All practice questions

Evaluate the algebraic limit:

\lim_{x \to 4} \frac{\sqrt{x} - 2}{x^2 - 16}

What is the exact value of this limit?

\frac{1}{32}

\frac{1}{16}

\frac{1}{8}

0

Show answer & explanation

Answer: $\frac{1}{32}$

Answer

The exact value of the limit is

\frac{1}{32}

.

The limit presents an indeterminate form

\frac{0}{0}

at

x = 4

. Factoring

x^2 - 16

into

(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)

allows the factor

(\sqrt{x} - 2)

to be cancelled from both the numerator and denominator. Substituting

x = 4

into the simplified expression

\frac{1}{(\sqrt{x} + 2)(x + 4)}

yields

\frac{1}{(2+2)(4+4)} = \frac{1}{32}

.

Step-by-Step Solution

1

Identify the form of the limit by direct substitution.

Substituting

x = 4

into

\frac{\sqrt{4} - 2}{4^2 - 16}

gives

\frac{0}{0}

, which is an indeterminate form.

Direct substitution yields

\frac{0}{0}

, requiring algebraic simplification.

2

Factor the denominator

x^2 - 16

.

x^2 - 16 = (x - 4)(x + 4)

Use the difference of two squares identity

a^2 - b^2 = (a - b)(a + b)

.

3

Factor

(x - 4)

as a difference of squares involving square roots.

x - 4 = (\sqrt{x})^2 - 2^2 = (\sqrt{x} - 2)(\sqrt{x} + 2)

This exposes the vanishing factor

(\sqrt{x} - 2)

in the denominator.

4

Cancel the common factor

(\sqrt{x} - 2)

and evaluate the limit.

\lim_{x \to 4} \frac{\sqrt{x} - 2}{(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)} = \lim_{x \to 4} \frac{1}{(\sqrt{x} + 2)(x + 4)} = \frac{1}{(\sqrt{4} + 2)(4 + 4)} = \frac{1}{4 \times 8} = \frac{1}{32}

Cancelling the factor removing the

\frac{0}{0}

condition allows direct evaluation.

Key Concept

Resolution of indeterminate limits of the form 0/0 using algebraic factorization and conjugate radical identities.

Estimated Time:2m 0s

Question 6962Question →

Find the value of

x

that satisfies the exponential equation

\frac{4^{x+1} + 4^x + 4^{x-1}}{2^{x+2} + 2^{x+1} + 2^x} = 24

Show answer & explanation

Answer: 5

Answer

5

Factoring

4^{x-1}

from the numerator yields

4^{x-1}(16 + 4 + 1) = 21 \cdot 4^{x-1}

. Factoring

2^x

from the denominator yields

2^x(4 + 2 + 1) = 7 \cdot 2^x

. Dividing the numerical coefficients gives

\frac{21}{7} = 3

. Substituting

4^{x-1} = 2^{2x-2}

into the ratio gives

3 \cdot \frac{2^{2x-2}}{2^x} = 3 \cdot 2^{x-2}

. Setting

3 \cdot 2^{x-2} = 24

leads to

2^{x-2} = 8 = 2^3

, which gives

x - 2 = 3

and therefore

x = 5

.

Step-by-Step Solution

1

Factor out common terms from the numerator and denominator

Numerator:

4^{x-1}(4^2 + 4^1 + 1) = 21 \cdot 4^{x-1}

. Denominator:

2^x(2^2 + 2^1 + 1) = 7 \cdot 2^x

.

Grouping power terms simplifies sums of exponential expressions.

2

Divide the numerator by the denominator and convert bases

\frac{21 \cdot 4^{x-1}}{7 \cdot 2^x} = 3 \cdot \frac{(2^2)^{x-1}}{2^x} = 3 \cdot \frac{2^{2x-2}}{2^x}

Simplifying

\frac{21}{7} = 3

and expressing base 4 in base 2 allows applying laws of indices.

3

Apply the quotient rule of indices:

\frac{a^m}{a^n} = a^{m-n}

3 \cdot 2^{(2x-2) - x} = 3 \cdot 2^{x-2}

Subtracting exponents of like bases simplifies the fractional index expression.

4

Set the simplified expression equal to 24 and solve for

x

3 \cdot 2^{x-2} = 24 \implies 2^{x-2} = 8 = 2^3 \implies x - 2 = 3 \implies x = 5

Equating the exponents when bases are identical yields the linear equation

x - 2 = 3

.

Key Concept

Factoring sums of exponential terms and applying the quotient rule of indices

Question 6963Question →

The viscous force $F$ acting on a small sphere of radius $r$ moving with velocity $v$ through a liquid is given by Stokes' law, $F = 6\pi \eta r v$ , where $\eta$ is the coefficient of viscosity. Which of the following expressions represents the base dimensions of $\eta$ ?

M L^{-1} T^{-1}

M L T^{-1}

M L^{-1} T^{-2}

M L^{-2} T^{-1}

Show answer & explanation

Answer: $M L^{-1} T^{-1}$

Answer

M L^{-1} T^{-1}

Rearranging Stokes' law gives

\eta = \frac{F}{6\pi r v}

. Substituting base dimensions

[F] = M L T^{-2}

,

[r] = L

, and

[v] = L T^{-1}

yields

[\eta] = \frac{M L T^{-2}}{L^2 T^{-1}} = M L^{-1} T^{-1}

.

Step-by-Step Solution

1

Express Stokes' law in terms of the coefficient of viscosity

\eta = \frac{F}{6\pi r v}

Isolating

\eta

allows substitution of fundamental dimensions.

