the salt solution is not contaminated with acid

To prevent the zinc tetraoxosulfate(VI) salt from undergoing hydrolysis in the aqueous solution

To act as a heterogeneous catalyst to increase the rate of crystallization

To convert the resulting normal salt into a basic salt by excess metal accumulation

Show answer & explanation

Answer: To ensure all the acid is completely reacted so that the salt solution is not contaminated with acid

Answer

Excess zinc metal is added to ensure all the acid is completely reacted so that the salt solution is not contaminated with acid.

When preparing a soluble salt from an acid and an insoluble solid (such as a metal, oxide, or carbonate), the solid is added in excess to ensure that every molecule of acid reacts completely. Because both the acid and the formed salt are soluble in water, any unreacted acid would remain mixed with the salt during crystallization and contaminate the product. The excess solid metal is easily removed by filtration before evaporating the filtrate.

Step-by-Step Solution

1

Identify the type of reaction and salt being prepared

Zinc tetraoxosulfate(VI) (

ZnSO_4

) is a soluble salt prepared by reacting a dilute acid (

H_2SO_4

) with a moderately reactive insoluble metal (

Zn

).

Soluble salts of reactive metals are prepared by acid-metal reactions.

2

Analyze the separation and purification step

If acid remains in excess, it cannot be separated from the soluble salt solution by filtration or simple evaporation.

Both the salt and acid are soluble in water, so unreacted acid would contaminate the final crystalline product.

3

Determine the role of adding excess insoluble solid reactant

Adding excess zinc metal guarantees 100% consumption of the acid. The excess insoluble zinc can easily be filtered off, leaving a pure aqueous salt solution.

Insoluble solids are easy to separate from solutions via filtration.

Key Concept

Method of preparing soluble salts by reacting dilute acid with an excess of an insoluble metal, base, or trioxocarbonate(IV).

Estimated Time:1m 0s

Question 6845Question →

Arrange the following atomic subshells in order of increasing energy according to the Aufbau principle and the $(n+l)$ rule, starting with the subshell of lowest energy:

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Answer

The correct order of subshells from lowest to highest energy is

5s \rightarrow 4d \rightarrow 5p \rightarrow 4f

.

According to the Aufbau principle and the

(n+l)

rule, subshells fill in order of increasing

(n+l)

value.

5s

has

(n+l) = 5+0 = 5

, making it lowest in energy. Both

4d

(

(n+l) = 4+2 = 6

) and

5p

(

(n+l) = 5+1 = 6

) have a sum of

6

, but

4d

has lower energy than

5p

because its principal quantum number

n=4

is smaller.

4f

has

(n+l) = 4+3 = 7

, placing it highest in energy. Thus, the correct sequence is

5s \rightarrow 4d \rightarrow 5p \rightarrow 4f

.

Step-by-Step Solution

1

Calculate the

(n+l)

value for each atomic subshell

For

5s

:

n=5, l=0 \implies n+l = 5

.
For

4d

:

n=4, l=2 \implies n+l = 6

.
For

5p

:

n=5, l=1 \implies n+l = 6

.
For

4f

:

n=4, l=3 \implies n+l = 7

.

According to Madelung's rule, orbitals fill in order of increasing

(n+l)

values.

2

Order subshells by increasing

(n+l)

sum

5s

(

(n+l)=5

) is lowest in energy, while

4f

(

(n+l)=7

) is highest.

Subshells with smaller

(n+l)

values are filled before those with larger

(n+l)

values.

3

Break ties for subshells with identical

(n+l)

values (

4d

and

5p

)

4d

(

n=4

) has lower energy than

5p

(

n=5

).

When two subshells share the same

(n+l)

value, the subshell with the smaller principal quantum number

n

is lower in energy.

4

Assemble the complete sequence from lowest to highest energy

5s < 4d < 5p < 4f

Combines the

(n+l)

rule and the tie-breaking principal quantum number rule.

