All practice questions

Question 8441Question →

Match each substance or chemical reagent related to iron extraction, corrosion, and qualitative analysis on the left with its corresponding chemical role or property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Coke matches with reacting with CO2 to generate CO; Calcium silicate matches with forming molten slag to prevent re-oxidation; Hydrated iron(III) oxide matches with the chemical composition of rust; Potassium hexacyanoferrate(III) matches with testing for Fe(II) ions.

Each substance is matched directly to its chemical function: Coke supplies carbon to generate carbon(II) oxide gas; Calcium silicate acts as slag floating atop molten iron; Hydrated iron(III) oxide is the exact chemical composition of rust; Potassium hexacyanoferrate(III) is the standard bench reagent for detecting iron(II) ions in qualitative testing.

Step-by-Step Solution

1

Analyze the blast furnace chemical reactions involving carbon input.

Coke (

\text{C}

) reacts with ascending

\text{CO}_2

(

\text{C} + \text{CO}_2 \rightarrow 2\text{CO}

) to produce carbon(II) oxide, which acts as the chief reducing agent for haematite.

Identify the role of Coke in blast furnace extraction.

2

Analyze slag formation and its function in the blast furnace.

Lime (

\text{CaO}

) combines with silica (

\text{SiO}_2

) to yield calcium silicate (

\text{CaSiO}_3

), a molten waste slag that floats above molten iron.

Identify the function of calcium silicate in molten iron isolation.

3

Identify the chemical identity of rust.

Atmospheric corrosion of iron in the presence of oxygen and water forms reddish-brown hydrated iron(III) oxide (

\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}

).

Match the rust formula with its physical phenomenon.

4

Recall qualitative analysis reagents for iron oxidation states.

Potassium hexacyanoferrate(III) reacts with

\text{Fe}^{2+}

ions to give a characteristic dark blue precipitate.

Match the analytical reagent with its specific ion test.

Key Concept

Extraction of Iron in the Blast Furnace, Rusting Mechanism, and Qualitative Analysis of Iron Ions

Question 8442Question →

Match each chemical transformation involving alkanols listed on the left with the appropriate reagent, enzyme, or catalyst required on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Glucose is fermented to ethanol using the enzyme zymase; dehydration of ethanol to ethene uses excess concentrated

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

at

170^\circ\text{C}

; ethanol is oxidized to ethanoic acid using acidified

\text{K}_2\text{Cr}_2\text{O}_7

under reflux; and industrial synthesis of ethanol from ethene uses steam with a phosphoric acid (

\text{H}_3\text{PO}_4

) catalyst at high temperature and pressure.

Each chemical process matches its unique catalyst or reaction conditions: zymase catalyzes glucose fermentation to ethanol, excess concentrated

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

at

170^\circ\text{C}

dehydrates ethanol to ethene, acidified

\text{K}_2\text{Cr}_2\text{O}_7

under reflux oxidizes ethanol to ethanoic acid, and phosphoric acid (

\text{H}_3\text{PO}_4

) on silica catalyzes the industrial hydration of ethene to ethanol.

Step-by-Step Solution

1

Identify the biological catalyst for sugar fermentation

Fermentation of glucose (

\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2

) is catalyzed specifically by the enzyme zymase.

Yeast produces zymase, which converts simple hexose sugars directly into ethanol.

2

Identify the reagent and temperature for elimination/dehydration

Heating ethanol with excess concentrated

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

at

170^\circ\text{C}

yields ethene via removal of a water molecule.

Concentrated tetraoxosulfate(VI) acid acts as a dehydrating agent; high temperature (

170^\circ\text{C}

) favors ethene formation over ethoxyethane formation.

3

Identify the oxidizing conditions for full alkanol oxidation

Primary alkanols undergo two-stage oxidation: first to an alkanal, then under reflux with acidified

\text{K}_2\text{Cr}_2\text{O}_7

to an alkanoic acid.

Acidified potassium heptaoxodichromate(VI) is a strong oxidizing agent capable of carrying the oxidation of ethanol fully to ethanoic acid.

4

Identify the industrial catalytic addition reaction conditions

Direct hydration of ethene (

\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}

) uses a phosphoric acid catalyst.

The reversible addition of steam across the double bond of ethene requires a solid phosphoric acid catalyst at

300^\circ\text{C}

and high pressure.

Key Concept

Reagents, enzymes, and conditions for alkanol preparation and reactions

Question 8443Question →

For a chemical reaction carried out at $27^\circ\text{C}$ , the standard Gibbs free energy change ( $\Delta G^\circ$ ) is $-54.0\text{ kJ mol}^{-1}$ . Given that the standard entropy change ( $\Delta S^\circ$ ) for the reaction is $-120.0\text{ J K}^{-1}\text{ mol}^{-1}$ , calculate the standard enthalpy change ( $\Delta H^\circ$ ) in $\text{kJ mol}^{-1}$ .

Show answer & explanation

Answer: -90

Answer

The standard enthalpy change ($\Delta H^\circ$) for the reaction is $-90.0\text{ kJ mol}^{-1}$.