2

Substitute fundamental dimensions for force, radius, and velocity

[\eta] = \frac{M L T^{-2}}{L \cdot L T^{-1}}

The constant

6\pi

is dimensionless, while

[F] = M L T^{-2}

,

[r] = L

, and

[v] = L T^{-1}

.

3

Simplify the powers of base quantities

M

,

L

, and

T

[\eta] = M L^{1 - 2} T^{-2 - (-1)} = M L^{-1} T^{-1}

Applying algebraic rules of exponents simplifies the expression.

Key Concept

Dimensions of Physical Quantities and Dimensional Analysis

Question 6964Question →

A charged oil droplet of mass $3.2 \times 10^{-15}\text{ kg}$ remains stationary in a vacuum between two horizontal charged plates where there is a uniform vertical electric field of strength $2.0 \times 10^4\text{ N C}^{-1}$ . Taking $g = 10\text{ m s}^{-2}$ and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , determine the number of excess electrons on the droplet.

Show answer & explanation

Answer: 10

Answer

The number of excess electrons on the droplet is 10.

The droplet is in mechanical equilibrium under two equal and opposite forces: the downward gravitational force

W = mg

and the upward electric force

F_e = qE

. Setting

qE = mg

gives

q = \frac{mg}{E} = 1.6 \times 10^{-18}\text{ C}

. By the quantization of charge (

q = Ne

), dividing this charge by the elementary charge

e = 1.6 \times 10^{-19}\text{ C}

gives exactly 10 excess electrons.

Step-by-Step Solution

1

Calculate the gravitational force (weight) acting on the droplet.

W = mg = (3.2 \times 10^{-15}\text{ kg}) \times (10\text{ m s}^{-2}) = 3.2 \times 10^{-14}\text{ N}

.

For stationary equilibrium, weight provides the downward vertical force.

2

Apply the equilibrium condition to find the electric force.

F_e = W = 3.2 \times 10^{-14}\text{ N}

.

The net vertical force must be zero for the droplet to remain suspended.

3

Determine the charge

q

using

F_e = qE

.

q = \frac{F_e}{E} = \frac{3.2 \times 10^{-14}\text{ N}}{2.0 \times 10^4\text{ N C}^{-1}} = 1.6 \times 10^{-18}\text{ C}

.

Electric field strength relates force and charge.

4

Calculate the number of elementary charges using charge quantization

q = Ne

.

N = \frac{q}{e} = \frac{1.6 \times 10^{-18}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 10

.

Electric charge exists in discrete integer multiples of the elementary charge

e

.

Key Concept

Equilibrium between electrostatic and gravitational forces combined with charge quantization.

Question 6965Question →

During the smelting stage of copper extraction, copper pyrites (

CuFeS_2

) is roasted in air according to the following balanced equation:

2CuFeS_{2(s)} + 4O_{2(g)} \rightarrow Cu_2S_{(s)} + 2FeO_{(s)} + 3SO_{2(g)}

If

0.50\text{ mole}

of

CuFeS_2

is reacted with

0.80\text{ mole}

of

O_2

, which reagent is limiting, what volume of

SO_2

gas is produced at STP, and what is the oxidation state of iron in the resulting

FeO

?
[Molar volume of gas at STP =

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

]

O_2

is limiting;

13.44\text{ dm}^3

of

SO_2

; iron oxidation state is

+2

CuFeS_2

is limiting;

16.80\text{ dm}^3

of

SO_2

; iron oxidation state is

+2

O_2

is limiting;

14.40\text{ dm}^3

of

SO_2

; iron oxidation state is

+2

O_2

is limiting;

13.44\text{ dm}^3

of

SO_2

; iron oxidation state is

0

Show answer & explanation

Answer: $O_2$ is limiting; $13.44\text{ dm}^3$ of $SO_2$ ; iron oxidation state is $+2$

Answer

Oxygen gas (

O_2

) is the limiting reactant, yielding

13.44\text{ dm}^3

of

SO_2

gas at STP, and iron in

FeO

has an oxidation state of

+2

.

The correct answer identifies oxygen gas (

O_2

) as the limiting reactant because

0.50\text{ mole}

of

CuFeS_2

requires

1.00\text{ mole}

of

O_2

for complete reaction, but only

0.80\text{ mole}

is provided. Utilizing

O_2

to calculate product yield gives

0.60\text{ mole}

of

SO_2

, which translates to

13.44\text{ dm}^3

at STP (

0.60 \times 22.4

). Iron in

FeO

carries an oxidation state of

+2

.

Step-by-Step Solution

1

Determine the limiting reactant by comparing the mole ratio of available reactants to stoichiometric coefficients

From the balanced equation,

2\text{ moles}

of

CuFeS_2

require

4\text{ moles}

of

O_2

(ratio

1:2

).

0.50\text{ mole}

of

CuFeS_2

requires

1.00\text{ mole}

of

O_2

. Since only

0.80\text{ mole}

of

O_2

is available,

O_2

is the limiting reactant.

The reaction extent is governed entirely by the reactant that is fully consumed first.

2

Calculate the moles of

SO_2

produced using the limiting reactant

According to the stoichiometric ratio,

4\text{ moles}

of

O_2

produce

3\text{ moles}

of

SO_2

. Therefore,

0.80\text{ mole}

of

O_2

produces

0.80 \times \frac{3}{4} = 0.60\text{ mole}

of

SO_2

.

The yield of product depends directly on the moles of the limiting reactant.