Key Concept

Aufbau Principle and the (n+l) Rule

Question 6846Question →

Under identical conditions of temperature and pressure, oxygen gas ( $O_2$ ) diffuses through a fine orifice at a rate of $15\text{ cm}^3/\text{s}$ . What is the rate of diffusion of hydrogen gas ( $H_2$ ) in $\text{cm}^3/\text{s}$ under the same conditions? (Relative atomic masses: $H = 1$ , $O = 16$ )

Show answer & explanation

Answer: 60; 60 cm^3/s; 60 cm³/s; 60 cm3/s

Answer

60 cm³/s

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}}

). Molar mass of

O_2 = 32\text{ g/mol}

and molar mass of

H_2 = 2\text{ g/mol}

. The ratio

\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

. Therefore,

r_{H_2} = 4 \times 15\text{ cm}^3/\text{s} = 60\text{ cm}^3/\text{s}

.

Step-by-Step Solution

1

Calculate the molar masses of oxygen (

O_2

) and hydrogen (

H_2

).

M(O_2) = 2 \times 16 = 32\text{ g/mol}

and

M(H_2) = 2 \times 1 = 2\text{ g/mol}

.

Graham's law relates rate of diffusion directly to the inverse square root of molar mass.

2

Apply Graham's Law of Diffusion ratio formula:

\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M(O_2)}{M(H_2)}}

.

\frac{r_{H_2}}{15} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

.

The lighter gas diffuses faster by a factor equal to the square root of the ratio of their molar masses.

3

Solve for the rate of diffusion of hydrogen (

r_{H_2}

).

r_{H_2} = 4 \times 15\text{ cm}^3/\text{s} = 60\text{ cm}^3/\text{s}

.

Multiply the rate of diffusion of oxygen by the calculated relative rate ratio.

Key Concept

Graham's Law of Diffusion

Estimated Time:1m 0s

Question 6847Question →

Match each type of warehouse used in commercial trade with its corresponding primary operational function or feature.

Click a left item, then click its matching right item

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Answer

Bonded Warehouse matches with storing imported goods on which customs duties are unpaid; Public Warehouse matches with providing rentable storage space open to the public; Private Warehouse matches with being owned and operated exclusively by a single producer or trader; State Warehouse matches with storing confiscated goods, uncleared imports, or government strategic reserves.

Each warehouse type is correctly matched according to commercial law and trade practice: Bonded warehouses handle uncleared dutiable goods under customs bond; Public warehouses offer commercial storage rental to the public; Private warehouses serve individual firms exclusively; and State warehouses store confiscated items or government emergency stocks.

Step-by-Step Solution

1

Identify the primary regulatory characteristic of a Bonded Warehouse.

It is controlled by customs authorities to hold dutiable imported goods prior to duty payment.

Bonded warehouses serve customs enforcement and entrepôt trade facilitation.

2

Distinguish between Public and Private Warehouses.

Public warehouses are open for rent to anyone, while private warehouses are exclusively owned and used by a single company.

Ownership structure and commercial access distinguish public from private facilities.

3

Determine the function of a State Warehouse.

It holds seized contraband, unclaimed cargo, or state strategic reserves managed by public port or customs authorities.

State warehouses operate under government administration rather than commercial profit motives.

Key Concept

Warehousing Types and Functions

Question 6848Question →

Match each aqueous salt solution ( $0.1\text{ mol dm}^{-3}$ ) to its corresponding pH nature resulting from salt hydrolysis.

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Answer

Potassium ethanoate matches with Alkaline solution (pH > 7) due to anion hydrolysis; Ammonium chloride matches with Acidic solution (pH < 7) due to cation hydrolysis; Sodium chloride matches with Neutral solution (pH = 7) with negligible hydrolysis; Ammonium ethanoate matches with Neutral solution (pH ≈ 7) due to equal mutual hydrolysis of both ions.

Potassium ethanoate (

CH_3COOK

) undergoes anion hydrolysis to yield hydroxide ions, producing an alkaline solution (

pH > 7

). Ammonium chloride (

NH_4Cl

) undergoes cation hydrolysis to yield hydroxonium ions, producing an acidic solution (

pH < 7

). Sodium chloride (

NaCl

) contains spectator ions from a strong acid and strong base, leading to negligible hydrolysis and a neutral solution (

pH = 7

). Ammonium ethanoate (

CH_3COONH_4

) undergoes mutual hydrolysis of both cation and anion to equal extents, resulting in a neutral solution (

pH \approx 7

).

