All practice questions

Question 6821Question →

Chief Kalu insured his commercial supermarket building, valued at $\text{N}50,000,000$ , against fire for a sum of $\text{N}35,000,000$ . The policy contains an average clause. If a fire outbreak damages the building causing a loss of $\text{N}10,000,000$ , what amount in Naira will the insurance company pay as indemnity?

Show answer & explanation

Answer: 7000000

Answer

7,000,000 Naira

Under the average clause in fire insurance policies, if a property is under-insured (insured for less than its true value), the insured is deemed to be their own insurer for the uninsured portion. Here, the building was insured for 70% of its full value (N35,000,000 / N50,000,000 = 0.70). Therefore, the insurance company pays only 70% of the actual loss incurred (70% of N10,000,000 = 7,000,000 Naira).

Step-by-Step Solution

1

Identify the property values given in the problem statement.

Actual Value = N50,000,000; Sum Insured = N35,000,000; Actual Loss = N10,000,000.

These figures are required to calculate indemnity under under-insurance.

2

Apply the average clause compensation formula.

Compensation = (Sum Insured / Actual Value) * Actual Loss

When property is insured for less than its full value and has an average clause, the insurer pays only a proportionate share of any loss.

3

Calculate the compensation payable.

Compensation = (35,000,000 / 50,000,000) * 10,000,000 = 7,000,000 Naira.

Since the owner insured 70% of the building's value, the insurer pays 70% of the actual loss suffered.

Key Concept

Average Clause in Insurance

Question 6822Question →

Complete the statement below by calculating the missing pOH and pH values for the given alkaline solution.

Fill in the blanks below

A solution of potassium hydroxide (

\text{KOH}

) has a hydroxide ion concentration of

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

at

25^\circ\text{C}

. The pOH of this solution is and its pH is .

Show answer & explanation

Answer

The pOH of the potassium hydroxide solution is 2 and its pH is 12.

Potassium hydroxide (

\text{KOH}

) dissociates completely in water to yield a hydroxide ion concentration of

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

. The pOH is calculated as

-\log_{10}(10^{-2}) = 2

. Since

\text{pH} + \text{pOH} = 14

at

25^\circ\text{C}

, subtracting 2 from 14 gives a pH of 12.

Step-by-Step Solution

1

Calculate the pOH from the given hydroxide ion concentration

[\text{OH}^-]

.

\text{pOH} = -\log_{10}(0.01) = -\log_{10}(10^{-2}) = 2

\text{KOH}

is a strong monobasic base that fully dissociates in water, giving

[\text{OH}^-] = 0.01\text{ mol dm}^{-3}

.

2

Calculate the pH using the relationship

\text{pH} + \text{pOH} = 14

at

25^\circ\text{C}

.

\text{pH} = 14 - \text{pOH} = 14 - 2 = 12

The sum of pH and pOH for any aqueous solution at standard temperature (

25^\circ\text{C}

) equals 14.

Key Concept

Determination of pOH and pH for strong base solutions using the ion product constant of water.

Estimated Time:1m 0s

Question 6823Question →

Which commercial banking payment service is described in the statement below?

Fill in the blanks below

An arrangement where a business owner authorizes their bank to allow a creditor to collect variable amounts of money directly from the business account at agreed intervals is known as a .

Show answer & explanation

Answer

Direct debit

Direct debit is the commercial banking service that enables a creditor, with prior consent from the account holder, to pull variable sums of money directly from the debtor's account.

Step-by-Step Solution

1

Analyze the operational features of the payment system described.

The payment arrangement allows a third party (the creditor) to initiate collection of varying amounts from the debtor's account after obtaining prior authorization.

Differentiating banking services relies on knowing who initiates the payment request and whether the payment amount is fixed or variable.

2

Compare direct debits with other bank transfer methods such as standing orders.

A standing order is initiated by the account holder for fixed amounts, whereas a direct debit is initiated by the creditor for variable or recurring amounts.

Business traders use direct debits for fluctuating settlement needs such as variable utility bills or inventory purchases.

Key Concept

A direct debit is an authorization given by an account holder to a commercial bank, empowering a named creditor to claim variable payments directly from the account at specified intervals.

Estimated Time:1m 0s

Question 6824Question →

A sealed rigid reaction flask contains a sample of argon gas at a pressure of $120\text{ kPa}$ when the temperature is $30^\circ\text{C}$ . If the flask is heated to a temperature of $333^\circ\text{C}$ while maintaining a constant volume, what is the final pressure of the gas?

60\text{ kPa}

240\text{ kPa}

1332\text{ kPa}

132\text{ kPa}

Show answer & explanation

Answer: $240\text{ kPa}$

Answer

The final pressure of the gas is

240\text{ kPa}

.

According to the Pressure Law (Gay-Lussac's Law), the pressure of a fixed mass of gas is directly proportional to its absolute temperature when volume remains constant (

P_1/T_1 = P_2/T_2

). Converting temperatures to Kelvin yields

T_1 = 30 + 273 = 303\text{ K}

and

T_2 = 333 + 273 = 606\text{ K}

. Substituting these into

P_2 = 120 \times (606 / 303)

gives

240\text{ kPa}

.

Step-by-Step Solution

1

Convert given temperatures from Celsius to Kelvin.

T_1 = 30 + 273 = 303\text{ K}

and

T_2 = 333 + 273 = 606\text{ K}

.

Gas law equations require absolute temperature measured on the Kelvin scale.

2

State the Pressure Law formula relating initial and final values at constant volume.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

.

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

3

Substitute the values into the equation to calculate final pressure

P_2

.

P_2 = 120\text{ kPa} \times \frac{606\text{ K}}{303\text{ K}} = 120\text{ kPa} \times 2 = 240\text{ kPa}

.

Since absolute temperature doubles from

303\text{ K}

to

606\text{ K}

, the pressure must also double.

Key Concept

Pressure Law (Gay-Lussac's Law)

Question 6825Question →

At a magnetic observation station, the horizontal component of the Earth's magnetic field is $40\text{ }\mu\text{T}$ and the vertical component is $30\text{ }\mu\text{T}$ . What is the total magnetic field intensity of the Earth at this station in $\mu\text{T}$ ?

Show answer & explanation

Answer: 50

Answer

The total magnetic field intensity of the Earth at this station is

50\text{ }\mu\text{T}

.

The horizontal component (

B_h

) and vertical component (

B_v

) of the Earth's magnetic field act at right angles to each other. Therefore, the resultant total magnetic field intensity (

B

) is calculated using vector addition:

B = \sqrt{B_h^2 + B_v^2} = \sqrt{40^2 + 30^2} = 50\text{ }\mu\text{T}

.

Step-by-Step Solution

1

Identify the vector relationship between the horizontal and vertical components of the Earth's magnetic field.

B = \sqrt{B_h^2 + B_v^2}

, where

B_h = 40\text{ }\mu\text{T}

and

B_v = 30\text{ }\mu\text{T}

.

The horizontal and vertical components of the Earth's magnetic field are mutually perpendicular vector components.

2

Substitute the values into the formula and solve for total magnetic field intensity

B

.

B = \sqrt{(40)^2 + (30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ }\mu\text{T}

.

Applying the Pythagorean theorem yields the magnitude of the resultant magnetic field vector.

Key Concept

Resolution of Earth's magnetic field into horizontal (

B_h

) and vertical (

B_v

) components.

Estimated Time:1m 0s

Question 6826Question →

A central bank issues a new currency series printed with precise physical specifications, ensuring that every note of a given denomination has identical appearance, weight, and value. Which characteristic of money does this standardized production primarily maintain?

Homogeneity

Divisibility

Durability

Portability

Show answer & explanation

Answer: Homogeneity

Answer

Homogeneity is the characteristic that guarantees every unit of money within a particular denomination is uniform and indistinguishable from another of equal face value.

Homogeneity (or uniformity) requires that every unit of currency within the same denomination must be identical in form, design, and purchasing power so that no unit is preferred over another of the same face value.

Step-by-Step Solution

1

Analyze the core requirement in the stem

The stem describes producing banknotes of equal denomination that are completely identical in size, appearance, and value.

Identifying the key attribute being described helps isolate the exact characteristic of money.

2

Evaluate the definition of homogeneity in economics

Homogeneity (or uniformity) dictates that all units of a particular denomination must be standard so traders treat them as identical substitutes.

This prevents discrimination between individual notes of the same face value.

Key Concept

Homogeneity as a key characteristic of money

Question 6827Question →

A merchant insured his shop inventory strictly against loss caused by fire. Subsequently, a heavy flood submerged the warehouse and damaged the inventory, but no fire occurred. The insurance company denied the merchant's claim for compensation. Which principle of insurance supports the insurer's decision?

Proximate cause

Indemnity

Subrogation

Utmost good faith

Show answer & explanation

Answer: Proximate cause

Answer

Proximate cause

Under the principle of proximate cause (causa proxima), an insurer is obligated to pay compensation only when the direct, immediate, and dominant cause of the loss is the specific risk insured against. Because the inventory was damaged by flooding (an uninsured peril) rather than fire (the insured peril), the insurer is legally justified in rejecting the claim.

Step-by-Step Solution

1

Identify the insured peril and the actual cause of damage

The insured peril was fire, whereas the actual cause of loss was flooding.

To determine policy coverage, the active event leading to the damage must be identified.

2

Apply the relevant insurance principle to the facts

The principle of proximate cause (causa proxima) dictates that compensation is only payable if the direct and effective cause of loss is the peril insured against.

Because flooding was not included in the fire policy, the loss falls outside the insurer's liability.

Key Concept

Principle of Proximate Cause (Causa Proxima)

Question 6828Question →

Match each money characteristic or function on the left with its corresponding operational description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Portability matches with allowing purchasing power to be carried around conveniently; Divisibility matches with facilitating transactions of small and varying denominations; Standard of deferred payment matches with providing an agreed unit for settling future credit obligations; Store of value matches with enabling income to be saved and preserved for future consumption.

Each feature of money uniquely satisfies a distinct commercial need: portability enables physical ease of carrying currency; divisibility permits payments of various exact sizes; standard of deferred payment facilitates credit and deferred settlements; and store of value allows saving purchasing power for future periods.

Step-by-Step Solution

1

Analyze physical characteristics of money (left_1 and left_2)

Portability guarantees ease of mobility (right_2), while divisibility allows money to exist in fractional units for exact change (right_1).

Physical characteristics determine how convenient money is to handle in everyday commercial exchanges.

2

Analyze secondary functions of money (left_3 and left_4)

Standard of deferred payment serves future debt settlements (right_4), while store of value enables wealth accumulation over time (right_3).

Secondary functions support economic planning, credit structures, and capital accumulation.

Key Concept

Functions and Characteristics of Money

Question 6829Question →

A solution is prepared by dissolving $0.04\text{ g}$ of sodium hydroxide ( $\text{NaOH}$ ) in distilled water to make $1.0\text{ dm}^3$ of solution at $25^\circ\text{C}$ . What is the $\text{pH}$ of the resulting solution? [Molar mass of $\text{NaOH} = 40\text{ g mol}^{-1}$ ]

3.0

11.0

1.4

12.6

Show answer & explanation

Answer: $11.0$

Answer

The pH of the resulting solution is

11.0

.

The solution contains

0.04\text{ g}

of

\text{NaOH}

in

1.0\text{ dm}^3

, which corresponds to

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

or

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

. Since

\text{NaOH}

is a strong base,

[\text{OH}^-] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

, giving a

\text{pOH}

of

3.0

. Using the relationship

\text{pH} + \text{pOH} = 14

, the

\text{pH}

is

14.0 - 3.0 = 11.0

.

Step-by-Step Solution

1

Calculate the molar concentration of sodium hydroxide (

\text{NaOH}

).

\text{Molar concentration} = \frac{0.04\text{ g dm}^{-3}}{40\text{ g mol}^{-1}} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Concentration must be in

\text{mol dm}^{-3}

before calculating ion concentration and

\text{pOH}

.

2

Determine the hydroxide ion concentration

[\text{OH}^-]

.

[\text{OH}^-] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

\text{NaOH}

is a strong monobasic base that completely dissociates into

\text{Na}^+

and

\text{OH}^-

ions.

3

Calculate the

\text{pOH}

of the solution.

\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1.0 \times 10^{-3}) = 3.0

.

\text{pOH}

is defined as the negative base-10 logarithm of the hydroxide ion concentration.

4

Convert

\text{pOH}

to

\text{pH}

.

\text{pH} = 14.0 - \text{pOH} = 14.0 - 3.0 = 11.0

.

At

25^\circ\text{C}

,

\text{pH} + \text{pOH} = 14.0

.

Key Concept

Calculating pH from the mass concentration of a strong base via pOH conversion

Estimated Time:1m 30s

Question 6830Question →

A fixed mass of an ideal gas occupies a volume of $3.0\text{ dm}^3$ at $27^\circ\text{C}$ and a pressure of $1.0\text{ atm}$ . What is the final volume of the gas in $\text{dm}^3$ when the temperature is increased to $127^\circ\text{C}$ while maintaining constant pressure?

Show answer & explanation

Answer: 4.0; 4; 4.0 dm3; 4 dm3; 4.0 dm³; 4 dm³

Answer

The final volume of the gas is

4.0\text{ dm}^3

.

First convert temperatures to Kelvin:

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

and

T_2 = 127 + 273 = 400\text{ K}

. At constant pressure,

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

. Rearranging to solve for

V_2

gives

V_2 = \frac{3.0 \times 400}{300} = 4.0\text{ dm}^3

.

Step-by-Step Solution

1

Convert both temperatures from degrees Celsius to Kelvin

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

and

T_2 = 127 + 273 = 400\text{ K}

Gas law calculations strictly require absolute temperature measured in Kelvin.

2

Apply Charles's Law derived from the ideal gas equation (

PV = nRT

) at constant pressure

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

Volume is directly proportional to absolute temperature when pressure and amount of gas are constant.

3

Substitute the values into the formula and solve for

V_2

V_2 = \frac{V_1 \times T_2}{T_1} = \frac{3.0 \times 400}{300} = 4.0\text{ dm}^3

Calculates the new gas volume.

Key Concept

General Gas Law and Absolute Temperature Conversion

Question 6831Question →

A trader holds Treasury bills, bills of exchange, and savings bank deposits. Although these assets serve as a reliable store of value and can easily be converted into cash, they cannot be spent directly as a medium of exchange to purchase goods. Under which form of money are these financial assets classified?

Near money

Legal tender

Fiat money

Token money

Show answer & explanation

Answer: Near money

Answer

Near money

Near money (or quasi-money) refers to highly liquid assets such as Treasury bills, bills of exchange, and savings deposits. They perform the store of value function of money and can quickly be turned into cash, but cannot be used directly as a medium of exchange for purchasing goods.

Step-by-Step Solution

1

Examine the properties of the financial assets listed in the stem (Treasury bills, bills of exchange, savings deposits).

These assets possess high liquidity and act as a store of wealth, but cannot be used directly to pay for goods in everyday transactions.

Forms of money are categorized by their liquidity level and their ability to serve directly as a medium of exchange.

2

Match these properties against the standard definitions of forms of money in Commerce.

Assets that must be converted into cash or bank deposits before spending are categorized as near money or quasi-money.

Near money acts as a close substitute for money without having immediate medium-of-exchange status.