The standard enthalpy change is calculated using the rearranged Gibbs free energy equation

\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ

. Converting

27^\circ\text{C}

to

300\text{ K}

and

-120.0\text{ J K}^{-1}\text{ mol}^{-1}

to

-0.120\text{ kJ K}^{-1}\text{ mol}^{-1}

yields

\Delta H^\circ = -54.0 + (300 \times -0.120) = -90.0\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Convert the temperature from degrees Celsius to Kelvin

T = 300 K

Thermodynamic calculations require absolute temperature in Kelvin: T = 27 + 273 = 300 K.

2

Convert the standard entropy change units from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹

ΔS° = -0.120 kJ K⁻¹ mol⁻¹

Since ΔG° is given in kJ mol⁻¹, ΔS° must be converted to kJ K⁻¹ mol⁻¹ by dividing by 1000.

3

Rearrange the Gibbs free energy equation to express ΔH° and substitute the given values

ΔH° = -90.0 kJ mol⁻¹

From ΔG° = ΔH° - TΔS°, rearranging gives ΔH° = ΔG° + TΔS° = -54.0 + (300 × -0.120) = -90.0 kJ mol⁻¹.

Key Concept

Gibbs Free Energy Equation and Unit Consistency

Question 8444Question →

What is the IUPAC name of the organic compound formed by the reduction of butan-2-one using lithium tetrahydridoaluminate(III) ( $LiAlH_4$ )?

Show answer & explanation

Answer: butan-2-ol; 2-butanol; butan 2 ol; 2-Butanol; Butan-2-ol

Answer

butan-2-ol

The reduction of alkanones (ketones) with reducing agents such as

LiAlH_4

converts the carbonyl group (

C=O

) into a secondary alcohol group (

-CH(OH)-

). Since the carbonyl carbon in butan-2-one is at position 2, the resulting hydroxyl group is also at position 2, yielding the secondary alcohol butan-2-ol.

Step-by-Step Solution

1

Identify the functional group and carbon chain length of the reactant.

Butan-2-one is a 4-carbon alkanone (ketone) with the carbonyl group at carbon-2 (

CH_3-CO-CH_2-CH_3

).

Determining the reactant structure establishes the expected reduction product.

2

Apply the reduction reaction mechanism for alkanones.

Reducing agents such as

LiAlH_4

or

NaBH_4

reduce alkanones to secondary alcohols by adding hydrogen across the

C=O

double bond.

The carbonyl group (

C=O

) is converted to a secondary alcohol group (

-CH(OH)-

).

3

Name the resulting alcohol using IUPAC nomenclature.

CH_3-CH(OH)-CH_2-CH_3

is named butan-2-ol.

The hydroxyl group (

-OH

) remains on carbon-2 of the 4-carbon parent chain.

Key Concept

Reduction of alkanones to secondary alcohols

Estimated Time:1m 0s

Question 8445Question →

Match each reaction condition modification on the left with its corresponding microscopic kinetic mechanism under collision theory on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Replacing zinc granules with zinc dust increases the surface area and total collision frequency; increasing temperature increases average kinetic energy and the fraction of particles with energy equal to or greater than activation energy (

E \ge E_a

); ultraviolet illumination provides photon energy to form free radicals; and adding solid vanadium(V) oxide provides a catalyst that lowers the activation energy pathway.

Each condition matches its microscopic collision theory explanation directly: surface area increases total collision frequency, temperature increases the fraction of molecules with

E \ge E_a

, UV light supplies energy for bond cleavage/radical formation, and catalysts lower the activation energy pathway.

Step-by-Step Solution

1

Analyze the particle size modification for zinc granules vs zinc dust.

Zinc dust has a much greater surface area per unit mass than a solid granule, which allows more collisions per second between

Zn(s)

and

H^+(aq)

ions.

Surface area dictates the contact zone for heterogeneous reactions.

2

Evaluate the effect of temperature increase on kinetic energy distribution.

An increase in temperature shifts the Maxwell-Boltzmann distribution toward higher kinetic energies, exponentially increasing the fraction of effective collisions (

E \ge E_a

).

Reaction rate depends exponentially on temperature via the Arrhenius relationship.

3

Examine the role of light in photochemical reactions.

Ultraviolet light absorbs photon quanta (

E = h\nu

) to homolytically split

Cl_2

into reactive chlorine radicals.

Light intensity and frequency act as energy sources to overcome bond dissociation energy.

4

Assess the function of vanadium(V) oxide (

V_2O_5

).

V_2O_5

acts as a catalyst in the Contact Process, lowering the activation energy barrier for the oxidation of

SO_2

to

SO_3

.

Catalysts alter the reaction pathway to accelerate both forward and reverse rates equally.

Key Concept

Collision Theory and Factors Affecting Rates of Reaction

Question 8446Question →

A sample of pure potassium trioxonitrate(V), $\text{KNO}_3$ , contains $1.806 \times 10^{23}$ oxygen atoms. What is the mass, in grams, of this sample of $\text{KNO}_3$ ? $[\text{K} = 39, \text{N} = 14, \text{O} = 16, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

Show answer & explanation

Answer: 10.1

Answer

The mass of the potassium trioxonitrate(V) sample is 10.1 g.

