Calculation Questions Collection – Models 1–10

• Graph (X): The product curve (rising from zero) finishes at a higher concentration than the reactant curve. Since [Products] > [Reactants], the math tells us Kc > 1. This means the forward reaction is predominant.

• Graph (Y): The reactants remain higher than the products at equilibrium, yielding a Kc < 1 (backward reaction predominant).

• Withdrawing Water Vapor: According to Le Chatelier's Principle, removing a product causes the system to shift forward (to the right) to replace the lost water vapor.

• The Result: This forward shift naturally generates more products, leading to a direct increase in the mass of solid CaO.

• Note: Altering the amounts of solids (choices a and c) has zero effect on the equilibrium position.

• Step 1: Dilution Formula (M₁V₁ = M₂V₂)
  V₁ = 50 mL, M₁ = 0.2 M.
  V₂ = 50 + 450 = 500 mL.
  M₂ = (0.2 × 50) / 500 = 0.02 M.

• Step 2: Calculate α
  Using the formula: α = √(K_b / C)
  α = √(1.8 × 10⁻⁵ / 0.02) = √(9 × 10⁻⁴) = 0.03.

• Find Partial Pressures:
  P_Total = P(CO₂) + P(CO)
  40 atm = P(CO₂) + 31.6 atm
  P(CO₂) = 40 - 31.6 = 8.4 atm.

• Apply Kp Formula:
  Kp = (P(CO))² / P(CO₂)
  Kp = (31.6)² / 8.4
  Kp = 998.56 / 8.4 ≈ 118.876.

• Anode (Oxidation): Cl⁻ is oxidized to Cl₂. The provided potential (+1.36 V) is for reduction. Thus, the oxidation potential is -1.36 V.

• Cathode (Reduction): MnO₄⁻ (Mn⁷⁺) is reduced to Mn²⁺. The provided potential (-1.52 V) is an oxidation potential (from 2+ to 7+). Thus, its reduction potential is +1.52 V.

• EMF Calculation: E_cell = E_ox(anode) + E_red(cathode) = -1.36 V + 1.52 V = +0.16 V.

• Analyze the Given Cells:
  For Cell 1: E_ox(X) + E_red(Ag) = 0.80 V.
  For Cell 2: E_ox(Y) + E_red(Ag) = 1.56 V.

• Find the Difference: Subtracting Cell 1 from Cell 2 cancels out the Silver (Ag) component:
  E_ox(Y) - E_ox(X) = 1.56 - 0.80 = 0.76 V.

• Construct New Cell (X, Y): Since Y has a higher oxidation potential, Y will act as the anode, and X will be forced to act as the cathode. EMF = E_ox(Y) + E_red(X) = 0.76 V.

• Activity Series: A (-0.44) > B (+0.34) > C (+0.80). Metal C is the least active.

• Protection Type: When a metal is coated with a less active metal, the coating acts as a physical barrier. If scratched, the underlying (more active) metal becomes the anode and rusts faster. This setup (coating is less active) is strictly defined as Cathodic Protection.

• Initial Moles:
  Moles H₂SO₄ = 0.2 L × 0.2 M = 0.04 mol.
  Moles Ca(OH)₂ = 0.3 L × 0.2 M = 0.06 mol.

• Reaction & Excess: The balanced equation shows a 1:1 reacting ratio. Thus, 0.04 mol of acid reacts with exactly 0.04 mol of base. The unreacted base left over is: 0.06 - 0.04 = 0.02 mol Ca(OH)₂.

• New Concentration: The final mixture volume is 200 mL + 300 mL = 500 mL (0.5 L).
  New Molarity = Moles / Volume = 0.02 mol / 0.5 L = 0.04 M.

• Base Solubility at 25°C:
  Ksp = 10^-21. Solubility (S) = √Ksp = 3.16 × 10^-11 mol/L.

• Concentration that Precipitated:
  Mass = 1.53 × 10^-5 g. Moles = (1.53 × 10^-5) / 97 = 1.577 × 10^-7 mol.
  Precipitated Molarity = Moles / 5 L ≈ 3.15 × 10^-8 mol/L.

• Total Solubility at 60°C:
  Solubility at 60°C = Solubility at 25°C + Concentration precipitated.
  Since 10^-11 is mathematically negligible compared to 10^-8, the total solubility is essentially 3.15 × 10^-8 mol/L.

• Calculate Ksp at 60°C:
  Ksp = S² = (3.15 × 10^-8)² ≈ 9.9 × 10^-16, which closely rounds to 1 × 10^-15.

• Determine Equivalent Mass: For Silver (Ag⁺), valency is 1. Equivalent mass = 108 / 1 = 108 g.

• Set Up the Equation:
  26.25 = (25 × Time × 108) / 96500.