Key Concept

Types and Forms of Money - Near Money (Quasi-Money)

Question 6832Question →

A chemist needs to prepare a pure, hydrated sample of copper(II) tetraoxosulfate(VI) ( $CuSO_4 \cdot 5H_2O$ ) starting from insoluble copper(II) oxide ( $CuO$ ). Which of the following experimental procedures is correct for this preparation?

Add excess copper(II) oxide to dilute tetraoxosulfate(VI) acid, filter the mixture, concentrate the filtrate, and allow it to cool for crystallization.

Mix aqueous copper(II) nitrate with aqueous sodium tetraoxosulfate(VI), filter off the precipitate formed, and dry it in an oven.

Titrate copper(II) oxide solution against dilute tetraoxosulfate(VI) acid using phenolphthalein, then evaporate the liquid to complete dryness.

Dissolve copper(II) oxide in hot distilled water, add sodium chloride solution, and collect the resulting salt crystals.

Show answer & explanation

Answer: Add excess copper(II) oxide to dilute tetraoxosulfate(VI) acid, filter the mixture, concentrate the filtrate, and allow it to cool for crystallization.

Answer

The correct procedure is to react excess insoluble copper(II) oxide with dilute tetraoxosulfate(VI) acid, filter out the unreacted solid, concentrate the filtrate by gentle heating, and cool to form crystals.

The standard method for preparing a soluble salt from an insoluble oxide involves adding excess oxide to hot dilute acid to ensure complete acid neutralization, filtering out the unreacted solid, and partially evaporating the filtrate so that hydrated crystals form upon cooling.

Step-by-Step Solution

1

Identify the solubility of the product salt and reactants

Copper(II) tetraoxosulfate(VI) (

CuSO_4

) is a soluble salt, while copper(II) oxide (

CuO

) is an insoluble base.

Soluble salts prepared from insoluble bases require the excess insoluble reactant method.

2

Neutralize dilute acid with excess insoluble base

Warm

H_2SO_4(aq)

reacts with excess

CuO(s)

according to:

CuO(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + H_2O(l)

.

Using excess base ensures all acid is fully consumed.

3

Separate unreacted solid and crystallize

Filter off unreacted

CuO

, heat the filtrate to crystallizing point, and cool slowy to obtain hydrated crystals (

CuSO_4 \cdot 5H_2O

).

Filtration removes insoluble excess, and gentle cooling preserves water of crystallization.

Key Concept

Preparation of soluble salts using dilute acids and insoluble bases

Estimated Time:1m 30s

Question 6833Question →

A rigid container holds a gas mixture containing $4.0\text{ g}$ of methane ( $\text{CH}_4$ ) and $14.0\text{ g}$ of nitrogen ( $\text{N}_2$ ). If the partial pressure exerted by methane in the mixture is $125\text{ kPa}$ , what is the total pressure of the gas mixture in $\text{kPa}$ ? [Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{N} = 14$ ]

Show answer & explanation

Answer: 375

Answer

The total pressure of the gas mixture is

375\text{ kPa}

.

The total pressure of

375\text{ kPa}

is determined by calculating the moles of methane (

0.25\text{ mol}

) and nitrogen (

0.50\text{ mol}

), giving a total of

0.75\text{ mol}

. Methane constitutes one-third (

1/3

) of the total moles, so its partial pressure is one-third of the total pressure. Dividing the partial pressure of methane (

125\text{ kPa}

) by its mole fraction (

1/3

) yields a total pressure of

375\text{ kPa}

.

Step-by-Step Solution

1

Calculate the amount of moles of each gas present in the mixture.

n_{\text{CH}_4} = \frac{4.0\text{ g}}{16\text{ g/mol}} = 0.25\text{ mol}

and

n_{\text{N}_2} = \frac{14.0\text{ g}}{28\text{ g/mol}} = 0.50\text{ mol}

.

Molar masses are determined from relative atomic masses:

\text{CH}_4 = 12 + 4(1) = 16\text{ g/mol}

and

\text{N}_2 = 2(14) = 28\text{ g/mol}

.