3

Convert moles of

SO_2

gas to volume at STP

Volume of

SO_2 = 0.60\text{ mole} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 13.44\text{ dm}^3

.

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

.

4

Determine the oxidation state of iron in

FeO

Oxygen has an oxidation state of

-2

. For neutral

FeO

,

\text{Fe} + (-2) = 0 \Rightarrow \text{Fe} = +2

.

The sum of oxidation states in a neutral compound equals zero.

Key Concept

Copper extraction roasting reaction stoichiometry and limiting reactant calculations

Estimated Time:2m 30s

Question 6966Question →

Supply the correct literary term in the blank provided to complete the statement regarding dramatic subgenres.

Fill in the blanks below

A tragic subgenre that centers on the emotional struggles, marital conflicts, and downfall of ordinary middle-class or working-class protagonists in everyday private settings, rather than heroic figures of royal lineage, is called .

Show answer & explanation

Answer

The correct term is domestic tragedy (or domestic drama).

Domestic tragedy is a dramatic subgenre that developed to depict serious, fatal conflicts among ordinary, middle-class or working-class individuals in household or local settings, departing from the classical tradition of noble protagonists.

Step-by-Step Solution

1

Examine the class status of the characters and setting described in the statement.

Noted that the characters are ordinary middle-class or working-class individuals in familiar private settings, contrasting with classical tragic heroes.

Classical tragedy requires royal or aristocratic figures of high stature, whereas domestic tragedy shifts the scope to common citizens.

2

Identify the subgenre of drama defined by this specific social focus and thematic concern.

Determined that drama depicting personal and household misfortunes of non-aristocratic figures is classified as domestic tragedy.

The term 'domestic' highlights the private, household setting and social rank of the protagonist within tragic drama.

Key Concept

Domestic Tragedy

Estimated Time:1m 0s

Question 6967Question →

A curve has the equation $y = \frac{1}{3}x^3 - 2x^2 + 3x + 1$ . What is the positive $x$ -coordinate of the point on the curve where the tangent line is parallel to the line $y = 8x - 5$ ?

Show answer & explanation

Answer: 5

Answer

The positive

x

-coordinate of the point where the tangent is parallel to the line is

5

.

The gradient of the line

y = 8x - 5

is

8

. Differentiating

y = \frac{1}{3}x^3 - 2x^2 + 3x + 1

gives

\frac{dy}{dx} = x^2 - 4x + 3

. Setting

\frac{dy}{dx} = 8

leads to

x^2 - 4x - 5 = 0

, which factors as

(x - 5)(x + 1) = 0

. The solutions are

x = 5

and

x = -1

. Selecting the positive value gives

x = 5

.

Step-by-Step Solution

1

Find the gradient (slope) of the given straight line.

The line equation is in slope-intercept form

y = mx + c

, where the slope

m = 8

.

Parallel lines have equal slopes, so the gradient of the tangent to the curve must equal 8.

2

Differentiate the curve equation to find the gradient function

\frac{dy}{dx}

.

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - 2x^2 + 3x + 1\right) = x^2 - 4x + 3

.

The first derivative of a curve represents the gradient of the tangent at any point

x

.

3

Equate the derivative to the slope of the line and solve for

x

.

x^2 - 4x + 3 = 8 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0

. The roots are

x = 5

and

x = -1

.

Solving the quadratic equation gives all

x

-values where the tangent line has a slope of 8.

4

Select the positive

x

-coordinate as requested.

x = 5

.

The question specifically asks for the positive value among the solutions.

Key Concept

Finding points on a curve where the tangent is parallel to a given line

Question 6968Question →

If $\log_4 x + \log_x 16 = 3$ , what is the sum of the possible values of $x$ ?

20

12

48

3

Show answer & explanation

Answer: 20

Answer

The sum of the possible values of

x

is 20.

Applying the change of base rule converts

\log_x 16

to

\frac{\log_4 16}{\log_4 x} = \frac{2}{\log_4 x}

. Setting

y = \log_4 x

transforms the equation into

y + \frac{2}{y} = 3

, which rearranges to

y^2 - 3y + 2 = 0

. The roots of this quadratic equation are

y = 1

and

y = 2

. Converting back to

x

yields

x = 4^1 = 4

and

x = 4^2 = 16

. Summing these solutions gives

4 + 16 = 20

.

Step-by-Step Solution

1

Apply the change of base formula to express

\log_x 16

in base 4.

\log_x 16 = \frac{\log_4 16}{\log_4 x} = \frac{2}{\log_4 x}

.

Logarithmic bases must be unified to combine terms.

2

Substitute

y = \log_4 x

into the original equation.

y + \frac{2}{y} = 3

.

Simplifies the equation into a quadratic form in terms of

y

.

3

Multiply by

y

and solve the quadratic equation

y^2 - 3y + 2 = 0

.

(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2

.

Finds the exponential power values.

4

Solve for

x

using

x = 4^y

and calculate the sum.

For

y = 1

,

x = 4^1 = 4

; for

y = 2

,

x = 4^2 = 16

. Sum =

4 + 16 = 20

.

Converts back from logarithmic space to solve for

x

and finds the requested sum.

Key Concept

Change of base rule and solving logarithmic quadratic equations

Question 6969Question →

A closed vessel initially contains air saturated with water vapour at $27^\circ\text{C}$ under a total pressure of $1.04 \times 10^5\text{ Pa}$ . Given that the saturated vapour pressure of water at $27^\circ\text{C}$ is $0.04 \times 10^5\text{ Pa}$ , what is the final total pressure inside the vessel if its volume is compressed to half of its initial volume while maintaining the temperature at $27^\circ\text{C}$ ?