• Solve for Time (in seconds):
  Time = (26.25 × 96500) / (25 × 108) = 2533125 / 2700 = 938.19 seconds.

• Convert to Minutes:
  938.19 / 60 ≈ 15.63 minutes.

• Cell 1 Analysis: The green arrow shows electrons flowing from electrode A to B. Therefore, Activity A > B.

• Cell 2 Analysis: The green arrow shows electrons flowing from electrode C to B. Therefore, Activity C > B.

• Surface Area: Powdered magnesium reacts much faster than a solid ribbon.

• Concentration: Higher molarity (0.2 M) of H₂SO₄ results in a faster rate than 0.1 M.

• Temperature: Elevated temperature (35°C) yields a faster rate than room temperature.

Given [H+] = 1.0 × 10-9 M:

Since pH > 7, the solution is basic (a base) with pH = 9.

The total pressure is the sum of all partial pressures:

• E°_ox(X) = +0.76 V

• E°_ox(Y) = +1.66 V

Reaction: H2SO4 + 2KOH → K2SO4 + 2H2O

• Moles of BaSO₄:
  Molar Mass of BaSO4 = 137.3 + 32 + 64 = 233.3 g/mol
Moles = 1.165 g / 233.3 g/mol = 0.005 mol

• Mass of Anhydrous ZnSO₄:
  Molar Mass of ZnSO4 = 65.4 + 32 + 64 = 161.4 g/mol
Mass of ZnSO4 = 0.005 mol × 161.4 g/mol = 0.807 g

• Mass of Water:
  Mass of H2O = 1.437 g - 0.807 g = 0.630 g
Moles of H2O = 0.630 g / 18 g/mol = 0.035 mol

• Determine X:
  X = Moles of H2O / Moles of ZnSO4 = 0.035 / 0.005 = 7

Mg(OH)2 ⇌ Mg2+ + 2OH-

• Since electrons flow from left to right in standard galvanic configurations, we find:
  E°ox(A) > E°ox(C) with difference of 1.2 V.
  E°ox(C) > E°ox(B) with difference of 0.8 V.

• Thus, the activity/oxidation potential order is: A > C > B.

• In a cell combining A and B: A is the natural anode and B is the natural cathode with spontaneous EMF = +2.0 V.

• If forced to run with B as anode and A as cathode, the reaction is reversed and becomes non-spontaneous with EMF = -2.0 V.

• Both tubes are at the same temperature (50°C) with equal masses of zinc.

• Tube A has HCl at 1M (higher concentration) while Tube B has HCl at 0.5M (lower concentration).

• Higher concentration → higher reaction rate → zinc disappears faster in tube A. ✓

• Upon complete dissolution of equal masses of zinc, the same amount of H₂ is produced in both tubes (since it depends on the mass of zinc, not the acid concentration, assuming excess acid).

• HCl, NaOH, HNO₃ are all strong electrolytes → they fully dissociate (α ≈ 1). ✓

• Boric acid (H₃BO₃) is a very weak acid → it barely dissociates in water (α ≪ 1). ✗

• A sacrificial electrode must be more reactive (higher oxidation potential) than iron → it should be the anode (electrons flow FROM the sacrificial metal TO iron).

• A → Fe: Electrons flow from A to Fe → A is the anode (more reactive than Fe). Cell potential = 1.4 V ✓

• C → Fe: C is also the anode. Cell potential = 0.5 V ✓ (but weaker protection)

• Fe → B and Fe → D: Fe is the anode here → B and D are less reactive than Fe → cannot be sacrificial. ✗

• Between A (1.4V) and C (0.5V), A has higher reactivity difference → A is preferred for best cathodic protection.

Step 1: Molar mass of Na₂CO₃·10H₂O = 2(23) + 12 + 3(16) + 10(18) = 46 + 12 + 48 + 180 = 286 g/mol

Step 2: Moles = 14.3/286 = 0.05 mol in 500 mL → Molarity = 0.05/0.5 = 0.1 M

Step 3: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Moles of Na₂CO₃ in 25 mL = 0.1 × 0.025 = 0.0025 mol

Moles of HCl needed = 2 × 0.0025 = 0.005 mol

Concentration of HCl = 0.005/0.025 = 0.2 M

Initial moles of NaOH = 0.01 × 0.2 = 0.002 mol

Final moles needed = 0.7 × 0.2 = 0.14 mol (assuming volume stays ≈ 200 mL since solid is added)

Additional moles = 0.14 − 0.002 = 0.138 mol

Mass = 0.138 × 40 = 5.52 g ✓

Initial: H₂ = 1 mol, I₂ = 1 mol, HI = 0

At equilibrium: H₂ = 0.3 mol, I₂ = 0.3 mol

Reacted = 1 − 0.3 = 0.7 mol each → HI formed = 2 × 0.7 = 1.4 mol

Concentrations (V = 2L): [H₂] = 0.3/2 = 0.15 M, [I₂] = 0.15 M, [HI] = 1.4/2 = 0.7 M

K_c = [HI]²/([H₂][I₂]) = (0.7)²/(0.15 × 0.15) = 0.49/0.0225 = 21.78 ✓

NH₄OH is a weak base.