First, the number of moles of oxygen atoms is found by dividing

1.806 \times 10^{23}

atoms by Avogadro's number (

6.02 \times 10^{23}\text{ mol}^{-1}

), yielding

0.3\text{ mol}

of oxygen. Since one mole of

\text{KNO}_3

contains three moles of oxygen atoms, the moles of

\text{KNO}_3

in the sample is

0.3 / 3 = 0.1\text{ mol}

. Multiplying

0.1\text{ mol}

by the molar mass of

\text{KNO}_3

(

101\text{ g/mol}

) gives the correct mass of

10.1\text{ g}

.

Step-by-Step Solution

1

Calculate the moles of oxygen atoms in the sample

n(\text{O}) = 0.3\text{ mol}

Divide the total number of oxygen atoms by Avogadro's constant (

N_A = 6.02 \times 10^{23}\text{ mol}^{-1}

).

2

Determine the moles of potassium trioxonitrate(V),

\text{KNO}_3

n(\text{KNO}_3) = 0.1\text{ mol}

Each formula unit of

\text{KNO}_3

contains 3 oxygen atoms, so divide the moles of oxygen atoms by 3.

3

Calculate the molar mass of

\text{KNO}_3

M(\text{KNO}_3) = 101\text{ g/mol}

Sum the relative atomic masses:

39\text{ (K)} + 14\text{ (N)} + 3 \times 16\text{ (O)} = 101\text{ g/mol}

.

4

Multiply moles of

\text{KNO}_3

by its molar mass to get total mass

\text{Mass} = 10.1\text{ g}

\text{Mass} = \text{moles} \times \text{molar mass} = 0.1\text{ mol} \times 101\text{ g/mol} = 10.1\text{ g}

.

Key Concept

Relationship between particle count, mole quantity of constituent atoms, and molar mass

Question 8447Question →

Brass is an alloy widely utilized in making musical instruments, door handles, and decorative fittings due to its strength, acoustic properties, and resistance to corrosion. Which pair of metals constitutes brass?

Copper and zinc

Copper and tin

Lead and tin

Iron and carbon

Show answer & explanation

Answer: Copper and zinc

Answer

Copper and zinc

Brass is an alloy specifically produced by melting together copper and zinc. The addition of zinc improves the mechanical strength and workability relative to pure copper.

Step-by-Step Solution

1

Identify the standard elemental composition of the specified metallic alloy.

Brass is a substitutional metallic alloy composed primarily of copper (Cu) and zinc (Zn).

Alloys are mixtures of metals or a metal with another element; brass specifically pairs copper as the primary metal with zinc.

Key Concept

Elemental compositions of common copper alloys (brass vs bronze)

Estimated Time:45s

Question 8448Question →

Under specified laboratory conditions, $120\text{ cm}^3$ of hydrogen gas ( $H_2$ ) diffuses through a porous membrane in a given time interval. What volume of oxygen gas ( $O_2$ ) will diffuse through the same membrane under identical conditions during the same time interval? [ $H = 1, O = 16$ ]

7.5\text{ cm}^3

30\text{ cm}^3

60\text{ cm}^3

480\text{ cm}^3

Show answer & explanation

Answer: $30\text{ cm}^3$

Answer

The volume of oxygen gas that will diffuse in the same time interval is

30\text{ cm}^3

.

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. Since oxygen (

O_2

,

32\text{ g/mol}

) is

16

times heavier than hydrogen (

H_2

,

2\text{ g/mol}

), its rate of diffusion is

\sqrt{16} = 4

times slower. In the same time interval, the volume of oxygen that diffuses is

\frac{120\text{ cm}^3}{4} = 30\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the molar masses of hydrogen gas (

H_2

) and oxygen gas (

O_2

).

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

, and

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

.

Graham's Law relates the rate of diffusion to the inverse square root of molecular masses.

2

Apply Graham's Law of Diffusion to find the ratio of rates of diffusion.

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

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

3

Relate the rate of diffusion to the volume of gas effused over a constant time interval (

t

).

Since

r = \frac{V}{t}

and time

t

is constant,

\frac{V_{H_2}}{V_{O_2}} = 4 \implies V_{O_2} = \frac{V_{H_2}}{4} = \frac{120\text{ cm}^3}{4} = 30\text{ cm}^3

.

The lighter gas diffuses 4 times faster than the heavier gas, so only a quarter of the volume of oxygen diffuses in the same time.

Key Concept

Graham's Law of Diffusion

Question 8449Question →

When

6.62\text{ g}

of lead(II) trioxonitrate(V),

\text{Pb(NO}_3)_2

, is heated strongly until complete decomposition occurs according to the equation:

2\text{Pb(NO}_3)_2(s) \rightarrow 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)

What is the total volume of gaseous products evolved at standard temperature and pressure (STP)?
[

M_{\text{r}}(\text{Pb(NO}_3)_2) = 331\text{ g mol}^{-1}

; Molar volume of gas at STP

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

]

1.12\text{ dm}^3

1.20\text{ dm}^3

0.896\text{ dm}^3

0.448\text{ dm}^3

Show answer & explanation

Answer: $1.12\text{ dm}^3$

Answer

The total volume of gaseous products evolved at STP is

1.12\text{ dm}^3

.