2.04 \times 10^5\text{ Pa}

2.08 \times 10^5\text{ Pa}

2.00 \times 10^5\text{ Pa}

1.08 \times 10^5\text{ Pa}

Show answer & explanation

Answer: $2.04 \times 10^5\text{ Pa}$

Answer

The final total pressure inside the vessel is

2.04 \times 10^5\text{ Pa}

.

The total pressure is the sum of the partial pressures of dry air and saturated water vapour. Under isothermal compression to half volume, dry air follows Boyle's law and its partial pressure doubles from

1.00 \times 10^5\text{ Pa}

to

2.00 \times 10^5\text{ Pa}

. The saturated water vapour pressure remains constant at

0.04 \times 10^5\text{ Pa}

because liquid condenses out. Summing these yields

2.04 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Calculate the initial partial pressure of the dry air.

P_{\text{air, 1}} = P_{\text{total, 1}} - P_{\text{vapour, 1}} = 1.04 \times 10^5\text{ Pa} - 0.04 \times 10^5\text{ Pa} = 1.00 \times 10^5\text{ Pa}

.

According to Dalton's law of partial pressures, the total pressure of a gas mixture is the sum of the partial pressures of its individual components.

2

Determine the final partial pressure of dry air after isothermal compression.

P_{\text{air, 2}} = P_{\text{air, 1}} \times \frac{V_1}{V_2} = 1.00 \times 10^5\text{ Pa} \times 2 = 2.00 \times 10^5\text{ Pa}

.

Dry air behaves as an ideal gas and follows Boyle's law (

P_1 V_1 = P_2 V_2

) at constant temperature.

3

Determine the final partial pressure of the saturated water vapour.

P_{\text{vapour, 2}} = 0.04 \times 10^5\text{ Pa}

.

Saturated vapour pressure depends strictly on temperature. When compressed at constant temperature, excess vapour condenses into liquid, keeping the partial pressure constant at the saturated value.

4

Sum the final partial pressures to find the new total pressure.

P_{\text{total, 2}} = P_{\text{air, 2}} + P_{\text{vapour, 2}} = 2.00 \times 10^5\text{ Pa} + 0.04 \times 10^5\text{ Pa} = 2.04 \times 10^5\text{ Pa}

.

The total final pressure is the sum of the new dry air pressure and the unchanged saturated vapour pressure.

Key Concept

Saturated Vapour Pressure and Gas Law Applications

Estimated Time:2m 0s

Question 6970Question →

A solid binary compound of oxygen and lead, $\text{PbO}_2$ , reacts with concentrated hydrochloric acid to yield lead(II) chloride, water, and chlorine gas, but fails to produce hydrogen peroxide when treated with cold dilute tetraoxosulfate(VI) acid. Based on this chemical behavior, which of the following statements correctly classifies $\text{PbO}_2$ and distinguishes it from a peroxide such as sodium peroxide ( $\text{Na}_2\text{O}_2$ )?

\text{PbO}_2

is a dioxide containing

\text{O}^{2-}

ions with lead in the

+4

oxidation state, whereas

\text{Na}_2\text{O}_2

is a peroxide containing

\text{O}_2^{2-}

ions.

\text{PbO}_2

is an amphoteric oxide that acts purely as a neutral compound, whereas

\text{Na}_2\text{O}_2

is an acidic anhydride.

\text{PbO}_2

is classified as a peroxide because its formula contains two oxygen atoms, whereas

\text{Na}_2\text{O}_2

is a basic monoxide.

\text{PbO}_2

contains the

\text{O}_2^{2-}

ion which oxidizes chloride ions to chlorine gas, whereas

\text{Na}_2\text{O}_2

contains

\text{O}^{2-}

ions.

Show answer & explanation

Answer: $\text{PbO}_2$ is a dioxide containing $\text{O}^{2-}$ ions with lead in the $+4$ oxidation state, whereas $\text{Na}_2\text{O}_2$ is a peroxide containing $\text{O}_2^{2-}$ ions.

Answer

Lead(IV) oxide (

\text{PbO}_2

) is classified as a dioxide containing

\text{O}^{2-}

ions with lead in the

+4

oxidation state, whereas sodium peroxide (

\text{Na}_2\text{O}_2

) is a peroxide containing the

\text{O}_2^{2-}

ion.

Lead(IV) oxide (

\text{PbO}_2

) is classified as a dioxide because it contains simple oxide ions (

\text{O}^{2-}

) with lead in the

+4

oxidation state. It acts as an oxidizing agent by reacting with concentrated

\text{HCl}

to liberate chlorine gas, but does not produce hydrogen peroxide when treated with cold dilute acids. In contrast, sodium peroxide (

\text{Na}_2\text{O}_2

) contains the peroxide ion (

\text{O}_2^{2-}

) and produces

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

upon reaction with cold dilute acids.

Step-by-Step Solution

1

Analyze the structural difference between peroxides and dioxides.

Peroxides contain the peroxide anion

(\text{O}_2^{2-})

where oxygen has an oxidation state of

-1

. Dioxides contain standard oxide anions

(\text{O}^{2-})

where oxygen has an oxidation state of

-2

and the metal is in the

+4

oxidation state.

Chemical classification depends on the specific ionic species present in the oxide crystal lattice.

2

Evaluate the reaction of

\text{PbO}_2

with dilute acid.