α = √(K_b/C) = √(1.8×10⁻⁵/0.1) = √(1.8×10⁻⁴) = 0.0134

[OH⁻] = α × C = 0.0134 × 0.1 = 1.34 × 10⁻³ M

pOH = −log(1.34×10⁻³) = 2.87

pH = 14 − 2.87 = 11.13

Since NH₄OH is a base → pH > 7 ✓ (option B)

Faraday's Law: m = (M × I × t) / (n × F)

Where: M = 23, I = 10 A, t = 10 × 3600 = 36000 s, n = 1 (Na⁺ + e⁻ → Na), F = 96500 C/mol

m = (23 × 10 × 36000) / (1 × 96500) = 8,280,000/96500 = 85.8 g ✓

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: $P_{\text{total}} = P_{HI} + P_{H_2} + P_{I_2} \Rightarrow 2.0 = P_{HI} + 0.4 + 0.4 \Rightarrow P_{HI} = 1.2\text{ atm}$.
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. $\text{pH} = 12.0 \Rightarrow \text{pOH} = 14.0 - 12.0 = 2.0$.
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. $\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.005\text{ mol} \times 40\text{ g/mol} = 0.2\text{ g}$.

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: $\frac{8.05}{2.01} \approx 4\text{ cells}$ 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 = $\frac{63.5}{2} = 31.75\text{ g}$.

• 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 ($M_1 V_1 = M_2 V_2$): $$0.2 \times X = 0.12 \times (X + 20)$$ $$0.2X = 0.12X + 2.4 \Rightarrow 0.08X = 2.4 \Rightarrow X = 30\text{ mL}$$

• (1) AgCl (AB type): $S = \sqrt{1.0 \times 10^{-10}} = 10^{-5}\text{ M}$. Total ion concentration = $2 \times 10^{-5}\text{ M}$.

• (3) AgBr (AB type): $S = \sqrt{1.0 \times 10^{-8}} = 10^{-4}\text{ M}$. Total ion concentration = $2 \times 10^{-4}\text{ M}$.

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

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

• $[A] = 4$

• $[C] = 3$

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.

Moles of Ca(OH)₂ = 0.1 L × 0.5 M = 0.05 moles.
Moles of HCl = 0.2 L × 0.6 M = 0.12 moles.
Reaction ratio is 1:2.

0.05 moles Ca(OH)₂ require exactly 0.10 moles of HCl.

Thus, HCl is in excess by (0.12 - 0.10) = 0.02 moles.
Total final volume = 100 mL + 200 mL = 300 mL (0.3 L).
Concentration of excess = 0.02 moles / 0.3 L ≈ 0.0667 M.

Moles of hydrated acid in 250 mL = 0.25 L × 0.08 M = 0.02 moles.
Molar mass of hydrated acid = 2.52 g / 0.02 moles = 126 g/mol.
Mass of water of hydration = 126 - 90 = 36 g/mol.

Since H₂O is 18 g/mol, X = 36 / 18 = 2.
Titration: 25 mL acid contains 0.002 moles.

Moles NaOH = 0.16 g / 40 g/mol = 0.004 moles.

Ratio of Acid:Base is 0.002:0.004 = 1:2.

The acid is diprotic.

Since pH = 3 for both, [H⁺] = 10^-3 M for both.
Using K_a ≈ [H⁺]² / C_a:
For HA: K_a = (10^-3)² / 0.1 = 10^-5
For HB: K_a = (10^-3)² / 0.01 = 10^-4
10^-4 > 10^-5, so acid HB has the greater K_a value.

The positive EMF (+1.6 V) indicates a spontaneous reaction, making it a Galvanic cell.

In the reaction, Y goes from oxidation state 0 to +1.

Oxidation occurs at the anode, so the Y-electrode is the anode.

First, harmonize standard potentials to Reduction Potentials (E°red):
X: -0.44V, Y: +1.5V, Z: +0.34V (flipped from ox), L: -1.18V (flipped from ox).
Greatest EMF = E°red(Cathode) - E°red(Anode).

Maximum gap is between highest (+1.5V for Y) and lowest (-1.18V for L).
L must oxidize (anode), Y must reduce (cathode): L + Y²⁺ → L²⁺ + Y.
EMF = 1.5 - (-1.18) = +2.68 V.

