Substitution alloys are formed when atoms of the host metal are replaced by atoms of another transition metal of similar size, crystal structure, and chemical properties without chemical bonding.

If they react easily with each other, they form intermetallic compounds instead of a substitution alloy.

• Element (X): Vanadium (V), as V₂O₅ is used as a catalyst for superconducting magnets. Vanadium ($[Ar] 3d³ 4s²$) has 3 unpaired electrons.

• Element (Y): Contains twice as many unpaired electrons ($3 \times 2 = 6$). This is Chromium (Cr, $[Ar] 3d⁵ 4s¹$), which has 6 unpaired electrons.

• The maximum oxidation state of Cr (Y) is +6, which is higher than that of V (X), which is +5.

Copper (Cu) is a transition element with a completely filled d-sublevel ($[Ar] 3d^{10} 4s^1$).

In the compound X₂O (Cu₂O), the copper ion is $Cu⁺$, meaning it loses its single s-electron, leaving the electronic configuration as [Ar] 3d¹⁰.

During roasting, moisture and carbon dioxide are driven off as gases/vapors.

Limonite dehydrates to hematite (changing color from yellow to red) but is not oxidized as the iron is already in the +3 state ($Fe_2O_3$).

Siderite ($FeCO_3$) is oxidized to hematite.

Therefore, not all iron ores undergo oxidation.

The $Mn^{3+}$ ion ($3d^4$) is highly unstable and is easily reduced to the much more stable $Mn^{2+}$ state, which features a half-filled, symmetric $3d^5$ configuration.

Adding HCl to potassium sulphide ($K_2S$) releases hydrogen sulphide ($H_2S$) gas.

When $H_2S$ is passed through acidified copper nitrate, it precipitates black copper sulphide ($CuS$).

Dilute hydrochloric acid does not react with either magnesium sulphate ($MgSO_4$) or magnesium chloride ($MgCl_2$), meaning it cannot be used to distinguish them.

Aluminum hydroxide Al(OH)₃ is amphoteric and dissolves in excess sodium hydroxide to form soluble sodium meta-aluminate (NaAlO₂).

Iron(II) hydroxide Fe(OH)₂ is basic and does not react or dissolve in NaOH, allowing separation by filtration.

• Salt X (NaCl): Reacts with conc. H₂SO₄ to evolve colorless HCl gas.

• Salt Y (NaI): Reacts with conc. H₂SO₄ to evolve violet iodine ($I_2$) vapors, which turn paper wet with starch blue.

• First, add Hydrochloric acid (HCl): This precipitates $Pb^{2+}$ selectively as insoluble white $PbCl_2$, which is separated by filtration.

• Next, add Sulphuric acid (H₂SO₄) to the filtrate: This precipitates $Ca^{2+}$ as insoluble white $CaSO_4$.

• The $Cu^{2+}$ remains dissolved as soluble $CuSO_4$ in the solution, achieving full separation.

Strongly heating hydrated barium chloride ($BaCl_2 \cdot 2H_2O$) in an open container drives off water of crystallization into the atmosphere.

Because the gaseous water escapes, the process is completely irreversible and cannot establish dynamic chemical equilibrium.

The moles of Zn added are $\frac{16.25}{65} = 0.25\text{ moles}$.
The reaction equation is: $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$.
To react completely, 0.25 moles of Zn require $2 \times 0.25 = 0.50\text{ moles}$ of HCl.
- Both solution A (1.0 mole HCl) and solution B (0.50 mole HCl) have enough acid to react with all of the zinc.

Thus, the total volume/moles of hydrogen gas produced will be exactly equal (0.25 mol $H_2$).
- Since solution A has a higher acid concentration (1.0 M) than B (0.5 M), its initial rate of reaction is faster.

This reaction is highly endothermic (ΔH > 0).

According to Le Chatelier's Principle, lowering the temperature shifts the equilibrium in the backward (exothermic) direction, which decreases the concentration of products and therefore decreases K_c.

Additionally, cooling always decreases reaction rates, hence the rate of water dissociation decreases.