Decomposition of

0.02\text{ mol}

of lead(II) trioxonitrate(V) yields

0.04\text{ mol}

of

\text{NO}_2

and

0.01\text{ mol}

of

\text{O}_2

, making

0.05\text{ mol}

of total gas. At STP,

0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount in moles of

\text{Pb(NO}_3)_2

decomposed

\text{Moles of Pb(NO}_3)_2 = \frac{6.62\text{ g}}{331\text{ g mol}^{-1}} = 0.02\text{ mol}

Dividing given mass by relative formula mass gives the mole amount.

2

Determine total moles of gaseous products using stoichiometric ratios

From the balanced equation,

2\text{ mol of Pb(NO}_3)_2

yields

4\text{ mol of NO}_2(g) + 1\text{ mol of O}_2(g) = 5\text{ mol of gas}

. Total gas moles

= 0.02 \times \frac{5}{2} = 0.05\text{ mol}

.

Both

\text{NO}_2

and

\text{O}_2

are gases at STP, so their mole quantities must be summed.

3

Calculate total gas volume at STP

\text{Volume}

= 0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

Multiplying total gaseous moles by the molar volume at STP (

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

) gives the required volume.

Key Concept

Thermal decomposition of metallic trioxonitrate(V) salts and gas stoichiometry at STP

Estimated Time:1m 30s

Question 8450Question →

In atmospheric chemistry, air pollutants are classified based on whether they are released directly from source emissions or formed via atmospheric reactions. Which of the following is a secondary pollutant produced during the formation of photochemical smog?

Peroxyacetyl nitrate (

\text{PAN}

)

Carbon(II) oxide (

\text{CO}

)

Sulfur(IV) oxide (

\text{SO}_2

)

Methane (

\text{CH}_4

)

Show answer & explanation

Answer: Peroxyacetyl nitrate ( $\text{PAN}$ )

Answer

Peroxyacetyl nitrate (

\text{PAN}

) is a secondary air pollutant formed through photochemical reactions between unburnt hydrocarbons and nitrogen oxides in the presence of sunlight.

Peroxyacetyl nitrate (

\text{PAN}

) is classified as a secondary pollutant because it is not emitted directly from any industrial or natural source; instead, it is synthesized in ambient air through complex photochemical reactions between nitrogen oxides and unburnt hydrocarbons driven by sunlight.

Step-by-Step Solution

1

Differentiate between primary and secondary air pollutants.

Primary pollutants are emitted directly from sources (e.g.,

\text{CO}

,

\text{SO}_2

,

\text{CH}_4

), whereas secondary pollutants form in the atmosphere through chemical transformations of primary pollutants.

Classification depends on whether the chemical is directly emitted or formed chemically in ambient air.

2

Identify the constituents and origin of photochemical smog.

Photochemical smog forms when sunlight drives reactions between nitrogen oxides (

\text{NO}_x

) and volatile organic compounds (hydrocarbons), forming ozone (

\text{O}_3

) and peroxyacetyl nitrate (

\text{PAN}

).

\text{PAN}

is a toxic component of smog created exclusively by atmospheric chemical processes.

Key Concept

Secondary Air Pollutants and Photochemical Smog Formation

Estimated Time:1m 0s

Question 8451Question →

Compound $W$ is a four-carbon carbonyl compound that yields a negative result when tested with ammoniacal silver nitrate solution. Complete reduction of compound $W$ using lithium aluminium hydride ( $\text{LiAlH}_4$ ) produces an alcohol, compound $Z$ . What is the IUPAC name of compound $Z$ ?

Show answer & explanation

Answer: butan-2-ol; 2-butanol; Butan-2-ol; 2-Butanol

Answer

The IUPAC name of compound

Z

is butan-2-ol.

Compound

W

does not react with Tollen's reagent (ammoniacal silver nitrate), which confirms it is an alkanone rather than an alkanal. The only four-carbon alkanone is butan-2-one. Reducing butan-2-one with

\text{LiAlH}_4

yields the secondary alcohol butan-2-ol.

Step-by-Step Solution

1

Determine the functional group of compound

W

based on the distinction test.

Compound

W

is an alkanone (ketone).

Alkanals (aldehydes) reduce ammoniacal silver nitrate (Tollen's reagent) to silver metal (silver mirror), whereas alkanones (ketones) give a negative result.

2

Identify the specific structure and IUPAC name of compound

W

.

Compound

W

is butan-2-one (

\text{CH}_3\text{COCH}_2\text{CH}_3

).

Since compound

W

is a four-carbon alkanone, its only structural isomer is butan-2-one.

3

Determine the product of the reduction reaction.

Compound

Z

is a secondary alcohol, butan-2-ol (

\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3

).

Reduction of alkanones with

\text{LiAlH}_4

converts the carbonyl group (

\text{C=O}

) into a secondary alcohol group (

\text{CH-OH}

).

