Question

a)in an experiment involving the exothermic dissolution of a salt, the salt did not dissolve completely how would this affect the calculated heat of solution

b)after ensuring all the product had precipitated out the beakers containing the precipitate were heated on a hotplate for 20 minutes. What is this process called and why is it used?

c)a student used 15.37ml of 0.1055m naoh to reach the end point. If he used 5ml of vinegar in this titration calculate the mass % of acetic acid in vinegar. Assume the density of vinegar to be 1.00g/ml

d)if 32.7 cal of heat is removed from 17.8g of copper, the temp decreased to 15.2C. what was the inital temp of the copper. the specific heat of copper is 0.0908cal/gC

Answer 1

Answer on Question #56492 – Chemistry - General Chemistry

Question:

a) in an experiment involving the exothermic dissolution of a salt, the salt did not dissolve completely how would this affect the calculated heat of solution

b) after ensuring all the product had precipitated out the beakers containing the precipitate were heated on a hotplate for 20 minutes. What is this process called and why is it used?

c) A student used 15.37 ml of 0.1055 M NaOH to reach the end point. If he used 5 ml of vinegar in this titration calculate the mass % of acetic acid in vinegar. Assume the density of vinegar to be 1.00 g/ml.

d) If 32.7 cal of heat is removed from 17.8 g of copper, the temp decreased to 15.2°C. What was the initial temp of the copper. The specific heat of copper is 0.0908 cal/gC.

Answer:

a) Heat output from dissolution process will be lower than the calculated one.

b) Precipitate heat treatment is the most widespread technique to increase the yield of precipitate. The process is called "precipitate hardening".

c)

$$v = \frac{m}{M}$$ $$C_M = \frac{v}{V}$$ $$\rho = \frac{m}{V}$$ $$\mathrm{CH_3COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O}$$ $$v(\mathrm{NaOH}) = C_M V = 0.1055 \cdot 15.37 = 1.62 \ \mathrm{mmol}$$

According to the reaction,

$$v(\mathrm{NaOH}) = v(\mathrm{CH_3COOH})$$ $$m(\mathrm{CH_3COOH}) = v(\mathrm{CH_3COOH}) \cdot M(\mathrm{CH_3COOH})$$ $$M(\mathrm{CH_3COOH}) = 48.04 \ \mathrm{g/mol}$$ $$m(\mathrm{CH_3COOH}) = 0.00162 \cdot 48.04 = 0.08 \ \mathrm{g}$$ $$\%(\mathrm{CH_3COOH}) = \frac{m(\mathrm{CH_3COOH})}{m(\mathrm{solution})} \cdot 100$$ $$\%(\mathrm{CH_3COOH}) = \frac{0.08}{5 \cdot 1.00} \cdot 100 = 1.6\%$$

d)

$$Q = mC\Delta T$$ $$32.7 = 17.8 \cdot 0.0908 \cdot (x_1 - 15.2)$$ $$x_1 = \frac{32.7}{17.8 \cdot 0.0908} + 15.2 = 35.5^\circ C$$

www.AssignmentExpert.com