Step-by-Step Solution

1

Classify each salt based on the strengths of its parent acid and parent base.

Potassium ethanoate comes from a weak acid (

CH_3COOH

) and strong base (

KOH

). Ammonium chloride comes from a weak base (

NH_3

) and strong acid (

HCl

). Sodium chloride comes from a strong acid (

HCl

) and strong base (

NaOH

). Ammonium ethanoate comes from a weak acid (

CH_3COOH

) and weak base (

NH_3

).

Only ions originating from weak electrolytes undergo significant reaction with water (hydrolysis).

2

Determine the hydrolyzing species and the resulting ion produced in water.

For

CH_3COOK

,

CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-

(alkaline). For

NH_4Cl

,

NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+

(acidic). For

NaCl

, no hydrolysis occurs. For

CH_3COONH_4

, both ions hydrolyze equally.

Anion hydrolysis increases

OH^-

concentration, whereas cation hydrolysis increases

H_3O^+

concentration.

3

Correlate each salt to its solution pH characteristics.

CH_3COOK \rightarrow pH > 7

,

NH_4Cl \rightarrow pH < 7

,

NaCl \rightarrow pH = 7

, and

CH_3COONH_4 \rightarrow pH \approx 7

.

The nature of the solution depends on which ion hydrolyzes or whether both hydrolyze to equal extents.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Question 6849Question →

A consignment of imported goods remains unclaimed at the port beyond the statutory grace period and is subsequently impounded by customs authorities. In which facility are such goods deposited until customs duties are settled or the items are auctioned?

State warehouse

Bonded warehouse

Public warehouse

Private warehouse

Show answer & explanation

Answer: State warehouse

Answer

The facility where unclaimed or customs-impounded goods are deposited is the State warehouse.

State warehouses are government-owned facilities controlled by customs authorities (such as the Nigeria Customs Service) to keep custody of seized contraband, forfeited cargo, or goods left unclaimed beyond legal port clearance deadlines.

Step-by-Step Solution

1

Analyze the status of the goods in the scenario.

The goods are unclaimed imported merchandise impounded by government customs officials.

Commercial warehouses serve voluntary private traders, whereas government-seized goods fall under customs enforcement authority.

2

Identify the warehouse specifically operated by the government for impounded or unclaimed cargo.

State warehouses (or Customs Warehouses) are designated by law to store seized, contraband, or overstayed port goods pending auction or release.

This separates government statutory control from commercial bonded, public, or private warehousing functions.

Key Concept

Functions and features of a State Warehouse in customs administration

Estimated Time:1m 0s

Question 6850Question →

Element $Q$ occurs naturally as two isotopes, $^{35}Q$ and $^{37}Q$ . If the relative atomic mass of element $Q$ is $35.5$ , what is the percentage abundance of the heavier isotope $^{37}Q$ ?

25%

75%

50%

35.5%

Show answer & explanation

Answer: 25%

Answer

The percentage abundance of

^{37}Q

is 25%.

The relative atomic mass (

35.5

) lies between

35

and

37

, closer to

35

. Setting up the weighted average equation

35.5 = \frac{35(100 - x) + 37x}{100}

yields

x = 25\%

for

^{37}Q

, while

^{35}Q

accounts for the remaining

75\%

.

Step-by-Step Solution

1

Set up the relative atomic mass equation using percentage abundances.

Let the percentage abundance of

^{37}Q

be

x\%

. Therefore, the abundance of

^{35}Q

is

(100 - x)\%

.

The sum of natural abundances for all isotopes of an element must equal 100%.

2

Substitute the isotopic mass numbers and relative atomic mass into the weighted average formula.

35.5 = \frac{35(100 - x) + 37x}{100}

Relative atomic mass is calculated as the weighted average of the mass numbers of naturally occurring isotopes.

3

Solve the linear equation for

x

.

3550 = 3500 - 35x + 37x \implies 50 = 2x \implies x = 25

Multiplying by 100 and rearranging terms isolates

x = 25

.

Key Concept

Calculation of isotopic abundances from relative atomic mass

Question 6851Question →

Real gases such as carbon dioxide ( $\text{CO}_2$ ) exhibit maximum deviation from ideal gas behavior under conditions of high pressure and low temperature because intermolecular attractive forces become significant and the physical volume occupied by gas molecules is no longer negligible.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. Real gases deviate most significantly from ideal behavior under conditions of high pressure and low temperature.

At low temperatures, gas molecules move slowly enough for intermolecular attractive forces to pull them together, reducing wall collision pressure. At high pressures, gas particles are forced closely together, meaning their individual molecular volumes are no longer negligible compared to the total volume occupied by the gas.

Step-by-Step Solution

1

Recall the fundamental assumptions of the kinetic theory for ideal gases.

An ideal gas assumes that gas molecules have negligible volume and experience no intermolecular forces of attraction or repulsion.

These assumptions allow the ideal gas equation (

PV = nRT

) to accurately predict gas behavior.

2

Analyze how high pressure affects real gas behavior.

High pressure compresses the gas into a smaller volume, making the actual physical volume occupied by the gas molecules a significant fraction of the container volume.

The volume term (

V

) in the ideal gas equation must be corrected using the van der Waals volume parameter (

b

).

3

Analyze how low temperature affects real gas behavior.

Low temperature decreases the average kinetic energy of gas molecules, allowing intermolecular attractive forces to pull molecules together.

Attractive forces reduce the force of impact against container walls, causing observed pressure to be lower than ideal pressure.

Key Concept

Conditions Causing Real Gas Deviation from Ideality

Question 6852Question →

Match each hydration phenomenon or property on the left with its corresponding vapor pressure behavior or physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Deliquescence matches with the condition where the saturated solution's vapor pressure is lower than ambient water vapor pressure, causing complete dissolution; Efflorescence matches with the condition where the crystal's vapor pressure exceeds ambient water vapor pressure, causing loss of hydration water; Hygroscopy matches with the absorption of moisture without dissolving; Water of Crystallization matches with the definite stoichiometric water molecules bound in the crystal lattice.

Each phenomenon is correctly paired based on its vapor pressure relationship and physical characteristics: Deliquescence requires the vapor pressure of the saturated solution to be lower than ambient water vapor pressure, causing liquefaction. Efflorescence requires the crystal's vapor pressure to exceed ambient water vapor pressure, leading to water loss. Hygroscopy involves moisture absorption without dissolving, and water of crystallization represents the fixed stoichiometric water bound in the crystal lattice.

Step-by-Step Solution

1

Analyze the thermodynamic driving force for deliquescence.

A substance deliquesces when it absorbs water from the air until it dissolves. This occurs because the saturated solution formed has a vapor pressure lower than the partial pressure of water vapor in the atmosphere.

Water spontaneously condenses into a system with lower vapor pressure.

2

Analyze the thermodynamic driving force for efflorescence.

A hydrated salt effloresces when its internal hydration vapor pressure is greater than the atmospheric water vapor pressure, causing water molecules to escape into the atmosphere.

Water leaves the crystal lattice when the internal vapor pressure is higher than ambient humidity.

3

Distinguish hygroscopy from deliquescence.

Hygroscopic substances absorb moisture from the atmosphere, but unlike deliquescent substances, they do not dissolve to form liquid solutions.

Hygroscopic materials absorb moisture into their bulk or surface without forming a solution.

4

Define water of crystallization.

Water of crystallization refers to water molecules chemically trapped in a fixed mole ratio inside the crystal lattice of salts.

It determines the specific hydrated chemical formula of the salt.

Key Concept

Atmospheric behavior of hydrated and anhydrous salts based on vapor pressure relationships.

Question 6853Question →

The solubility of a sparingly soluble salt $MX_2$ (molar mass $= 200\text{ g mol}^{-1}$ ) in water at $25^\circ\text{C}$ is $0.20\text{ g dm}^{-3}$ . What is the solubility product ( $K_{sp}$ ) of $MX_2$ at this temperature?