Key Concept

Reduction of alkanones to secondary alcohols and distinction between alkanals and alkanones using Tollen's reagent

Question 8452Question →

Match each soil pollutant or waste management technique in Column A with its corresponding chemical mechanism, bio-environmental effect, or operational principle in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Arsenic contamination in agricultural soil matches the disruption of cellular respiration by binding to sulfhydryl groups and mimicking phosphate ions. Organophosphate pesticides match the inhibition of acetylcholinesterase enzymes with shorter environmental persistence. Phytoremediation matches the extraction and bioconcentration of soil heavy metals into plant biomass. Pyrolysis matches thermal decomposition in the absence of oxygen to produce bio-oil, gases, and char.

Each soil contaminant and waste treatment procedure is correctly linked to its fundamental biochemical pathway or thermodynamic condition: Arsenic disrupts sulfhydryl enzymes and phosphate pathways; Organophosphates selectively inhibit acetylcholinesterase; Phytoremediation relies on plant bioaccumulation of soil metals; and Pyrolysis achieves thermal decomposition under anaerobic conditions.

Step-by-Step Solution

1

Analyze the biochemical toxicity mechanism of Arsenic

Arsenic (

\text{As}

) forms covalent bonds with enzyme sulfhydryl (

-\text{SH}

) groups and acts as a structural analog to phosphate, inhibiting ATP synthesis.

Heavy metals disrupt metabolic pathways by binding to functional group residues on key metabolic enzymes.

2

Examine the mode of action and biodegradability of Organophosphates

Organophosphates inhibit acetylcholinesterase, leading to acetylcholine buildup. They are biodegradable compared to persistent chlorinated hydrocarbons like DDT.

Synthetic pesticide classification relies on chemical structure, biological targets, and environmental degradation rates.

3

Identify the eco-friendly soil cleanup method utilizing plants

Phytoremediation uses hyperaccumulators to take up heavy metal pollutants from contaminated ground into plant tissues.

Biological soil remediation relies on bio-uptake processes to remove heavy metals without chemical soil destruction.

4

Distinguish Pyrolysis from other thermal waste treatment methods

Pyrolysis is anaerobic thermal degradation yielding bio-oil, syngas, and char, contrasting with oxygen-rich incineration.

Thermal waste processing methods differ in operating atmosphere (presence vs. absence of

\text{O}_2

) and end-products.

Key Concept

Mechanisms of Soil Pollutants, Agrochemicals, Heavy Metal Toxicity, and Modern Waste Processing Techniques

Question 8453Question →

Given the following standard reduction potentials at

25^\circ\text{C}

:

\text{Al}^{3+}_{\text{(aq)}} + 3\text{e}^- \rightarrow \text{Al}_{\text{(s)}} \quad E^\circ = -1.66\text{ V}

\text{Cu}^{2+}_{\text{(aq)}} + 2\text{e}^- \rightarrow \text{Cu}_{\text{(s)}} \quad E^\circ = +0.34\text{ V}

Calculate the standard electromotive force ( $E^\circ_{\text{cell}}$ ), in volts, of the galvanic cell formed by coupling these two half-cells.

Show answer & explanation

Answer: 2

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) of the galvanic cell is

2.00\text{ V}

.

The standard electromotive force of a galvanic cell is defined as

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

. Since copper has a higher standard reduction potential (

+0.34\text{ V}

) than aluminium (

-1.66\text{ V}

), reduction occurs at the copper electrode (cathode) and oxidation occurs at the aluminium electrode (anode). Evaluating the potential difference yields

E^\circ_{\text{cell}} = 0.34\text{ V} - (-1.66\text{ V}) = 2.00\text{ V}

.

Step-by-Step Solution

1

Determine which half-cell undergoes reduction (cathode) and which undergoes oxidation (anode).

Copper is the cathode (

E^\circ = +0.34\text{ V}

) and aluminium is the anode (

E^\circ = -1.66\text{ V}

).

The half-cell with the higher standard reduction potential spontaneously undergoes reduction at the cathode.

2

Apply the formula for calculating standard electromotive force (

E^\circ_{\text{cell}}

).

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

The cell EMF is the standard potential difference between the reduction half-reaction and the oxidation half-reaction.

3

Perform the subtraction to evaluate

E^\circ_{\text{cell}}

.

E^\circ_{\text{cell}} = 0.34 - (-1.66) = 2.00\text{ V}

Subtracting a negative quantity is mathematically equivalent to adding its positive value.

Key Concept

Calculation of standard cell potential (

E^\circ_{\text{cell}}

) from standard electrode reduction potentials.

Question 8454Question →

A sample of $4.80\text{ g}$ of pure rhombic sulfur is completely burned in excess oxygen gas at room temperature and pressure (RTP). What is the volume of sulfur(IV) oxide gas liberated at RTP, and what is the change in the oxidation state of sulfur during this combustion process?

