Calculate the empirical formula for a compound that is 40.0% C , 6.7% H , and 53.3% O by mass.

Assuming 100g sample: moles of C = 3.33, moles of H = 6.67, moles of O = 3.33. Dividing by the smallest value (3.33) yields a ratio of 1:2:1. The empirical formula is $CH_2O$ .

Given 10.0 g of $CH_3OH$ reacting with excess $PBr_3$, calculate the theoretical yield of $CH_3Br$ if the reaction stoichiometry is $3CH_3OH + PBr_3 \rightarrow 3CH_3Br + H_3PO_3$.

Moles of $CH_3OH$ = $\frac{10.0 \text{ g}}{32.04 \text{ g/mol}} \approx 0.312 \text{ mol}$. Since the ratio is 1:1, theoretical yield of $CH_3Br$ = $0.312 \text{ mol} \times 94.94 \text{ g/mol} = 29.6 \text{ g}$.

Determine the pH of a solution formed by mixing 50.0 mL of 0.10 M HCl with 50.0 mL of 0.10 M NaOH .

The reaction $H^+ + OH^- \rightarrow H_2O$ results in complete neutralization. At the equivalence point for strong acid/strong base, the solution is neutral, so pH = 7.00 .

Calculate the enthalpy change ($ΔH°_{rxn}$)** for a reaction using bond dissociation enthalpies (BDE) given $\sum BDE_{reactants} = 2500 \text{ kJ/mol}$ and $\sum BDE_{products} = 2800 \text{ kJ/mol}$.

The enthalpy of reaction is calculated as $\Delta H_{rxn} = \sum BDE_{broken} - \sum BDE_{formed}$. Thus, $\Delta H = 2500 - 2800 = -300 \text{ kJ/mol}$.

Determine the temperature at which a reaction with $\Delta H = 150 \text{ kJ/mol}$ and $\Delta S = 0.500 \text{ kJ/(mol·K)}$ becomes spontaneous .

A reaction is spontaneous when $\Delta G 300 K .

For a first-order reaction with a rate constant $k = 0.050 \text{ s}^{-1}$ , calculate the half-life ($t_{1/2}$).

For a first-order reaction, $t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{0.050 \text{ s}^{-1}} = 13.86 \text{ s}$.

Derive the rate law for a mechanism where the first step $A + B \rightleftharpoons C$ (fast) is followed by $C \rightarrow D$ (slow).

The rate is determined by the slow step: $Rate = k_2[C]$. Using the equilibrium constant $K_{eq} = \frac{[C]}{[A][B]}$, substitute $[C] = K_{eq}[A][B]$ into the rate law to get $Rate = k'[A][B]$ .

Calculate the equilibrium constant ($K_p$) for $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ if partial pressures are $P_{N_2O_4} = 0.50 \text{ atm}$ and $P_{NO_2} = 0.20 \text{ atm}$.

$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(0.20)^2}{0.50} = \frac{0.04}{0.50} = 0.08$.

Determine the pH of a buffer solution containing 0.20 M $CH_3COOH$ and 0.10 M $CH_3COONa$ , given $pK_a = 4.74$.

Using the Henderson-Hasselbalch equation : $pH = pK_a + \log(\frac{[base]}{[acid]}) = 4.74 + \log(\frac{0.10}{0.20}) = 4.74 + (-0.30) = 4.44$.

Calculate the molar solubility of $PbCl_2$ in water if $K_{sp} = 1.6 \times 10^{-5}$.

The equilibrium is $PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-$. $K_{sp} = (s)(2s)^2 = 4s^3$. Thus, $s = \sqrt[3]{\frac{1.6 \times 10^{-5}}{4}} = \sqrt[3]{4.0 \times 10^{-6}} \approx 0.0159 \text{ M}$.

Use the Nernst equation to calculate the cell potential $E_{cell}$ for a cell where $Q = 0.01$ and $E^\circ_{cell} = 1.10 \text{ V}$ at 298 K ($n=2$).

$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n}\log(Q) = 1.10 - \frac{0.0592}{2}\log(0.01) = 1.10 - (0.0296 \times -2) = 1.10 + 0.059 = 1.159 \text{ V}$.

Calculate the mass of Copper deposited in 1 hour if a current of 2.00 A is passed through a $CuSO_4$ solution ($M = 63.55 \text{ g/mol}$, $n=2$, $F = 96485 \text{ C/mol}$).

Charge $Q = I \times t = 2.00 \text{ A} \times 3600 \text{ s} = 7200 \text{ C}$. Moles of Cu = $\frac{Q}{nF} = \frac{7200}{2 \times 96485} \approx 0.0373 \text{ mol}$. Mass = $0.0373 \times 63.55 \approx 2.37 \text{ g}$.

Identify the hybridization and molecular geometry of the central atom in $XeOF_4$.

The central Xenon atom has 5 bonding pairs and 1 lone pair, totaling 6 electron domains. This corresponds to $sp^3d^2$ hybridization and square pyramidal geometry.

Determine the formal charge of the Nitrogen atom in the nitrite ion ($NO_2^-$) using the standard Lewis structure.

In the standard Lewis structure, N has one double bond (2 electrons), one single bond (1 electron), and one lone pair (2 electrons). Formal charge = (5 valence e⁻) - (2 + 1 + 2) = 0.

Which mechanism, $S_N1$ or $S_N2$ , is favored for the substitution of a primary alkyl halide with a strong nucleophile?

Primary alkyl halides have minimal steric hindrance, making them highly reactive toward $S_N2$ mechanisms, which occur in a single concerted step.

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Mastering Quantitative Foundations and Analytical Techniques for Chemistry Olympiad Exams

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Q: Calculate the empirical formula for a compound that is 40.0% C, <s...

Q: Given 10.0 g of $CH_3OH$ reacting with excess $PBr_3$, calculate the theore...

Q: Determine the pH of a solution formed by mixing 50.0 mL of <strong...

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