2

Calculate total moles and the mole fraction of methane.

n_{\text{total}} = 0.25 + 0.50 = 0.75\text{ mol}

;

X_{\text{CH}_4} = \frac{0.25\text{ mol}}{0.75\text{ mol}} = \frac{1}{3}

.

Mole fraction is the ratio of the number of moles of a specific gas component to the total number of moles in the gas mixture.

3

Apply Dalton's Law of Partial Pressures to find total pressure.

P_{\text{total}} = \frac{P_{\text{CH}_4}}{X_{\text{CH}_4}} = \frac{125\text{ kPa}}{1/3} = 375\text{ kPa}

.

According to Dalton's Law, the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure (

P_i = X_i \times P_{\text{total}}

).

Key Concept

Dalton's Law of Partial Pressures and Mole Fraction

Estimated Time:1m 30s

Question 6834Question →

What is the ground-state electronic configuration of the chromium(III) ion, $Cr^{3+}$ ? (Atomic number of $Cr = 24$ )

1s^2 2s^2 2p^6 3s^2 3p^6 3d^3

1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2

1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^1

1s^2 2s^2 2p^6 3s^2 3p^6 3d^4

Show answer & explanation

Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$

Answer

1s^2 2s^2 2p^6 3s^2 3p^6 3d^3

Neutral chromium (

Z = 24

) has the ground-state electron configuration

1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1

. When forming the

Cr^{3+}

ion, three electrons must be removed. Outer shell (

4s

) electrons are lost first, followed by

(n-1)d

electrons. Removing one electron from

4s

and two from

3d

gives the configuration

1s^2 2s^2 2p^6 3s^2 3p^6 3d^3

.

Step-by-Step Solution

1

Determine the electronic configuration of neutral chromium (

Cr

)

Ground state

Cr

(

Z = 24

) is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1

Chromium exhibits an anomalous configuration due to the extra stability of a half-filled

3d

subshell.

2

Apply the rule for cation formation in transition metals

Electrons in the outermost shell (

n=4

,

4s

subshell) are lost before electrons in the inner subshell (

n=3

,

3d

subshell).

The

4s

electrons experience lower effective nuclear attraction in the ionized state and are situated at a higher principal energy level.

3

Remove 3 electrons to form

Cr^{3+}

Remove 1 electron from

4s

and 2 electrons from

3d

:

(3d^5 4s^1) - 3e^- = 3d^3 4s^0

Removing 1 electron from

4s

leaves

3d^5

; removing two more from

3d

leaves

3d^3

.

Key Concept

Electronic configuration of d-block transition metal cations

Estimated Time:1m 0s

Question 6835Question →

Match each mode of transport or specialized logistical system on the left with its primary operational feature or commercial function in trade on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pipeline Transport matches continuous automated fluid movement; Containerization matches standardized intermodal unit loads; Tramp Steamer Shipping matches unscheduled ocean charters for bulk cargo; Inland Waterway Barges match low-cost river and canal transport for heavy bulk goods.

Each mode of transport is uniquely suited to specific cargo types and operational conditions. Pipeline transport provides automated, continuous flow for liquids and gases; containerization standardizes unit loading across different transit modes; tramp steamers offer flexible unscheduled bulk ocean carriage; and inland waterway barges afford economical bulk movement along navigable waterways.

Step-by-Step Solution

1

Analyze the operational characteristics of each transport mode and logistical system.

Identify that pipelines convey fluids continuously, containerization standardizes unit loading, tramp steamers operate on unscheduled charters, and inland barges carry heavy bulk goods along rivers.

Matching transport modes to commercial functions requires aligning technical capabilities with trade requirements.

2

Pair each term on the left with its unique matching function on the right.

Pipeline matches automated fluid conveyance; Containerization matches intermodal unit loading; Tramp Steamer matches unscheduled ocean charters; Inland Waterway Barges match low-cost river bulk transport.

Each mode serves distinct commercial needs regarding cargo type, flexibility, route fixing, and cost efficiency.

Key Concept

Modes of transportation and specialized logistics systems in commercial trade

Question 6836Question →

Complete the following statement regarding laboratory methods of salt preparation and salt classification.

Fill in the blanks below

The preparation of anhydrous iron(III) chloride (

FeCl_3

) by reacting red-hot iron filings directly with dry chlorine gas is an example of , whereas potash alum,

KAl(SO_4)_2 \cdot 12H_2O

, is classified as a salt.

Show answer & explanation

Answer

The preparation method is direct combination (or synthesis), and potash alum is classified as a double salt.

Anhydrous binary salts like iron(III) chloride must be synthesized by direct combination of their elements in dry conditions to avoid hydrolysis. Potash alum is formed by equimolar crystallization of potassium sulfate and aluminium sulfate, yielding a double salt.

Step-by-Step Solution

1

Determine the preparation method for anhydrous iron(III) chloride.

Direct combination of elements (

2Fe_{(s)} + 3Cl_{2(g)} \rightarrow 2FeCl_{3(s)}

) is used because aqueous crystallization causes salt hydrolysis.

Reacting two elements directly to form a compound without water is direct combination/synthesis.

2

Classify potash alum based on its chemical formula and ionic behavior.

Potash alum contains two distinct metallic cations (

K^+

and

Al^{3+}

) combined with sulfate anions (

SO_4^{2-}

), making it a double salt.

Salts composed of two simple salts crystallized together in equimolar proportions that dissociate completely into constituent ions are double salts.

Key Concept

Salt Preparation Methods and Classifications

Question 6837Question →

A gas sample collected in a laboratory syringe occupies a volume of $600\text{ cm}^3$ at $27^\circ\text{C}$ and $1.0\text{ atm}$ pressure. What is the volume of the gas when the pressure is increased to $2.0\text{ atm}$ and the temperature is raised to $81^\circ\text{C}$ ?

354\text{ cm}^3

900\text{ cm}^3

708\text{ cm}^3

1416\text{ cm}^3

Show answer & explanation

Answer: $354\text{ cm}^3$

Answer

The final volume of the gas is

354\text{ cm}^3

.

Using the General Gas Law

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

with absolute temperatures

T_1 = 300\text{ K}

(

27^\circ\text{C}

) and

T_2 = 354\text{ K}

(

81^\circ\text{C}

), the calculation yields

V_2 = \frac{1.0 \times 600 \times 354}{2.0 \times 300} = 354\text{ cm}^3

.

Step-by-Step Solution

1

Convert temperatures from Celsius to Kelvin

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

and

T_2 = 81 + 273 = 354\text{ K}

Gas laws require absolute temperature values on the Kelvin scale.

2

Set up the General Gas Law equation

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

Pressure, volume, and temperature all change simultaneously for a fixed mass of gas.

3

Substitute the known values and calculate

V_2

V_2 = \frac{1.0\text{ atm} \times 600\text{ cm}^3 \times 354\text{ K}}{2.0\text{ atm} \times 300\text{ K}} = 354\text{ cm}^3

Simplifying

\frac{600}{2.0 \times 300} = 1

leaves

1 \times 354 = 354\text{ cm}^3

.

Key Concept

General Gas Law (Combined Gas Law)

Estimated Time:1m 15s

Question 6838Question →

At the Earth's magnetic poles, the horizontal component of the Earth's magnetic field reaches its maximum value while the vertical component is equal to zero.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. At the magnetic poles, the angle of dip is

90^\circ

, which means the horizontal component of the Earth's magnetic field is zero and the vertical component is at its maximum.

The statement is false because at the Earth's magnetic poles, the angle of inclination (dip) is

90^\circ

. Substituting this angle into the component resolution formulas yields

B_h = B \cos 90^\circ = 0

and

B_v = B \sin 90^\circ = B

. Therefore, the horizontal component is zero and the vertical component is maximum at the poles.

Step-by-Step Solution

1

Determine the angle of dip (

\theta

) at the Earth's magnetic poles.

The angle of dip at the poles is

\theta = 90^\circ

.

The Earth's magnetic flux lines enter or leave the surface vertically at the magnetic poles.

2

Calculate the horizontal component (

B_h

) using the magnetic field resolution formula.

B_h = B \cos(90^\circ) = 0

.

Since

\cos(90^\circ) = 0

, there is no horizontal magnetic field component at the poles.

3

Calculate the vertical component (

B_v

) using the magnetic field resolution formula.

B_v = B \sin(90^\circ) = B

.

Since

\sin(90^\circ) = 1

, the vertical component accounts for the entire total magnetic field intensity.

4

Compare the calculated component values against the given statement.

The statement claims

B_h

is maximum and

B_v = 0

, which contradicts the physical reality where

B_h = 0

and

B_v

is maximum.

Therefore, the statement is evaluated as false.

Key Concept

Earth's Magnetic Field Components at Magnetic Poles

Question 6839Question →

A pharmaceutical firm based in Lagos needs to urgently dispatch a small, high-value consignment of temperature-sensitive vaccines to a health facility in Abuja within six hours. Which mode of transportation is most appropriate for moving this cargo?

Air transport

Rail transport

Sea transport

Pipeline transport

Show answer & explanation

Answer: Air transport

Answer

Air transport is the most appropriate mode of transportation.

Air transport is the fastest mode of transportation in commerce. It is uniquely suited for high-value, light-weight, highly perishable, or time-critical shipments like vaccines, where speed and delicate handling outweigh higher freight charges.

Step-by-Step Solution

1

Analyze cargo characteristics and transit requirements

The shipment consists of high-value, low-weight, temperature-sensitive vaccines requiring emergency delivery across a long inland distance within six hours.

Mode selection in trade depends on speed, cost, safety, distance, and the physical characteristics of the cargo.

2

Evaluate transport options against constraints

Air transport is the only mode capable of fulfilling a strict six-hour transit window across long distances while ensuring maximum security for delicate cargo.

Rail and sea transport are too slow, and pipelines are restricted strictly to liquid/gas continuous bulk transfers.

Key Concept

Transportation mode selection based on speed, cargo perishability, value, and urgency

Estimated Time:1m 0s

Question 6840Question →

The solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water at $40^\circ\text{C}$ is $6.0\text{ mol dm}^{-3}$ . What mass of $\text{KNO}_3$ , in grams, is required to prepare a saturated solution in $250\text{ cm}^3$ of water at this temperature? (Molar mass of $\text{KNO}_3 = 101\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 151.5

Answer

The mass of

\text{KNO}_3

required to prepare a saturated solution in

250\text{ cm}^3

of water at

40^\circ\text{C}

is

151.5\text{ g}

.

To find the mass of solute required for saturation, convert the given volume of water to cubic decimeters (

250\text{ cm}^3 = 0.25\text{ dm}^3

). Multiply the volume by the molar solubility (

6.0\text{ mol dm}^{-3} \times 0.25\text{ dm}^3 = 1.5\text{ mol}

) to obtain the number of moles, then multiply by the molar mass (

1.5\text{ mol} \times 101\text{ g mol}^{-1} = 151.5\text{ g}

).

Step-by-Step Solution

1

Convert volume from

\text{cm}^3

to

\text{dm}^3

0.25\text{ dm}^3

Molar solubility is expressed per

\text{dm}^3

, so the volume of solvent must be in

\text{dm}^3

.

2

Determine moles of

\text{KNO}_3

needed for saturation

1.5\text{ mol}

Multiply molar solubility by the volume in

\text{dm}^3

.

3

Convert moles to mass in grams

151.5\text{ g}

Multiply moles by the molar mass of

\text{KNO}_3

(

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

).

Key Concept

Calculating solute mass for saturation using molar solubility and volume

Answer

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