According to Faraday's law, Equivalent Mass = Mass / Faradays = 18 g / 1.5 F = 12 g/eq.
We check the equivalent mass for each metal:
Na (23/1 = 23), Mg (24/2 = 12), Ca (40/2 = 20), K (39/1 = 39).
Magnesium (Mg) matches.

Molar mass of Al(OH)₃ = 27 + (3 × 17) = 78 g/mol.
Moles of precipitate = 2.03 g / 78 g/mol = 0.0260 mol of Al.
Mass of Al = 0.0260 mol × 27 g/mol = 0.702 g.
Percentage in 5 g sample = (0.702 / 5) × 100 ≈ 14.05%.

PbCl₂ dissociates into 3 ions, Ksp = 4S³ = 1.2 × 10^-5.
S = &root3(1.2 × 10^-5 / 4) ≈ 0.0144 M.
Moles of 0.1 g = 0.1 / 278 ≈ 3.6 × 10^-4 moles.
Volume = Moles / Molarity = 3.6 × 10^-4 / 0.0144 ≈ 0.025 L = 25 mL.

K_c = (0.500)² / (0.200 × 3.00) = 0.4167.
To increase [C] from 0.5 to 0.7, 0.2 moles of C must form, meaning 0.1 moles of A and B react.

New [B] = 3.0 - 0.1 = 2.9 M.
At new equilibrium: (0.700)² / ([A]_eq × 2.9) = 0.4167 → [A]_eq ≈ 0.405 M.
The total A required before reaction shift is 0.405 + 0.1 = 0.505 M.
We started with 0.200 M, so we must add 0.505 - 0.200 = 0.305 mol.

Moles of Zinc consumed (oxidized) = 6.5 g / 65 g/mol = 0.1 mol.
Since both Zn and Cu are divalent (exchange 2 e^-), 0.1 mol of Cu is deposited.
Mass of Cu = 0.1 mol × 63.5 g/mol = 6.35 g.
In a Galvanic (Daniell) cell, Copper acts as the Cathode, which is the POSITIVE electrode.

Because the temperature is constant, Kc remains the same before and after adding substance A.

Using the first set of equilibrium concentrations:

Kc = (3² × 4.5³) / (1.5² × 2.3³) = (9 × 91.125) / (2.25 × 12.167) = 820.125 / 27.37575 ≈ 29.958

Using the new equilibrium concentrations and the same Kc:

29.958 = (4² × 6³) / (2.5² × [B]³)

29.958 = (16 × 216) / (6.25 × [B]³)

29.958 = 3456 / (6.25 × [B]³)

[B]³ = 3456 / (6.25 × 29.958) = 3456 / 187.2375 ≈ 18.458

[B] = ∛18.458 ≈ 2.64 M

Reasoning: The ionization constant (Ka) is a thermodynamic value that depends only on temperature. Diluting a weak acid changes its degree of ionization (α), hydronium concentration, and pH, but it does NOT change the Ka value. Therefore, Ka of X equals Ka of Y.

Step 1: Determine EMF.
The reaction shows Cu²⁺ is reduced (Cathode) and Cr is oxidized (Anode).
Reduction potential of Cu = +0.340 V. Oxidation potential of Cr = +0.740 V.
EMF = E°(oxidation) + E°(reduction) = 0.740 V + 0.340 V = +1.08 V.

Step 2: Determine Cell Type.
Because the EMF is positive, the reaction is spontaneous. A spontaneous electrochemical cell is a Galvanic cell.

Correction Note: In the user-provided prompt 'Answer Key', the answer is marked as (C) for 19. Let's recalculate based on standard chemistry.

First cell: X is Anode. E_cell = E_ox(X) + E_red(H) => 0.23 = E_ox(X) + 0. So, Oxidation Potential of X = +0.23 V.

Second cell: Y is Cathode. E_cell = E_ox(H) + E_red(Y) => 0.80 = 0 + E_red(Y). So, Reduction Potential of Y = +0.80 V. (Oxidation Potential of Y = -0.80 V).

Comparing Oxidation Potentials: X (+0.23 V) > Y (-0.80 V).
Because X has a higher oxidation potential, X is the Anode and Y is the Cathode.

EMF = E_ox(Anode) + E_red(Cathode) = 0.23 V + 0.80 V = +1.03 V.

(Thus, statement (a) is chemically correct based on the provided text, despite what a raw key might say. X oxidizes, Y reduces.)

• Anodic Protection (Sacrificial Anode): Coating a metal with a more active metal. The coating has a higher oxidation potential.

• Cathodic Protection (Cathodic Coating): Coating a metal with a less active metal. The coating has a lower oxidation potential.

Step 1: Write the balanced chemical equation.

From the equation, 1 mole of acid reacts with 2 moles of base (nₐ = 1, n_b = 2).

Step 2: Use the titration formula.

(Mₐ × 32) / 1 = (M_b × 16) / 2

32 Mₐ = 8 M_b

Mₐ = (8 / 32) M_b = ¼ M_b

Therefore, the concentration of the sulphuric acid is a quarter of the concentration of the potassium hydroxide.

Step 1: Convert solubility to Molarity (mol/L).
Solubility given = 4.4 × 10⁻⁵ mol / 100 mL.
To find mol / 1000 mL (1 Liter), multiply by 10:
Molar Solubility (s) = 4.4 × 10⁻⁴ mol/L.

Step 2: Write the dissociation equation and Ksp expression.
MgCO₃(s) ⇌ Mg²⁺(aq) + CO₃²⁻(aq)
Ksp = [Mg²⁺][CO₃²⁻] = (s)(s) = s².

Step 3: Calculate Ksp.
Ksp = (4.4 × 10⁻⁴)² = 19.36 × 10⁻⁸ = 1.936 × 10⁻⁷.

The graph must show reactant [A] starting at 0.8 and dropping to level off at 0.4. Product [B] must start at 0 and rise to level off at 0.8. (Graph B illustrates this perfectly).

Kc = (0.8)² / (0.4) = 0.64 / 0.4 = 1.6.

Step 1: Electricity for Gold
Reaction: Au³⁺ + 3e⁻ → Au
To precipitate 1 mole of Au, we need 3 moles of electrons (3 Faradays).

Step 2: Determine valence of unknown metal
We have 3 Faradays of electricity available. We want to know which ion will yield 1.5 moles of precipitate with these 3 Faradays.
Moles of metal = (Faradays) / (Valence charge z)
1.5 = 3 / z
z = 3 / 1.5 = 2.

The unknown ion must have a +2 charge. Looking at the options, only Mg²⁺ is divalent (+2).

From the balanced half-reaction, producing 1 mole of MnO₄⁻ requires 5 moles of electrons (5 Faradays).

To produce 0.2 moles of MnO₄⁻, the moles of electrons required = 0.2 × 5 = 1 mole of electrons (1 Faraday).

We know that 1 Faraday is equal to 96500 Coulombs (C). Therefore, the required electricity is 96500 C.

Decreasing volume increases pressure, shifting equilibrium backward (toward fewer gas molecules), reducing C.

[H⁺] = 10⁻ᵖᴴ = 10⁻⁴·⁷ ≈ 2 × 10⁻⁵ M

Ka = [H⁺]² / C

C = [H⁺]² / Ka = (2 × 10⁻⁵)² / (6.2 × 10⁻¹⁰)

C = 4 × 10⁻¹⁰ / 6.2 × 10⁻¹⁰ ≈ 0.645 M

n = M × V = 0.645 × 0.300 = 0.193 ≈ 0.19 mol ✓

Stoichiometry: 1 N₂H₄ → 1 N₂ + 2 H₂

If [H₂] = 0.2 M → [N₂] = 0.1 M (from 2:1 ratio)

[N₂H₄] remaining = 0.2 − 0.1 = 0.1 M

ΔH < 0 → forward reaction is exothermic

Increasing T shifts equilibrium backward (Le Chatelier)

[N₂] must decrease below 0.1 M

Only option < 0.1 M is 0.08 M ✓

Zn(s) ⇌ Zn²⁺(aq) + 2e⁻

Zinc has high oxidation tendency:

• Zn atoms leave the rod as Zn²⁺ ions → solution contains positive Zn²⁺ ions

• Electrons remain on the rod → rod becomes negatively charged

The density of the electrolyte (H₂SO₄) indicates the battery's charge state:

Battery Y (charged) acts as the energy source and charges battery X (discharged).

During charging, reactions are reversed from discharge:

At − electrode (cathode in electrolytic cell):

Pb²⁺ + 2e⁻ → Pb (reduction) ✓

At + electrode (anode in electrolytic cell):

Pb²⁺ → Pb⁴⁺ + 2e⁻ (oxidation, forms PbO₂)

At + electrode (cathode in galvanic cell):

Pb⁴⁺ + 2e⁻ → Pb²⁺ (reduction) ✓

At − electrode (anode in galvanic cell):

Pb → Pb²⁺ + 2e⁻ (oxidation)

Option (d) correctly describes:

• X: Pb²⁺ → Pb at − electrode (reduction during charging) ✓

• Y: Pb⁴⁺ → Pb²⁺ at + electrode (reduction during discharge) ✓

Iron has higher oxidation potential (+0.409 V) → Fe is anode (oxidized)

Tin: Sn²⁺ + 2e⁻ → Sn (reduction) → Sn is cathode

EMF = E°(oxidation, anode) + E°(reduction, cathode)

EMF = 0.23 + 0.80 = +1.03 V

Positive EMF → reaction is spontaneous → Galvanic cell ✓

n(BaCl₂·2H₂O) = 45 / 244 = 0.1844 mol

M = 0.1844 / 0.500 L = 0.369 M

n(BaCl₂) = 0.369 × 0.0842 = 0.0311 mol

n(Na₂SO₄) = 0.0311 mol

M(hydrate) = 10 g / 0.0311 mol = 322 g/mol

322 = 142 + 18X

18X = 180 → X = 10 ✓

Formula: Na₂SO₄·10H₂O

X₂Y₃ → 2X³⁺ + 3Y²⁻

If s = 1×10⁻⁴: [X³⁺] = 2s; [Y²⁻] = 3s

Ksp = (2s)² × (3s)³ = 4s² × 27s³ = 108 s⁵

= 108 × (10⁻⁴)⁵ = 108 × 10⁻²⁰ = 1.08 × 10⁻¹⁸ ✓

This matches the given Ksp exactly, confirming the formula is X₂Y₃.

Y has higher Ka (5.1 × 10⁻⁴ > 1.8 × 10⁻⁵)

→ Y is the stronger acid → more ionization

• More ionization in Y → more ions → better electrical conductivity

• Higher [H⁺] in Y → lower pH

Au³⁺ + 3e⁻ → Au

Moles Au = (moles e⁻) / 3 = 0.5 / 3 = 0.1667 mol

Mass = 0.1667 × 197 = 32.83 g ≈ 32.8 g ✓

In electroplating, the anode is the same metal (Au) as the coating.

At anode: Au → Au³⁺ + 3e⁻ (replenishes ions)

At cathode: Au³⁺ + 3e⁻ → Au (deposits on spoon)

Ions consumed = ions produced → concentration remains constant ✓

n(NaOH) = 0.010 L × 1 M = 0.010 mol

V(HCl) needed = n / M = 0.010 / 2 = 0.005 L = 5 mL

Final burette reading = 6.5 + 5.0 = 11.5 mL

K_sp = [Ag⁺][Cl⁻] = S² → S = √(1.6×10⁻¹⁰) = 1.265×10⁻⁵ M

Total ions = [Ag⁺] + [Cl⁻] = 2S = 2 × 1.265×10⁻⁵ = 2.53×10⁻⁵ M

pOH = 14 − 12.3 = 1.7 → [OH⁻] = 10^−1.7 = 0.02 M

[Ba(OH)₂] = 0.02/2 = 0.01 M

n = 0.01 × 1.5 = 0.015 mol → m = 0.015 × 171 = 2.565 ≈ 2.56 g

K_p = P(N₂O₄) / [P(NO₂)]²

6.15 = P(N₂O₄) / (0.15)² = P(N₂O₄) / 0.0225

P(N₂O₄) = 6.15 × 0.0225 = 0.1383 atm

X is oxidised (anode); Cu²⁺ is reduced (cathode), E°(Cu²⁺/Cu) = +0.34 V.

E_cell = E°_cathode − E°_anode

+1.8 = +0.34 − E°(X) → E°(X) = 0.34 − 1.8 = −1.46 V

Reddish-brown precipitate = Fe(OH)₃ (iron(III) hydroxide).

Fe₃O₄ = FeO·Fe₂O₃ = contains both Fe²⁺ and Fe³⁺ ions.

(b) Oxidation first would work, but the question asks for the simplest correct sequence. Option (a) achieves the same result more directly by allowing air oxidation during the waiting period.

(c) Reduction would convert Fe³⁺ to Fe²⁺, making it harder to get Fe(OH)₃.

(d) Dilute HCl produces FeCl₂ and FeCl₃, but without the waiting period, Fe(OH)₂ (green) would also form.

MgCl₂ + H₂SO₄ → MgSO₄ + 2HCl

MgSO₄ + Na₂CO₃ → MgCO₃↓ (M = 84 g/mol) + Na₂SO₄

n(MgCO₃) = 7/84 = 0.08333 mol = n(MgCl₂)

m(MgCl₂) = 0.08333 × 95 = 7.916 g

% = (7.916/10) × 100 = 79.16%

Mass of FeCl_x = 5.34 × 0.6008 = 3.208 g

Mass of 6H₂O lost = 5.34 − 3.208 = 2.132 g → n(H₂O) = 2.132/18 = 0.11844 mol

n(FeCl_x) = 0.11844/6 = 0.01974 mol

M(FeCl_x) = 3.208/0.01974 = 162.5 g/mol

56 + 35.5x = 162.5 → 35.5x = 106.5 → x = 3

[NO] = 0.15 M, [O₂] = 0.15 M, [NO₂] = 0.25 M

Q = [NO₂] / ([NO][O₂]^½) = 0.25 / (0.15 × √0.15) = 0.25 / 0.0581 ≈ 4.30

Q (4.30) < K_c (6.33) → reaction proceeds forward to increase [NO₂] and reach equilibrium.

HCN ⇌ H⁺ + CN⁻

x = 0.2 − 0.167 = 0.033 M

[H⁺] = [CN⁻] = 0.033 M

K_a = [H⁺][CN⁻] / [HCN]

K_a = (0.033)(0.033) / 0.167

K_a = 0.001089 / 0.167

K_a ≈ 5.45 × 10⁻³

If K_a = 5.45 × 10⁻³:

x² / (0.2 − x) = 5.45 × 10⁻³

x² = 5.45 × 10⁻³ × 0.167 = 9.10 × 10⁻⁴

x = √(9.10 × 10⁻⁴) ≈ 0.0302 M ≈ 0.033 M ✓

n(NaCl) = 5.85/58.5 = 0.1 mol; requires 0.1 mol e⁻ = 0.1 F for complete decomposition.

Cathode: Na⁺ + e⁻ → Na → m(Na) = 0.1 × 23 = 2.3 g

Anode: 2Cl⁻ → Cl₂ + 2e⁻ → n(Cl₂) = 0.05 mol → m(Cl₂) = 0.05 × 71 = 3.55 g

0.1 F is exactly enough → complete decomposition.

The correct option is a) n = 2.

The correct option is b).

The diprotic acid at 0.5 M supplies 1 eq/L of H⁺; the monoacidic base KOH at 1 M supplies 1 eq/L of OH⁻ — so equal volumes neutralise exactly. ✓

• Surface area: Powder > strip > piece → higher rate.

• Concentration: 1 M > 0.1 M → more collisions.

• Temperature: 50°C > 25°C → more energy.

The correct option is b).

• Pipe = Z (−2.71 V): Most active → sacrificial anode for both tank and tap.

• Tap = Y (+0.30 V): Most noble → provides cathodic protection to the tank.

• Tank = X (−0.40 V): Protected cathodically by both. ✓

The correct option is a).

Hydrogen at 800 °C carries out the complete reduction of Fe₂O₃ to metallic iron (Fe); strong heating of iron in air then forms the magnetic oxide Fe₃O₄. ✓

The correct option is c).

Only K₂CO₃ reacts: K₂CO₃ + MgSO₄ → MgCO₃↓ + K₂SO₄ (KCl gives no precipitate).

Correction note: 0.0595 × 138 = 8.21 g (not 3.81 g) — this gives % KCl = 31.55%.

The correct option is c).

The correct option is a).

Chromium in Cr₂O₇²⁻ is +6, so each Cr atom needs 6 electrons. Dividing the total electrons by 4 (electrons per O₂) gives 4.36 mol O₂. ✓

The correct option is b).

🎯 The Core Chemical Rule: The activation energy and energy profiles of reactants or products are heavily governed by thermal factors. Changes in basal energy fields without shifting the absolute inner mechanism dimensions correspond cleanly to external kinetic variables.

✅ Precise Mathematical Curve Analysis:
• In Diagram (1): Reactant energy = 100 kJ, Product energy = 25 kJ, and Activated complex energy (peak) = 175 kJ. Thus, the activation energy for the forward pathway is $E_a = 175 - 100 = \text{\color{#4da3ff}75 kJ}$.
• In Diagram (2): Reactant energy = 225 kJ, Product energy = 150 kJ, and Activated complex energy (peak) = 300 kJ. Thus, the forward activation energy is $E_a = 300 - 225 = \text{\color{#ffb347}75 kJ}$.
• The mathematical activation energy ($E_a$) and enthalpy change ($\Delta H = 25 - 100 = -75\text{ kJ}$) remain completely identical across both figures. However, all absolute chemical energy values have been elevated symmetrically by exactly 125 kJ because supplying thermal motion raises the base internal kinetic energy of all chemical entities, indicating an **increase in temperature**.

❌ Why Other Options Fail:
• Adding a Catalyst (Option c): A catalyst creates an alternate pathway with a lower barrier, lowering only the peak (activated complex) while leaving the base horizontal energy lines for reactants and products completely unchanged.
• Increasing Surface Area or Concentration (Options b and d): These factors increase the collision frequency per unit time, enhancing the reaction rate, but do not alter the potential energy values along the vertical axis of a reaction profile diagram.

🎯 The Core Chemical Rule: The equilibrium constant ($K_c$) is determined by establishing the balanced chemical equation, constructing an ICE (Initial, Change, Equilibrium) table to track concentration variables, and inserting the final equilibrium values into the law of mass action expression.

✅ Step-by-Step ICE Table Construction:
• Balanced Dissociation Equation: $2\text{SO}_3(g) \rightleftharpoons 2\text{SO}_2(g) + \text{O}_2(g)$
• Initial Concentrations ($C = n/V$): Since the volume is $1\text{ L}$, initial $[\text{SO}_3] = 0.2\text{ M}$. Initial products are $0\text{ M}$.
• Change Count via Decomposition Degree: $10\%$ of the initial $\text{SO}_3$ decomposes: $0.2 \times 0.10 = 0.02\text{ M}$.
    - Change for $\text{SO}_3 = -0.02\text{ M}$
    - Change for $\text{SO}_2 = +0.02\text{ M}$ (due to $2:2$ molar ratio)
    - Change for $\text{O}_2 = +0.01\text{ M}$ (due to $2:1$ molar ratio)
• Equilibrium Concentrations:
    - $[\text{SO}_3]_{eq} = 0.2 - 0.02 = 0.18\text{ M}$
    - $[\text{SO}_2]_{eq} = 0.02\text{ M}$
    - $[\text{O}_2]_{eq} = 0.01\text{ M}$

📊 Calculating the Value of $K_c$:
The equilibrium mathematical expression is:
    $K_c = \frac{[\text{SO}_2]^2 \cdot [\text{O}_2]}{[\text{SO}_3]^2}$
    $K_c = \frac{(0.02)^2 \cdot 0.01}{(0.18)^2} = \frac{4 \times 10^{-4} \cdot 10^{-2}}{0.0324} = \frac{4 \times 10^{-6}}{3.24 \times 10^{-2}} \approx \mathbf{1.23 \times 10^{-4}}$

🎯 The Core Chemical Rule: For weak basic (alkaline) solutions, the concentration of hydroxide ions ([OH^-]) is determined by multiplying the initial molarity of the base (C_b) by its degree of dissociation (α). Taking the negative logarithm of this concentration yields the pOH value.

✅ Step-by-Step Mathematical Solution:
• 1. Base Concentration (C_b): Since the total solution volume is exactly 1 L, the base concentration matches its number of moles directly:
    C_b = n / V = 0.25 mol / 1 L = 0.25 M
• 2. Hydroxide Ion Concentration ([OH^-]): According to Ostwald's dilution law equations:
    [OH^-] = α × C_b
    [OH^-] = (2 × 10^-2) × 0.25 = 0.005 M = 5 × 10^-3 M
• 3. Computing the pOH: Apply the negative base-10 logarithm function:
    pOH = -log[OH^-]
    pOH = -log(5 × 10^-3) = 3 - log(5) = 3 - 0.7 = 2.3

📊 Analytical Insight: The direct calculation yields a pOH of 2.3, matching option (d). If the question had requested the pH value instead, it would be calculated as 14 - 2.3 = 11.7 (which corresponds to option a).

🎯 The Core Chemical Rule: According to Le Chatelier's principle, altering the volume or pressure shifts the equilibrium position toward the side that counteracts the change. Crucially, **the equilibrium constant ($K_c$) value is altered exclusively by changing the temperature**.

✅ Step-by-Step Reaction Analysis:
• The Fixed $K_c$ Constraint: The question specifies increasing the yield of products "without affecting the $K_c$ value". This condition immediately eliminates option (a), because changing temperature changes $K_c$.
• Counting Gaseous Moles:
    - Reactants side has $1\text{ mol}$ of gas ($\text{CO}_2$). Pure solid carbon ($\text{C}_{(s)}$) is excluded from gas laws.
    - Products side has $2\text{ mol}$ of gas ($\text{CO}$).
• Effect of Increasing Vessel Volume: By **increasing the volume of the reaction vessel**, the internal system pressure drops. The equilibrium counters this change by shifting forward to the side with the larger number of gaseous moles ($1 \to 2$). This forward shift successfully increases the amount of carbon monoxide ($\text{CO}$) gas while keeping $K_c$ unchanged.

❌ Why Other Options Fail:
• Option (b): Carbon is a pure solid; modifying its mass or surface area does not shift the equilibrium position or change gas ratios.
• Option (c): Decreasing the volume increases pressure, driving the system in the reverse direction toward fewer gaseous moles, which decreases the yield of $\text{CO}$.

n(HCl)=7.3/36.5=0.2 mol → n(CaCO₃)=0.1 mol → mass=10 g. Purity=80%, impurities=20%.

Exp1: acid:base = 2 mmol:1 mmol = 2:1. Exp2: base=5 mmol → acid=10 mmol → 10/40 = 0.250 M.

Ca₃(PO₄)₂: Ksp=(3S)³(2S)²=108S⁵=108×10⁻²⁰=1.08×10⁻¹⁸ ✔. Anion = phosphate.

n(Cu)=0.12 mol → e⁻=0.24 mol → Q=0.24×96500=23160 C → t=23160/96.5=240 sec.