(Relative atomic mass: $S = 32.0$ ; Molar volume of gas at RTP = $24.0\text{ dm}^3\text{ mol}^{-1}$ )

3.60\text{ dm}^3

and an increase from 0 to

+4

3.36\text{ dm}^3

and an increase from 0 to

+4

3.60\text{ dm}^3

and an increase from 0 to

+6

1.80\text{ dm}^3

and an increase from

-2

to

+4

Show answer & explanation

Answer: $3.60\text{ dm}^3$ and an increase from 0 to $+4$

Answer

The volume of sulfur(IV) oxide gas produced at RTP is

3.60\text{ dm}^3

, and the oxidation state of sulfur increases from 0 to

+4

.

Burning

4.80\text{ g}

(

0.15\text{ mol}

) of sulfur produces

0.15\text{ mol}

of

SO_2

gas. Multiplying by the molar volume at RTP (

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

) gives

3.60\text{ dm}^3

. In elemental rhombic sulfur, sulfur has an oxidation state of 0, which increases to

+4

in

SO_2

.

Step-by-Step Solution

1

Calculate the amount of sulfur reacted in moles.

\text{Moles of } S = \frac{4.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.15\text{ mol}

Dividing the mass of sulfur by its relative atomic mass gives the mole amount.

2

Write the balanced chemical equation and determine the mole ratio.

S(s) + O_2(g) \rightarrow SO_2(g)

(Mole ratio of

S : SO_2 = 1 : 1

, so

0.15\text{ mol}

of

SO_2

is formed).

Direct combustion of sulfur yields sulfur(IV) oxide.

3

Calculate the volume of

SO_2

gas produced at RTP.

\text{Volume} = 0.15\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 3.60\text{ dm}^3

At room temperature and pressure (RTP),

1\text{ mol}

of any gas occupies

24.0\text{ dm}^3

.

4

Determine the change in oxidation state of sulfur.

Elemental sulfur

S(s)

has an oxidation number of 0. In

SO_2

, oxygen is

-2

, so

x + 2(-2) = 0 \implies x = +4

. The change is from 0 to

+4

.

Free uncombined elements have an oxidation state of 0.

Key Concept

Stoichiometry of gas liberation at RTP and oxidation state changes during non-metal combustion

Question 8455Question →

Municipal solid waste containing large amounts of polyethene packaging materials is buried in a sanitary landfill. Over time, these materials persist in the soil without decomposing. Which chemical characteristic of synthetic addition polymers accounts for their resistance to biological waste degradation in soil?

They contain strong, non-polar carbon-carbon single bonds that soil microorganisms lack enzymes to cleave

They readily breakdown in moist soil to release carbon(IV) oxide gas that destroys stratospheric ozone

They react rapidly with soil moisture through reversible esterification to form soluble alkanols

They behave as biodegradable natural polymers that decompose quickly into soil nutrients when exposed to water

Show answer & explanation

Answer: They contain strong, non-polar carbon-carbon single bonds that soil microorganisms lack enzymes to cleave

Answer

Synthetic addition polymers resist biological breakdown in soil because they contain strong, non-polar carbon-carbon single bonds that microorganisms lack the necessary enzymes to cleave.

The correct option identifies that synthetic addition polymers like polyethene consist of unreactive, non-polar carbon-carbon single bonds. Naturally occurring soil microorganisms do not produce enzymes capable of cleaving these stable hydrocarbon backbones, causing the material to persist as non-biodegradable waste in soil.

Step-by-Step Solution

1

Analyze the chemical structure of synthetic addition polymers like polyethene

The polymer backbone consists of long, saturated hydrocarbon chains formed entirely of strong, non-polar

C-C

and

C-H

single bonds.

Understanding the chemical bonding of the polymer determines its chemical reactivity and biological stability.

2

Evaluate microbial decomposition mechanisms in soil waste management

Soil bacteria and fungi secrete enzymes that hydrolyze polar functional groups (such as esters or amides in natural polymers), but cannot digest long non-polar

C-C

alkane-like backbones.

Enzymatic specificity requires compatible functional groups for microbial degradation to occur.

3

Conclude the cause of non-biodegradability in landfill soil

Due to the absence of appropriate microbial enzymes and the high stability of non-polar carbon-carbon bonds, polyethene packaging persists unchanged in soil waste sites.

This structural stability is the primary chemical reason synthetic addition polymers are classified as non-biodegradable pollutants.

Key Concept

Chemical basis of non-biodegradable waste in soil pollution and waste management

Question 8456Question →

The standard reduction potentials for aluminium and nickel half-cells at

25^\circ\text{C}

are given below:

\text{Al}^{3+}(aq) + 3e^- \rightarrow \text{Al}(s) \quad E^\circ = -1.66\text{ V}

\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \quad E^\circ = -0.25\text{ V}

Calculate the standard electromotive force ( $E^\circ_{\text{cell}}$ ) in volts for the spontaneous reaction between these two half-cells.

Show answer & explanation

Answer: 1.41

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) for the spontaneous galvanic cell reaction is +1.41 V.