\text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow \text{Pb(SO}_4)_2 + 2\text{H}_2\text{O}

(no

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

formed).

True peroxides yield hydrogen peroxide

(\text{H}_2\text{O}_2)

upon treatment with cold dilute acids, whereas dioxides do not.

3

Evaluate the reaction of

\text{PbO}_2

with concentrated hydrochloric acid.

\text{PbO}_2 + 4\text{HCl} \rightarrow \text{PbCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2\uparrow

.

The

\text{Pb}^{4+}

ion acts as an oxidizing agent, oxidizing

\text{Cl}^-

to

\text{Cl}_2

gas, confirming its behavior as a higher dioxide.

Key Concept

Distinction between Peroxides and Dioxides

Estimated Time:1m 30s

Question 6971Question →

A step-down transformer connected to a $240\text{ V}$ AC mains supply operates a $12\text{ V}, 48\text{ W}$ lamp at its normal brightness rating. If the efficiency of the transformer is $80\%$ , what is the electric current drawn by the primary winding from the mains supply?

0.16\text{ A}

0.20\text{ A}

0.25\text{ A}

4.00\text{ A}

Show answer & explanation

Answer: $0.25\text{ A}$

Answer

0.25\text{ A}

The output power delivered to the lamp is

48\text{ W}

. Accounting for the transformer's

80\%

efficiency, the input power required at the primary winding is

\frac{48\text{ W}}{0.80} = 60\text{ W}

. Since the primary voltage is

240\text{ V}

, the primary current drawn is

I_p = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}

.

Step-by-Step Solution

1

Determine the power output at the secondary winding

P_s = 48\text{ W}

The lamp operates at its normal rating, so secondary power equals the lamp rating.

2

Calculate the required power input to the primary winding using transformer efficiency

P_p = \frac{P_s}{\eta} = \frac{48\text{ W}}{0.80} = 60\text{ W}

Efficiency is defined as

\eta = \frac{P_s}{P_p}

, meaning primary input power must exceed secondary output power due to losses.

3

Compute the primary current drawn from the supply

I_p = \frac{P_p}{V_p} = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}

Power in an AC primary circuit is given by

P_p = V_p I_p

assuming a purely resistive secondary load.

Key Concept

Transformer Efficiency and Power Transfer in Electromagnetic Induction

Question 6972Question →

Element $Y$ exists naturally as two isotopes, $^{63}Y$ and $^{65}Y$ . If the relative atomic mass of element $Y$ is $63.5$ , what is the percentage abundance of the lighter isotope, $^{63}Y$ ?

75\%

25\%

50\%

35\%

Show answer & explanation

Answer: $75\%$

Answer

The percentage abundance of the lighter isotope

^{63}Y

is

75\%

.

The percentage abundance is calculated using the weighted average formula:

\text{RAM} = \frac{(m_1 \times x_1) + (m_2 \times x_2)}{100}

. Substituting the given values yields

63.5 = \frac{63x + 65(100-x)}{100}

, which simplifies to

-2x = -150

, giving

x = 75\%

for

^{63}Y

.

Step-by-Step Solution

1

Define variables for isotopic abundances.

Let the abundance of

^{63}Y

be

x\%

. The abundance of

^{65}Y

is therefore

(100 - x)\%

.

The sum of all relative isotopic abundances for a natural element equals

100\%

.

2

Set up the weighted average formula for relative atomic mass.

\text{Relative Atomic Mass} = \frac{(63 \times x) + 65 \times (100 - x)}{100} = 63.5

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

3

Solve the equation for

x

.

63x + 6500 - 65x = 6350 \implies -2x = -150 \implies x = 75

Simplifying algebraic terms gives the percentage abundance of the lighter isotope.

Key Concept

Isotopic Abundance and Relative Atomic Mass Calculation

Estimated Time:1m 30s

Question 6973Question →

Match each condensed organic structural formula listed on the left with its correct functional group classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

\text{CH}_3\text{COCH}_3

matches with Alkanone;

\text{CH}_3\text{CH}_2\text{CHO}

matches with Alkanal;

\text{CH}_3\text{CH}_2\text{COOH}

matches with Alkanoic acid;

\text{HCOOCH}_3

matches with Alkanoate (Ester).

Each condensed formula corresponds directly to its functional group family:

\text{CH}_3\text{COCH}_3

is propanone (an alkanone),

\text{CH}_3\text{CH}_2\text{CHO}

is propanal (an alkanal),

\text{CH}_3\text{CH}_2\text{COOH}

is propanoic acid (an alkanoic acid), and

\text{HCOOCH}_3

is methyl methanoate (an ester / alkanoate).

Step-by-Step Solution

1

Identify the characteristic functional group in each condensed formula.

\text{CH}_3\text{COCH}_3

has a non-terminal carbonyl

\text{-CO-}

;

\text{CH}_3\text{CH}_2\text{CHO}

has a formyl

\text{-CHO}

;

\text{CH}_3\text{CH}_2\text{COOH}

has a carboxyl

\text{-COOH}

;

\text{HCOOCH}_3

has an ester linkage

\text{-COO-}

.

Functional groups determine the class of organic compounds.

2

Match each identified group to its IUPAC homologous series classification.

Non-terminal carbonyl (

\text{-CO-}

) = Alkanone; Formyl (

\text{-CHO}

) = Alkanal; Carboxyl (

\text{-COOH}

) = Alkanoic acid; Ester linkage (

\text{-COO-}

) = Alkanoate (Ester).