Find the initial partial pressure of HI: $P_{\text{total}} = P_{HI} + P_{H_2} + P_{I_2} \⇒ 2.0 = P_{HI} + 0.4 + 0.4 \⇒ P_{HI} = 1.2\text{ atm}$.
2.

Calculate $K_p$: $$K_p = \frac{P_{H_2} \times P_{I_2}}{(P_{HI})^2} = \frac{0.4 \times 0.4}{(1.2)^2} = \frac{0.16}{1.44} = 0.111$$ 3.

Since the moles of gaseous reactants (2 moles of HI) equal the moles of gaseous products (1 mole of $H_2$ + 1 mole of $I_2$), changing the container volume does not affect the position of equilibrium or change the value of $K_p$.

K_p remains 0.11.

• Diluting a weak acid (Sample A) decreases the concentration of $H^+$ ions in solution, which causes the pH to increase.

• Diluting a weak base (Sample B) decreases the concentration of $OH^-$ ions in solution, which causes the pOH to increase and therefore the pH to decrease.

1. $\text{pH} = 12.0 \⇒ \text{pOH} = 14.0 - 12.0 = 2.0$.
2. $[OH^-] = 10^{-\text{pOH}} = 10^{-2} = 0.01\text{ mol/L}$.
3.

Since NaOH is a strong base: $[\text{NaOH}] = [OH^-] = 0.01\text{ M}$.
4. $\text{Moles} = \text{Molar Concentration} \times \text{Volume (L)} = 0.01 \times 0.500 = 0.005\text{ moles}$.
5. $\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.005\text{ mol} \times 40\text{ g/mol} = 0.2\text{ g}$.

According to the electrochemical activity series, Magnesium (Mg) is more active than Zinc (Zn).

Therefore, Magnesium can spontaneously displace Zinc ions from solution.

All other options describe displacement of a more active metal by a less active metal, which is non-spontaneous.

Electrode X has the lower standard reduction potential (-1.670 V), so it acts as the anode.
2.

Electrode Y has the higher reduction potential (+0.34 V), so it acts as the cathode.
3.

The EMF of a single cell is: $$\text{EMF} = E^\circ_{\text{reduction}}(\text{cathode}) - E^\circ_{\text{reduction}}(\text{anode}) = 0.34 - (-1.670) = 2.01\text{ V}$$ 4.

To get a total EMF of 8.05 V, we need: $\frac{8.05}{2.01} \≈ 4\text{ cells}$ connected in series.

• W: $E^\circ_{\text{red}} = +1.42\text{ V}$ (Cathode relative to Zn)

• Z: $E^\circ_{\text{red}} = -2.375\text{ V}$ (Much more active than Zn)

• Y: $E^\circ_{\text{red}} = +1.2\text{ V}$ (Cathode relative to Zn)

• X: $E^\circ_{\text{red}} = +0.34\text{ V}$ (Cathode relative to Zn)

• Cell 3 (Electrode E): $Ni^{2+} \rightarrow$ Equivalent weight = $\frac{58.7}{2} = 29.35\text{ g}$.

• Cell 2 (Electrode C): $Ag^{+} \rightarrow$ Equivalent weight = $\frac{108}{1} = 108\text{ g}$.

• Cell 1 (Electrode A): $Cu^{2+} \rightarrow$ Equivalent weight = $\frac{63.5}{2} = 31.75\text{ g}$.

The spontaneous reaction of hydrogen with oxygen to form water is utilized in hydrogen fuel cells to generate clean electrical energy.

All other choices represent non-spontaneous electrolytic processes that consume electrical energy.

• Cell Type Shift: During charging, the battery switches from a galvanic cell to an electrolytic cell.

• Anode Identity: The positive plate behaves as the anode where oxidation occurs during the recharging phase.

• Oxidation State Change: Lead sulfate (\(\text{PbSO}_4\), where lead is \(\text{Pb}^{2+}\)) is oxidized into lead dioxide (\(\text{PbO}_2\), where lead is \(\text{Pb}^{4+}\)).

• Reaction Equation: \(\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^-\).

• Option b: The conversion of \(\text{Pb}^{2+}\) to metallic \(\text{Pb}\) is a reduction process, which occurs exclusively at the cathode (negative plate) during charging.