For a spontaneous electrochemical reaction, the standard cell electromotive force (

E^\circ_{\text{cell}}

) must be positive. The half-reaction with the more positive standard reduction potential (

\text{Ni}^{2+}/\text{Ni}

at

-0.25\text{ V}

) proceeds as a reduction at the cathode. The half-reaction with the less positive potential (

\text{Al}^{3+}/\text{Al}

at

-1.66\text{ V}

) proceeds as an oxidation at the anode. Calculating

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.25\text{ V} - (-1.66\text{ V}) = +1.41\text{ V}

. Because

E^\circ

is an intensive property, the stoichiometric coefficients used to balance electrons (

2\text{Al} + 3\text{Ni}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Ni}

) do not affect the numerical values of the half-cell potentials.

Step-by-Step Solution

1

Determine which electrode undergoes reduction (cathode) and which undergoes oxidation (anode)

Nickel ion reduction occurs at the cathode (

E^\circ = -0.25\text{ V}

), and aluminium metal oxidation occurs at the anode (

E^\circ = -1.66\text{ V}

).

A galvanic cell operates spontaneously (

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

) when the half-cell with the more positive standard reduction potential acts as the cathode.

2

State the equation for standard cell potential

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

The cell potential measures the potential difference between the reduction half-reaction and the oxidation half-reaction under standard conditions.

3

Substitute the reduction potential values into the equation

E^\circ_{\text{cell}} = -0.25\text{ V} - (-1.66\text{ V}) = +1.41\text{ V}

Standard electrode potentials are intensive properties; hence, balancing electron stoichiometry does not scale

E^\circ

values.

Key Concept

Calculating standard cell electromotive force (

E^\circ_{\text{cell}}

) and predicting spontaneity from standard reduction potentials

Question 8457Question →

In electrophilic aromatic substitution, benzene reacts with strong electrophiles generated by specific catalyst-reagent combinations. Match each benzene reaction system on the left with its corresponding active electrophile species generated during the reaction mechanism on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Nitration produces the nitronium ion (

\text{NO}_2^+

); Friedel-Crafts acylation generates the acylium ion (

\text{CH}_3\text{C}^+=\text{O}

); Catalytic bromination produces the bromonium ion (

\text{Br}^+

); Sulfonation generates neutral sulfur trioxide (

\text{SO}_3

).

Each benzene electrophilic substitution reaction relies on a specific reagent and catalyst mechanism to create a powerful electrophile capable of disrupting benzene's stable aromatic system. Nitration generates

\text{NO}_2^+

via acid-base protonation of nitric acid by sulfuric acid. Friedel-Crafts acylation forms the acylium ion

\text{CH}_3\text{C}^+=\text{O}

through chloride abstraction by the Lewis acid

\text{AlCl}_3

. Bromination generates a polarized

\text{Br}^+

complex using

\text{FeBr}_3

. Sulfonation relies on

\text{SO}_3

, which features an electron-deficient sulfur atom due to polar S=O bonds.

Step-by-Step Solution

1

Identify the electrophile in nitration

Concentrated

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

acts as an acid to protonate

\text{HNO}_3

. Loss of

\text{H}_2\text{O}

yields

\text{NO}_2^+

(nitronium ion).

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

is a stronger acid than

\text{HNO}_3

and forces

\text{HNO}_3

to act as a base.

2

Identify the electrophile in Friedel-Crafts acylation

The catalyst

\text{AlCl}_3

abstracts

\text{Cl}^-

from

\text{CH}_3\text{COCl}

, leaving the resonance-stabilized cations

\text{CH}_3\text{C}^+=\text{O}

.

\text{AlCl}_3

is an electron-deficient Lewis acid capable of coordinating chloride.

3

Identify the electrophile in bromination

\text{FeBr}_3

coordinates with a bromine atom of

\text{Br}_2

, polarising the bond to create an effective

\text{Br}^+

electrophile.

Benzene requires a Lewis acid catalyst to polarize halogen molecules sufficiently for reaction.

4

Identify the electrophile in sulfonation

Equilibrium in concentrated/fuming

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

produces neutral

\text{SO}_3

, which has a highly electron-deficient sulfur atom.

The three electronegative oxygen atoms in

\text{SO}_3

withdraw electron density from the central sulfur atom.

Key Concept

Generation of Electrophiles in Benzene Electrophilic Substitution

Question 8458Question →

During the industrial extraction of iron in a blast furnace, hematite ( $\text{Fe}_2\text{O}_3$ ) is reduced to molten iron in the upper reduction zone. Which chemical species serves as the chief reducing agent responsible for this reduction?

Carbon(II) oxide (

\text{CO}

)

Solid carbon (coke,

\text{C}

)

Calcium oxide (

\text{CaO}

)

Carbon(IV) oxide (

\text{CO}_2

)

Show answer & explanation

Answer: Carbon(II) oxide ( $\text{CO}$ )

Answer

Carbon(II) oxide (

\text{CO}

) is the chief reducing agent in the upper reduction zone of the blast furnace.

In the upper, lower-temperature region of the blast furnace, ascending carbon(II) oxide (

\text{CO}

) gas chemically reduces hematite (

\text{Fe}_2\text{O}_3

) to iron, producing carbon(IV) oxide (

\text{CO}_2

) gas as a byproduct.