1.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

2.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

3.2 \times 10^{-2}\text{ mol}^3\text{ dm}^{-9}

Show answer & explanation

Answer: $4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}$

Answer

The solubility product (

K_{sp}

) of

MX_2

at

25^\circ\text{C}

is

4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

.

To find the solubility product (

K_{sp}

), first convert the solubility from

\text{g dm}^{-3}

to molar solubility (

s

) in

\text{mol dm}^{-3}

by dividing by the molar mass:

s = \frac{0.20}{200} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. The dissociation equation

MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq)

yields

[M^{2+}] = s

and

[X^-] = 2s

. Substituting these into the solubility product expression gives

K_{sp} = [M^{2+}][X^-]^2 = 4s^3 = 4 \times (1.0 \times 10^{-3})^3 = 4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

.

Step-by-Step Solution

1

Convert solubility from

\text{g dm}^{-3}

to molar solubility (

s

) in

\text{mol dm}^{-3}

s = \frac{\text{Solubility in g dm}^{-3}}{\text{Molar Mass}} = \frac{0.20\text{ g dm}^{-3}}{200\text{ g mol}^{-1}} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

Solubility product calculations require concentration units in

\text{mol dm}^{-3}

.

2

Write the ionic dissociation equation and express equilibrium concentrations

MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq)

[M^{2+}] = s = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

[X^-] = 2s = 2.0 \times 10^{-3}\text{ mol dm}^{-3}

Each mole of

MX_2

yields 1 mole of

M^{2+}

and 2 moles of

X^-

upon dissolution.

3

Write the

K_{sp}

expression and calculate the numerical value

K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3 = 4 \times (1.0 \times 10^{-3})^3 = 4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

Substitute the molar equilibrium concentrations into the equilibrium constant expression.

Key Concept

Relationship between Molar Solubility and Solubility Product (

K_{sp}

)

Question 6854Question →

A naturally occurring sample of neon gas consists of $90\%$ $^{20}\text{Ne}$ and $10\%$ $^{22}\text{Ne}$ . What is the relative atomic mass of neon in this sample?

Show answer & explanation

Answer: 20.2

Answer

The relative atomic mass of neon in the sample is 20.2.

The relative atomic mass of an element is calculated by summing the products of the mass number of each isotope and its fractional abundance:

\text{RAM} = (20 \times 0.90) + (22 \times 0.10) = 18.0 + 2.2 = 20.2

.

Step-by-Step Solution

1

Calculate the weighted contribution of

^{20}\text{Ne}

20 \times 0.90 = 18.0

The isotope

^{20}\text{Ne}

accounts for

90\%

of the sample.

2

Calculate the weighted contribution of

^{22}\text{Ne}

22 \times 0.10 = 2.2

The isotope

^{22}\text{Ne}

accounts for

10\%

of the sample.

3

Sum the weighted contributions to find the relative atomic mass

18.0 + 2.2 = 20.2

The relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes.

Key Concept

Calculation of Relative Atomic Mass from Isotopic Abundance

Question 6855Question →

In an acid-base titration, $25.0\text{ cm}^3$ of a $0.10\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution required $12.5\text{ cm}^3$ of a tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution for complete neutralization. What is the concentration of the acid solution in $\text{g dm}^{-3}$ ? $[\text{H} = 1.0, \text{O} = 16.0, \text{S} = 32.0]$

9.80\text{ g dm}^{-3}

19.60\text{ g dm}^{-3}

4.90\text{ g dm}^{-3}

0.10\text{ g dm}^{-3}

Show answer & explanation

Answer: $9.80\text{ g dm}^{-3}$

Answer

The concentration of the tetraoxosulfate(VI) acid solution is

9.80\text{ g dm}^{-3}

.

The balanced chemical equation

\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

establishes a mole ratio (

n_a : n_b

) of

1 : 2

. Substituting the given values into

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}

gives

C_a = 0.10\text{ mol dm}^{-3}

. Multiplying this by the molar mass of

\text{H}_2\text{SO}_4

(

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

) yields

9.80\text{ g dm}^{-3}

.

Step-by-Step Solution

1

Write the balanced chemical equation to determine the mole ratio.

\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

, giving

n_a = 1

and

n_b = 2

.

Stoichiometric coefficients define the mole ratio required for complete neutralization.

2

Calculate the molar concentration of tetraoxosulfate(VI) acid (

C_a

) using the titration formula

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}

.

\frac{C_a \times 12.5}{0.10 \times 25.0} = \frac{1}{2} \implies C_a \times 12.5 = 1.25 \implies C_a = 0.10\text{ mol dm}^{-3}

.

Relates the volumes and concentrations of acid and base according to their stoichiometry.

3

Convert the molar concentration to mass concentration in

\text{g dm}^{-3}

.

Molar mass of

\text{H}_2\text{SO}_4 = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-1}

. Mass concentration

= 0.10\text{ mol dm}^{-3} \times 98.0\text{ g mol}^{-1} = 9.80\text{ g dm}^{-3}

.

Mass concentration equals molar concentration multiplied by relative molar mass.

Key Concept

Volumetric calculations involving diprotic acid-monoprotic base titrations and conversion between molarity and mass concentration.

Question 6856Question →

An element $X$ with atomic number $12$ combines with an element $Y$ with atomic number $17$ to form a solid compound. Which of the following statements correctly accounts for the electrical conductivity of this compound?

It conducts electricity in the molten state because the giant ionic lattice breaks down, allowing the ions to move freely.

It conducts electricity in the solid state because delocalized valence electrons migrate freely through the crystal lattice.

It does not conduct electricity when dissolved in water because polar solvent molecules neutralize ionic charges.

It conducts electricity in the solid state because covalent electron sharing forms continuous conductive pathways.

Show answer & explanation

Answer: It conducts electricity in the molten state because the giant ionic lattice breaks down, allowing the ions to move freely.

Answer

The compound conducts electricity in the molten state because the giant ionic lattice breaks down, allowing the ions to move freely.

In solid electrovalent compounds, ions are locked into fixed lattice coordinates by strong electrostatic attraction and cannot migrate. Heating the compound until it melts breaks down the lattice, liberating the cations and anions so they can move freely under an applied electrical potential.

Step-by-Step Solution

1

Determine the type of bonding present in the compound formed by

X

and

Y

.

Element

X

(atomic number 12) has an electronic configuration of

2,8,2

(metal). Element

Y

(atomic number 17) has an electronic configuration of

2,8,7

(non-metal). Metal

X

transfers 2 electrons to two atoms of non-metal

Y

, forming an electrovalent (ionic) compound

XY_2

consisting of

X^{2+}

and

Y^-

ions.

Electrovalent bonding occurs between electropositive metals and electronegative non-metals via full electron transfer.

2

Evaluate the state of charge carriers in the solid state versus the molten state.

In the solid state, strong electrostatic forces hold

X^{2+}

and

Y^-

ions in rigid, fixed lattice positions, so no charge carriers can move. In the molten state, heat breaks the lattice, enabling the ions to move freely toward oppositely charged electrodes.

Electrical conduction requires mobile charge carriers. In ionic substances, these carriers are mobile ions present only in liquid (molten) or aqueous states.

Key Concept

Ionic bonding and electrical conductivity of ionic compounds

Question 6857Question →

Magnesium oxide ( $MgO$ ) has a significantly higher melting point than sodium chloride ( $NaCl$ ) because the electrostatic forces of attraction between its divalent ions ( $Mg^{2+}$ and $O^{2-}$ ) are substantially stronger than those between the monovalent ions ( $Na^+$ and $Cl^-$ ).

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because the electrostatic force holding an electrovalent lattice together scales with the product of the ionic charges. Divalent magnesium (

Mg^{2+}

) and oxide (

O^{2-}

) ions form a lattice with much higher lattice energy than monovalent sodium (

Na^+

) and chloride (

Cl^-

) ions, giving magnesium oxide a much higher melting point.

Step-by-Step Solution

1

Determine the ionic charges of the component ions in both compounds

MgO

is composed of

Mg^{2+}

and

O^{2-}

ions (divalent), while

NaCl

is composed of

Na^+

and

Cl^-

ions (monovalent).

The magnitude of ionic charges directly dictates the strength of electrostatic forces in an electrovalent crystal lattice.

2

Apply Coulomb's Law to compare lattice attraction strength

The attraction force scales with the charge product: for

MgO

,

|(+2) \times (-2)| = 4

; for

NaCl

,

|(+1) \times (-1)| = 1

.

A fourfold increase in charge product produces significantly stronger ionic bonds and greater lattice energy.

3

Correlate lattice energy with melting point

Greater thermal energy is required to overcome the electrostatic forces in

MgO

than in

NaCl

, resulting in a vastly higher melting point.

Melting point directly reflects the energy required to break down the solid giant ionic lattice structure.

Key Concept

Lattice Energy and Ion Charge Dependency in Ionic Compounds

Question 6858Question →

Consider the redox reaction represented by the following equation:

\text{CuO}_{(s)} + \text{H}_{2(g)} \rightarrow \text{Cu}_{(s)} + \text{H}_2\text{O}_{(g)}

Which of the following statements correctly describes hydrogen (

\text{H}_2

) from both classical and modern perspectives of redox?

Classically, hydrogen is oxidized because it gains oxygen; modernly, it is oxidized because its oxidation state increases from 0 to +1.

Classically, hydrogen is reduced because it gains oxygen; modernly, it is oxidized because it gains electrons.

Classically, hydrogen is oxidized because it accepts electrons; modernly, its oxidation state decreases from +1 to 0.

Classically, hydrogen is reduced because it acts as an oxygen donor; modernly, it is reduced because its oxidation state increases from 0 to +1.

Show answer & explanation

Answer: Classically, hydrogen is oxidized because it gains oxygen; modernly, it is oxidized because its oxidation state increases from 0 to +1.

Answer

Classically, hydrogen is oxidized because it gains oxygen; modernly, it is oxidized because its oxidation state increases from 0 to +1.

In the given reaction, hydrogen gas (

\text{H}_2

) combines with oxygen to form water. Classically, gaining oxygen is defined as oxidation. Modernly, the oxidation number of hydrogen increases from 0 (in

\text{H}_2

) to +1 (in

\text{H}_2\text{O}

), which also represents oxidation (loss of electrons). Thus, both concepts agree that hydrogen undergoes oxidation.

Step-by-Step Solution

1

Analyze hydrogen under the classical concept of redox.

In

\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}

, elemental hydrogen (

\text{H}_2

) gains oxygen to form water (

\text{H}_2\text{O}

). Under the classical definition, oxidation is defined as the addition or gain of oxygen.

Classical redox concepts define oxidation as the addition of oxygen (or removal of hydrogen) to a substance.

2

Analyze hydrogen under the modern concept of redox.

The oxidation state of free elemental hydrogen (

\text{H}_2

) is 0. In water (

\text{H}_2\text{O}

), hydrogen has an oxidation state of +1. The oxidation state increases from 0 to +1, which corresponds to a loss of electrons.

Modern redox concepts define oxidation as a loss of electrons or an increase in oxidation state.

3

Synthesize classical and modern findings to select the correct description.

Hydrogen is oxidized under both classical (gain of oxygen) and modern (increase in oxidation number from 0 to +1) definitions.

Both definitions consistently classify hydrogen as undergoing oxidation in this reaction.

Key Concept

Comparison of Classical (Oxygen/Hydrogen Transfer) and Modern (Electron Transfer/Oxidation Number) Concepts of Redox

Question 6859Question →

Match each inorganic chemical formula on the left with its corresponding systematic IUPAC name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

\text{KMnO}_4

matches Potassium tetraoxomanganate(VII);

\text{KClO}_3

matches Potassium trioxochlorate(V);

\text{K}_2\text{CrO}_4

matches Potassium tetraoxochromate(VI);

\text{KNO}_2

matches Potassium dioxonitrate(III).

Each chemical formula is correctly matched to its systematic IUPAC name by evaluating the oxidation state of the central atom:

+7

for

\text{Mn}

in

\text{KMnO}_4

(Potassium tetraoxomanganate(VII)),

+5

for

\text{Cl}

in

\text{KClO}_3

(Potassium trioxochlorate(V)),

+6

for

\text{Cr}

in

\text{K}_2\text{CrO}_4

(Potassium tetraoxochromate(VI)), and

+3

for

\text{N}

in

\text{KNO}_2

(Potassium dioxonitrate(III)).

Step-by-Step Solution

1

Determine the oxidation state of the central atom in each chemical species using standard rules (Group 1 alkali metals

= +1

, Oxygen

= -2

, total neutral compound charge

= 0

).

Manganese in

\text{KMnO}_4

is

+7

; Chlorine in

\text{KClO}_3

is

+5

; Chromium in

\text{K}_2\text{CrO}_4

is

+6

; Nitrogen in

\text{KNO}_2

is

+3

.

The Roman numeral in IUPAC nomenclature for oxo-compounds directly corresponds to the calculated oxidation state of the central non-metal or transition metal.

2

Map each chemical formula to its IUPAC name based on the oxygen ligand count prefix and the central atom's Roman numeral oxidation state.

\text{KMnO}_4 \rightarrow

Potassium tetraoxomanganate(VII);

\text{KClO}_3 \rightarrow

Potassium trioxochlorate(V);

\text{K}_2\text{CrO}_4 \rightarrow

Potassium tetraoxochromate(VI);

\text{KNO}_2 \rightarrow

Potassium dioxonitrate(III).

IUPAC convention requires specifying oxygen count (dioxo-, trioxo-, tetraoxo-) followed by the central element suffix (-ate) and its oxidation state in parentheses.

Key Concept

Oxidation number determination and IUPAC nomenclature of inorganic redox species.

Question 6860Question →

When the following redox reaction is balanced in an acidic medium using the smallest whole-number coefficients:

\text{ClO}_3^-(\text{aq}) + a\text{Fe}^{2+}(\text{aq}) + b\text{H}^+(\text{aq}) \rightarrow \text{Cl}^-(\text{aq}) + c\text{Fe}^{3+}(\text{aq}) + d\text{H}_2\text{O}(\text{l})

What is the value of the coefficient

b

for

\text{H}^+

?

3

5

6

12

Show answer & explanation

Answer: 6

Answer

The stoichiometric coefficient for hydrogen ions is 6.

In the reduction half-reaction,

\text{ClO}_3^-

is reduced to

\text{Cl}^-

. Balancing the 3 oxygen atoms requires

3\text{ H}_2\text{O}

on the product side. Consequently,

6\text{ H}^+

ions are required on the reactant side to balance the 6 hydrogen atoms, giving

b = 6

.

Step-by-Step Solution

1

Determine the oxidation state change for chlorine

In

\text{ClO}_3^-

, chlorine has an oxidation state of

+5

. In

\text{Cl}^-

, chlorine has an oxidation state of

-1

. The total change is a gain of

6\text{ e}^-

.

Knowing the electron transfer per mole of chlorate ion establishes the electron requirement for the reduction half-reaction.

2

Balance oxygen atoms using water

The chlorate ion

\text{ClO}_3^-

contains 3 oxygen atoms, requiring

3\text{ H}_2\text{O}

on the product side.

In acidic redox balancing, oxygen atoms are balanced by adding water molecules to the side deficient in oxygen.

3

Balance hydrogen atoms using hydrogen ions

To balance the 6 hydrogen atoms in

3\text{ H}_2\text{O}

, add

6\text{ H}^+

to the reactant side.

Hydrogen atoms from the water molecules on the product side must originate from hydrogen ions in the acidic medium.

4

Verify overall mass and charge balance

The reduction half-reaction is

\text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}

. Adding the oxidation half-reaction

6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-

yields the overall balanced equation with

b = 6

.

Combining half-reactions confirms that electrons cancel out and mass and charge are conserved.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium

Answer

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