• Option c: Since charging is forced by an external power source, electrons are pumped from the positive electrode to the negative electrode via the external circuit.

• Option d: Within the internal solution of the cell, electrical charge is carried by moving ions (\(\text{H}^+\) and \(\text{SO}_4^{2-}\)), not free electrons.

• Compound X: Ethyne (Acetylene, C₂H₂) reacts explosively with chlorine with a flame, producing carbon soot and HCl. It belongs to the alkyne family, which obeys the general formula C_nH_2n-2.

• Compound Y: Gammaxane (C₆H₆Cl₆), used as an insecticide, is an alicyclic hydrocarbon derivative, not aromatic, which perfectly obeys the molecular relation C_nH_nCl_n.

Let's draw the structure of 2,3-dimethylbutane:

• Diethyl ether: CH₃-CH₂-O-CH₂-CH₃

• Methyl propyl ether: CH₃-O-CH₂-CH₂-CH₃

• Methyl isopropyl ether: CH₃-O-CH(CH₃)₂

Dry distillation of a sodium salt of a carboxylic acid with soda-lime yields an alkane with one less carbon atom than the parent acid. $$\text{CH}_3\text{-CH(CH}_3\text{)-CH}_2\text{-COONa} + \text{NaOH} \xrightarrow{\text{CaO, }\Delta} \text{CH}_3\text{-CH(CH}_3\text{)-CH}_3 + \text{Na}_2\text{CO}_3$$ Neutralizing 3-methylbutanoic acid yields sodium 3-methylbutanoate, which upon dry distillation yields 2-methylpropane (isobutane).

• Neutralization: React heptanoic acid with NaOH to obtain sodium heptanoate (C₆H₁₃COONa).

• Dry Distillation: Heat sodium heptanoate with soda-lime to yield hexane (C₆H₁₄).

• Catalytic Reforming: Pass hexane over a hot platinum catalyst to undergo cyclization and dehydrogenation, yielding benzene (C₆H₆).

The given compound is salicyl alcohol (2-hydroxybenzyl alcohol).

Complete oxidation of the primary alcohol group ($-CH_2OH$) yields a carboxylic acid group ($-COOH$), producing Salicylic acid.

Salicylic acid is widely used to prepare drugs like aspirin and methyl salicylate (for rheumatic pain).

• Acid Y (Citric acid): It is a tricarboxylic aliphatic acid containing 3 carboxyl groups (which react with 3 moles of NaOH) and 1 tertiary hydroxyl group (which reacts only with sodium metal, not NaOH).

Citric acid is widely used as a preservative for frozen fruits.

• Perform catalytic reforming of hexane to prepare benzene, which is then chlorinated to chlorobenzene.

• Perform alkaline hydrolysis of chlorobenzene to produce phenol.

• React phenol with formaldehyde via condensation polymerization to yield Bakelite.

• Ore Dressing: Roasting the ore to dry and oxidize impurities.

• Reduction Stage: Using a Midrex furnace (or Blast furnace) to reduce hematite to iron.

• Steel Production: Using an Oxygen converter (or electric/open-hearth furnace) to produce steel.

• Moles of $AgCl$ precipitated = $\frac{0.963\text{ g}}{143.5\text{ g/mol}} = 0.00671\text{ moles}$.

• Since 1 mole of $MCl$ yields 1 mole of $AgCl$, moles of $MCl = 0.00671\text{ moles}$.

• Molar mass of $MCl = \frac{0.5\text{ g}}{0.00671\text{ mol}} = 74.5\text{ g/mol}$.

• Atomic mass of $M = 74.5 - 35.5 (\text{Cl}) = 39\text{ g/mol}$ (which is Potassium, K).

Moles of $H_2SO_4$ in the titration = $\frac{0.1176\text{ g}}{98\text{ g/mol}} = 0.0012\text{ moles}$.
2.

Neutralization reaction: $2\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$.
3.

Moles of KOH in 20 mL of diluted solution = $2 \times 0.0012 = 0.0024\text{ moles}$.
4.