Step-by-Step Solution

1

Analyze the reactions occurring in the upper temperature zone (

400^\circ\text{C} - 700^\circ\text{C}

) of the blast furnace.

Ascending carbon(II) oxide gas contacts descending solid hematite ore.

Gaseous carbon(II) oxide provides effective gas-solid contact necessary for chemical reduction.

2

Write the balanced chemical equation for the reduction of hematite.

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(l) + 3\text{CO}_2(g)

Carbon(II) oxide removes oxygen from iron(III) oxide, reducing iron from the +3 oxidation state to metallic iron (0 oxidation state).

Key Concept

Iron Extraction in the Blast Furnace

Question 8459Question →

An unknown organic compound $Y$ forms a brick-red precipitate when warmed with Fehling's solution and also gives a yellow precipitate of triiodomethane ( $CHI_3$ ) when treated with iodine in sodium hydroxide solution. Which of the following compounds is $Y$ ?

Ethanal

Propan-2-one

Propanal

Methanal

Show answer & explanation

Answer: Ethanal

Answer

Ethanal

Ethanal (

CH_3CHO

) is an alkanal, so it reduces copper(II) ions in Fehling's solution to copper(I) oxide (

Cu_2O

), producing a brick-red precipitate. Furthermore, because its carbonyl group is bonded directly to a methyl group (

CH_3C=O

), it undergoes halogenation and cleavage in alkaline iodine to produce a yellow precipitate of triiodomethane (

CHI_3

).

Step-by-Step Solution

1

Analyze the Fehling's solution reaction.

Fehling's test distinguishes alkanals (aldehydes) from alkanones (ketones). A positive test (brick-red

Cu_2O

precipitate) indicates

Y

must be an alkanal.

Alkanals are easily oxidized to alkanoic acids, whereas alkanones resist mild oxidation.

2

Analyze the triiodomethane (iodoform) reaction.

A positive triiodomethane test (yellow

CHI_3

precipitate) requires a methyl carbonyl group (

CH_3C=O

) or a secondary alcohol structure (

CH_3CH(OH)-

).

Iodine in aqueous alkali oxidizes and iodinates compounds containing the methyl carbonyl structural unit.

3

Combine the structural requirements.

The compound must be both an alkanal (

-CHO

) and contain a methyl carbonyl group (

CH_3C=O

). Ethanal (

CH_3CHO

) is the only alkanal that possesses a

CH_3C=O

group.

Other alkanals like methanal (

HCHO

) or propanal (

CH_3CH_2CHO

) lack the

CH_3C=O

structural unit.

Key Concept

Distinction tests for carbonyl compounds: Fehling's test and Iodoform test

Estimated Time:1m 0s

Question 8460Question →

Graphite and diamond are two crystalline allotropes of carbon that exhibit vastly different physical properties. Which of the following statements correctly accounts for why graphite conducts electricity whereas diamond is an electrical insulator?

Graphite consists of carbon atoms that are structurally isomeric to diamond, allowing free molecules to move through the crystal layers.

Each carbon atom in graphite forms three covalent bonds using

sp^2

hybridization, leaving one delocalized electron per atom free to move along the hexagonal layers.

Diamond is composed of heavier isotopes of carbon whose extra neutrons trap valence electrons within the atomic nucleus.

Graphite contains mobile carbon ions that dissociate under an applied electric field, whereas diamond contains neutral molecules.

Show answer & explanation

Answer: Each carbon atom in graphite forms three covalent bonds using $sp^2$ hybridization, leaving one delocalized electron per atom free to move along the hexagonal layers.

Answer

Each carbon atom in graphite forms three covalent bonds using

sp^2

hybridization, leaving one delocalized electron per atom free to move along the hexagonal layers.

In graphite, each carbon atom forms three covalent

\sigma

-bonds via

sp^2

hybridization. The fourth valence electron resides in an unhybridized

p

-orbital and becomes delocalized across the hexagonal layer, allowing electrical current to flow. In contrast, diamond features

sp^3

hybridization where all four valence electrons are tightly bound in localized covalent bonds, leaving no mobile electrons.

Step-by-Step Solution

1

Analyze the bonding and hybridization of carbon in diamond.

In diamond, each carbon atom undergoes

sp^3

hybridization and forms four strong covalent bonds directed tetrahedrally toward adjacent carbon atoms.

All four valence electrons per carbon atom are involved in localized single covalent bonds, leaving no free electrons to conduct electricity.

2

Analyze the bonding and hybridization of carbon in graphite.

In graphite, each carbon atom undergoes

sp^2

hybridization to form three strong covalent bonds within a hexagonal planar sheet.

This leaves one unhybridized

p

-orbital electron per carbon atom, which forms a delocalized

\pi

-electron system capable of moving freely along the layers.

3

Correlate structural features to electrical conductivity.

The presence of delocalized valence electrons in graphite enables electrical conduction, whereas the localized

sp^3

bonding in diamond makes it an insulator.

Electrical conduction in solid non-metallic elements requires mobile charge carriers such as delocalized electrons.

Key Concept

Structural bonding and hybridization of carbon allotropes (Graphite vs Diamond)

Estimated Time:1m 0s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept