Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.
We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:
osmotic pressure = CRT
<span>C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L</span>
<span>moles of particles = C*V = 0.3189*0.250 =0.0797 mol </span>
<span>0.0797 = moles of sucrose + 2*moles of salt </span>
<span>x + 2y = 0.0797 </span>
<span>and </span>
<span>x(MMsucrose) + y(MMNaCl) = 10.2</span>
<span>342x + 58.5y = 10.2
</span>
<span>solve for x and y
</span>
<span>x = 0.0252 mol sucrose</span>
<span>y = 0.0273 mol NaCl
</span>
<span>mass Sucrose = 0.0252(342) = 8.6184 g </span>
<span>mass NaCl = 0.0273(58.5) = 1.5971 g </span>
<span>% NaCl = (1.5971 / 10.2)*100 = 15.66%</span>
2C4H10 + 13O2 → 8CO2 + 10H2O Since the equation is balanced, we can set up a proportion: 13 moles of O2 react with 2 moles of C4H10x moles of O2 react with 0.425 moles of C4H10 13 → 2x → 0.425 x = 13 * 0.425 / 2 = 2.7625 <span>2.7625 moles of O2 react with 0.425 moles of C4H10</span>