Concentration of the diluted KOH solution = $\frac{0.0024\text{ moles}}{0.020\text{ L}} = 0.12\text{ M}$.
5.

Using the dilution law ($M_1 V_1 = M_2 V_2$): $$0.2 \times X = 0.12 \times (X + 20)$$ $$0.2X = 0.12X + 2.4 \⇒ 0.08X = 2.4 \⇒ X = 30\text{ mL}$$

• (1) AgCl (AB type): S = √(1.0 × 10^-10) = 10^-5 M. Total ion concentration = 2 × 10^-5 M.

• (3) AgBr (AB type): S = √(1.0 × 10^-8) = 10^-4 M. Total ion concentration = 2 × 10^-4 M.

• (2) PbI₂ (AB₂ type): $4S^3 = 4.0 \times 10^{-12} \⇒ S = 10^{-4}\text{ M}$. Total ion concentration = 3 × 10^-4 M.

• (4) BiI₃ (AB₃ type): $27S^4 = 2.7 \times 10^{-15} \⇒ S = 10^{-4}\text{ M}$. Total ion concentration = 4 × 10^-4 M.

• $[A] = 4$

• $[C] = 3$

Since $E^\circ_{\text{cell}} = +1\text{ V}$ (positive), the cell is galvanic.
2.

In the cell reaction, metal M is oxidized ($M \rightarrow M^{3+}$), so M is the anode.
3.

Using the cell EMF equation: $$E^\circ_{\text{cell}} = E^\circ_{\text{reduction}}(\text{cathode}) - E^\circ_{\text{reduction}}(\text{anode})$$ $$+1.0\text{ V} = -0.76\text{ V} - E^\circ_{\text{reduction}}(M^{3+}/M)$$ $$E^\circ_{\text{reduction}}(M^{3+}/M) = -1.76\text{ V}$$ 4.

Standard oxidation potential is the negative of the reduction potential: $E^\circ_{\text{oxidation}} = +1.76\text{ V}$.

At the anode, chloride ions are oxidized to chlorine gas: $$2\text{Cl}^- \rightarrow \text{Cl}_{2(g)} + 2e^-$$ To evolve 1 mole of diatomic chlorine gas ($Cl_2$), 2 moles of electrons (2 Faradays) are required.

Therefore, passing 1 Faraday of electricity will evolve exactly 0.5 moles of chlorine gas.

• Acid (A) Identification: Carbolic acid (Phenol) acts as a very weak acid but does not react with sodium bicarbonate, meaning it fails the acidity test.

• Reduction Behavior: Carbolic acid lacks a carbonyl or carboxyl group, so it cannot be reduced by hydrogen over a copper(II) chromate catalyst.

• Acid (B) Identification: Salicylic acid is a bifunctional organic acid used directly to manufacture aspirin (acetylsalicylic acid), which treats headaches.

• Option a: Acetic acid is a carboxylic acid that passes the acidity test by reacting vigorously with sodium bicarbonate.

• Option c: Benzoic acid passes the acidity test, and it is not the primary compound used for making aspirin.

• Option d: Oxidizing para-methyl toluene produces terephthalic acid, not salicylic acid.

• Alkylation: Alkylating benzene (C₆H₆) yields toluene, C₆H₅CH₃ (X).

• Oxidation: Oxidizing toluene yields benzoic acid, C₆H₅COOH (Y).

• Esterification: Reacting benzoic acid with ethanol (C₂H₅OH) produces ethyl benzoate (Z).

• Compound X: Acetic acid reacts with methanol (CH₃OH) to produce methyl acetate, which has a fruity odor.

• Compound Y: 2-propanol is a secondary alcohol that oxidizes easily, causing the violet color of acidified KMnO₄ to disappear.

• Compound Z: Carbolic acid (phenol) reacts with bromine water to form a white precipitate of 2,4,6-tribromophenol.

• Alkaline Hydrolysis: Heat the ester with NaOH to obtain 1-propanol and sodium propanoate.

• Dehydration: Heat 1-propanol with concentrated $H_2SO_4$ at 180°C to obtain propene.

• Catalytic Hydrogenation: React propene with $H_2$ gas over a nickel catalyst to obtain propane.