IUPAC nomenclature categorizes compounds based on these specific functional group structures.

Key Concept

Classification of Organic Compounds by Functional Groups

Question 6974Question →

A municipal water treatment facility processes raw river water containing fine colloidal clay particles, dissolved hydrogen sulfide, soluble iron(II) compounds, and harmful microorganisms. During the treatment process, the water is subjected to aeration, addition of potash alum, sedimentation, sand filtration, and chlorination. Which of the following statements correctly distinguishes the primary chemical roles of aeration and potash alum in this treatment sequence?

Aeration oxidizes soluble iron(II) compounds and expels dissolved foul-smelling gases, whereas potash alum induces coagulation of fine colloidal clay particles.

Aeration precipitates dissolved calcium tetraoxosulfate(VI) to remove permanent hardness, whereas potash alum sterilizes harmful microorganisms.

Aeration adjusts the pH of acidic water by dissolving atmospheric carbon(IV) oxide, whereas potash alum removes dissolved magnesium ions to soften the water.

Aeration kills pathogenic bacteria through dissolved oxygen toxicity, whereas potash alum neutralizes organic foul odors and tastes.

Show answer & explanation

Answer: Aeration oxidizes soluble iron(II) compounds and expels dissolved foul-smelling gases, whereas potash alum induces coagulation of fine colloidal clay particles.

Answer

Aeration oxidizes soluble iron(II) compounds and expels dissolved foul-smelling gases, whereas potash alum induces coagulation of fine colloidal clay particles.

In municipal water purification, aeration serves to bubble air through raw water, oxidizing dissolved iron(II) salts to insoluble iron(III) precipitates and stripping out dissolved volatile gases (such as hydrogen sulfide and carbon(IV) oxide). Potash alum (hydrated potassium aluminium tetraoxosulfate(VI)) supplies aluminium ions that coagulate fine suspended colloidal particles into larger flocs that readily settle out during sedimentation.

Step-by-Step Solution

1

Analyze the primary chemical and physical effects of aeration in municipal water treatment.

Aeration increases dissolved oxygen, which oxidizes soluble

\text{Fe}^{2+}

ions to insoluble

\text{Fe(OH)}_3

precipitates, while simultaneously purging volatile dissolved gases such as

\text{H}_2\text{S}

and excess

\text{CO}_2

.

Aeration targets volatile odorous compounds and oxidizable dissolved inorganic species.

2

Analyze the role of potash alum (

\text{KAl(SO}_4\text{)}_2 \cdot 12\text{H}_2\text{O}

) during water purification.

Potash alum dissociates to yield

\text{Al}^{3+}

ions, which neutralize the negative surface charges on microscopic colloidal clay particles, allowing them to coagulate and settle during sedimentation.

Colloidal suspensions will not settle naturally under gravity without chemical coagulation.

3

Evaluate the choices to distinguish these functions from water softening and sterilization.

Sterilization is achieved by chlorination, and water softening requires removal of

\text{Ca}^{2+}

or

\text{Mg}^{2+}

ions via chemical precipitation or ion exchange, confirming the statement regarding oxidation/gas expulsion for aeration and coagulation for alum.

Distinguishing municipal water treatment for domestic supply (removal of turbidity and pathogens) from water softening avoids common conceptual confusion.

Key Concept

Distinct roles of aeration (gas removal and oxidation) and coagulation (potash alum flocculation) in municipal water treatment

Question 6975Question →

A fruit vendor has $7$ distinct types of fresh fruits on display. A customer wants to buy a gift basket containing exactly $4$ different types of fruits. How many different combinations of fruits can the customer choose?

Show answer & explanation

Answer: 35

Answer

The customer can choose the fruits in 35 different ways.

The number of ways to choose

r = 4

items from

n = 7

distinct items without regard to order is given by

^7C_4 = \frac{7!}{4!3!} = \frac{210}{6} = 35

.

Step-by-Step Solution

1

Identify total elements (

n

) and selected subset size (

r

).

n = 7

and

r = 4

.

Since the selection order inside the fruit basket does not matter, combinations (

nCr

) must be used.

2

Substitute values into the combination formula

^nC_r = \frac{n!}{r!(n-r)!}

.

^7C_4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!}

.

This evaluates the total ways to choose 4 items from 7 without repetition or ordering.

3

Simplify the factorials and compute the numerical result.

\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

.

Canceling out

4!

from numerator and denominator gives

(7 \times 6 \times 5) / 6 = 35

.

Key Concept

Combinations (

nCr

)

Estimated Time:45s

Question 6976Question →

An aqueous solution of hydrocyanic acid ( $\text{HCN}$ ), a weak monobasic acid, has a concentration of $0.40\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Given that the acid dissociation constant ( $K_a$ ) for $\text{HCN}$ is $4.9 \times 10^{-10}\text{ mol dm}^{-3}$ , calculate the hydrogen ion concentration, $[\text{H}^+]$ , in $\text{mol dm}^{-3}$ . Express your answer as the coefficient $A$ in the form $A \times 10^{-5}\text{ mol dm}^{-3}$ .

Show answer & explanation

Answer: 1.4

Answer

The coefficient A is 1.4, which corresponds to a hydrogen ion concentration of

1.4 \times 10^{-5}\text{ mol dm}^{-3}

.

For a weak monobasic acid, the hydrogen ion concentration is determined using the weak acid ionization relationship

[\text{H}^+] = \sqrt{K_a \cdot c}

. Substituting

K_a = 4.9 \times 10^{-10}\text{ mol dm}^{-3}

and

c = 0.40\text{ mol dm}^{-3}

yields

[\text{H}^+] = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}

, giving a coefficient of 1.4.

Step-by-Step Solution

1

Write the ionization reaction and equilibrium constant expression

\text{HCN}_{(aq)} \rightleftharpoons \text{H}^+_{(aq)} + \text{CN}^-_{(aq)}

, giving

K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}

Hydrocyanic acid is a weak monobasic acid that ionizes partially in water.

2

Apply weak acid approximations

Since

[\text{H}^+] = [\text{CN}^-]

and

K_a

is extremely small,

[\text{HCN}] \approx c = 0.40\text{ mol dm}^{-3}

. Thus,

K_a = \frac{[\text{H}^+]^2}{c}

.

The negligible ionization degree allows the equilibrium concentration of un-ionized acid to be approximated as its initial concentration.

3

Substitute values and solve for

[\text{H}^+]

[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{4.9 \times 10^{-10} \times 0.40} = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}

Multiplying

K_a

by the molar concentration gives the square of the hydrogen ion concentration.

4

Determine the coefficient A

A = 1.4

Matching

1.4 \times 10^{-5}\text{ mol dm}^{-3}

to the requested standard scientific notation form

A \times 10^{-5}\text{ mol dm}^{-3}

yields

A = 1.4

.

Key Concept

Weak Acid Ionization Equilibrium and Ka Calculations

Question 6977Question →

Complete the balanced reduction half-reaction equation for the conversion of nitrate ions to nitrogen monoxide gas in an acidic medium by providing the missing stoichiometric coefficients. What are the values of the coefficients for hydrogen ions and electrons?

Fill in the blanks below

$$\text{NO}_3^-(\text{aq}) + \text{H}^+(\text{aq}) + e^- \rightarrow \text{NO}(\text{g}) + 2\text{H}_2\text{O}(\text{l})$$

Show answer & explanation

Answer

The balanced reduction half-reaction requires 4 hydrogen ions (H⁺) and 3 electrons (e⁻).

The balanced half-reaction is NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l). Four hydrogen ions balance the four hydrogens present in the two water molecules on the right. Three electrons balance the net charge (+3 on the left reactant side versus 0 on the neutral product side), confirming that nitrogen is reduced from oxidation state +5 in NO₃⁻ to +2 in NO.

Step-by-Step Solution

1

Balance oxygen atoms using water molecules

One nitrate ion (NO₃⁻) has 3 oxygen atoms, and NO has 1 oxygen atom, requiring 2 H₂O molecules on the product side: NO₃⁻ → NO + 2 H₂O

In aqueous acid media, oxygen atoms are balanced by adding H₂O molecules to the oxygen-deficient side.

2

Balance hydrogen atoms using hydrogen ions

The right side has 4 hydrogen atoms in 2 H₂O, so 4 H⁺ ions must be added to the left side: NO₃⁻ + 4 H⁺ → NO + 2 H₂O

Hydrogen atoms in acidic media are balanced by adding H⁺ ions to the hydrogen-deficient side.

3

Balance electrical charge using electrons

Left side net charge = (-1) + 4(+1) = +3. Right side net charge = 0. Adding 3 electrons (3 e⁻) to the left side gives a net charge of 0 on both sides: NO₃⁻ + 4 H⁺ + 3 e⁻ → NO + 2 H₂O

Electrons are added to the side with the higher net positive charge to satisfy conservation of charge.

Key Concept

Half-Reaction Method for Balancing Redox Equations in Acidic Medium

Estimated Time:1m 30s

Question 6978Question →

An organic liquid $X$ with the molecular formula $C_4H_8O$ forms a yellow precipitate when warmed with iodine in alkaline solution, but shows no observable reaction with Tollen's reagent. When compound $X$ is reacted with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ ) in dry ether, it produces compound $Y$ . What is the IUPAC name of compound $Y$ ?

Butan-2-ol

Butan-1-ol

2-Methylpropan-2-ol

Butanoic acid

Show answer & explanation

Answer: Butan-2-ol

Answer

Butan-2-ol

The positive iodoform test shows the presence of a methyl ketone carbonyl group (

CH_3C=O

), while the negative Tollen's test confirms that compound X is a ketone (butan-2-one) rather than an aldehyde. Reducing a ketone with lithium tetrahydridoaluminate(III) (

LiAlH_4

) converts the carbonyl carbon to a secondary alcohol group, yielding butan-2-ol.

Step-by-Step Solution

1

Analyze the chemical test results to identify the functional group of compound

X

.

Compound

X

(

C_4H_8O

) gives a positive triiodomethane (iodoform) test (yellow precipitate) but a negative Tollen's test (no silver mirror). This confirms that compound

X

is a methyl ketone, specifically butan-2-one (

CH_3COCH_2CH_3

).

Alkanals reduce Tollen's reagent to silver metal, whereas alkanones do not. The positive iodoform test indicates the presence of a

CH_3C=O

carbonyl group.

2

Determine the reaction outcome when compound

X

is reduced using

LiAlH_4

.

Reduction of the ketone butan-2-one with

LiAlH_4

converts the carbonyl group (

C=O

) into a secondary alcohol group (

-CH(OH)-

).

LiAlH_4

acts as a hydride donor reducing agent, turning alkanones into secondary alkanols.

3

Deduce the IUPAC name of the resulting product compound

Y

.

Reducing

CH_3COCH_2CH_3

produces

CH_3CH(OH)CH_2CH_3

, which is named butan-2-ol.

The four-carbon chain with the hydroxyl group at carbon-2 is named butan-2-ol.

Key Concept

Distinction between alkanals and alkanones via Tollen's and Iodoform tests, and reduction of ketones to secondary alcohols.

Question 6979Question →

Agricultural soil adjacent to a battery recycling plant became severely contaminated with toxic lead compounds. Environmental engineers applied slaked lime, $\text{Ca(OH)}_2$ , to immobilize the heavy metal by precipitating it as insoluble lead(II) hydroxide, $\text{Pb(OH)}_2$ . However, excessive liming caused the soil pH to rise above $10.5$ , triggering a secondary leaching event where lead concentrations in groundwater rapidly increased. Which chemical reaction accounts for the unexpected re-mobilization of lead under strongly alkaline soil conditions?

Formation of soluble tetrahydroxoplumbate(II) complex ions,

[\text{Pb(OH)}_4]^{2-}

, due to the amphoteric behavior of lead(II) compounds in excess base

Precipitation of lead(II) hydroxide behaving as a purely basic compound that undergoes complete salt hydrolysis to yield free soluble lead cations

Oxidation of lead(II) ions by stratospheric ozone trapped in soil pores, converting insoluble lead compounds into highly soluble lead(IV) oxide

Condensation of lead ions with soil humic acids to form synthetic addition polymers that undergo rapid microbial degradation

Show answer & explanation

Answer: Formation of soluble tetrahydroxoplumbate(II) complex ions, $[\text{Pb(OH)}_4]^{2-}$ , due to the amphoteric behavior of lead(II) compounds in excess base

Answer

The re-mobilization of lead in strongly alkaline soil is caused by the formation of soluble tetrahydroxoplumbate(II) complex ions,

[\text{Pb(OH)}_4]^{2-}

, owing to the amphoteric character of lead(II) compounds.

Lead(II) compounds exhibit amphoterism. While moderate liming precipitates lead as insoluble

\text{Pb(OH)}_2

, over-liming creates a high concentration of

\text{OH}^-

ions (pH > 10.5). This causes the precipitate to dissolve by forming the soluble complex anion

[\text{Pb(OH)}_4]^{2-}

(tetrahydroxoplumbate(II)), leading to groundwater pollution.

Step-by-Step Solution

1

Analyze the chemical behavior of lead(II) hydroxide upon initial neutralization.

Adding moderate amounts of base causes

\text{Pb}^{2+}

ions to precipitate as insoluble lead(II) hydroxide:

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

.

This temporarily immobilizes heavy metal contamination in the soil matrix.

2

Determine the effect of excess hydroxide ions (pH > 10.5) on the precipitate.

Because lead is an amphoteric metal (similar to aluminium and zinc), its oxide and hydroxide react with excess alkali:

\text{Pb(OH)}_{2(\text{s})} + 2\text{OH}^-_{(\text{aq})} \rightarrow [\text{Pb(OH)}_4]^{2-}_{(\text{aq})}

.

The formation of the complex tetrahydroxoplumbate(II) anion brings lead back into aqueous solution, increasing heavy metal leaching into groundwater.

Key Concept

Amphoteric behavior of heavy metals in soil remediation and waste management

Estimated Time:2m 0s

Question 6980Question →

A filamentous cyanobacterium performing oxygenic photosynthesis in an aquatic habitat is also capable of fixing atmospheric nitrogen. Which cellular adaptation enables this organism to fix nitrogen efficiently without inactivating the oxygen-sensitive enzyme nitrogenase?

Isolation of nitrogenase within specialized thick-walled cells called heterocysts that lack oxygen-evolving photosystem II

Synthesis of a protective cell wall composed of cellulose to completely prevent oxygen gas entry into all vegetative cells

Secretion of a protein capsid around the nuclear region to compartmentalize respiratory and nitrogen-fixing enzymes

Direct conversion of atmospheric nitrogen into nitrates within vegetative cells using Nitrosomonas enzymes

Show answer & explanation

Answer: Isolation of nitrogenase within specialized thick-walled cells called heterocysts that lack oxygen-evolving photosystem II

Answer

Isolation of nitrogenase within specialized thick-walled cells called heterocysts that lack oxygen-evolving photosystem II

Nitrogenase is an enzyme that reduces atmospheric nitrogen to ammonia but is destroyed by molecular oxygen. Cyanobacteria overcome this by compartmentalizing nitrogenase inside specialized cells called heterocysts. These heterocysts develop thick walls that impede gas diffusion and inactivate Photosystem II, preventing oxygen generation while allowing nitrogen fixation to take place safely.

Step-by-Step Solution

1

Analyze the biochemical conflict in cyanobacteria

Oxygenic photosynthesis produces oxygen as a byproduct, whereas the enzyme nitrogenase (required for nitrogen fixation) is rapidly and irreversibly inactivated by free oxygen.

Understanding why nitrogen fixation requires physiological or spatial separation from oxygen production.

2

Identify the cellular adaptation in Kingdom Monera (Cyanobacteria)

Filamentous cyanobacteria differentiate specialized cells called heterocysts. Heterocysts feature thickened cell walls to limit oxygen diffusion and selectively lack Photosystem II (which photolyzes water to yield oxygen), creating an anaerobic environment for nitrogenase activity.

Heterocysts allow simultaneous photosynthesis in vegetative cells and nitrogen fixation in heterocysts.

Key Concept

Cyanobacterial cell specialization and heterocyst function

Estimated Time:1m 30s

Answer

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