Question

a chemist has synthesized a greenish - yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/l at 36°c and 2.88 atm. calculate the molar mass of the compound. (the unit is g/mol and your answer should have 3 significant figures).
add your answer
integer, decimal, or e notation allowed
question 2
4 points
a sample of a gas occupies 1.40 ` 10³ ml at 25°c and 760 mmhg. what volume will it occupy at the same temperature and 380 mmhg? ( your answer should have 4 significant figures and the unit is ml)
add your answer
integer, decimal, or e notation allowed
question 3
4 points
aluminum metal has a specific heat of 0.900 j/g·°c. calculate the amount of heat in kj required to raise the temperature of 10.5 moles of al from 30.5°c to 225°c. (your answer should have 3 significant figures).
add your answer
integer, decimal, or e notation allowed

Explanation:

Question 1

Step1: Convert temperature to Kelvin

$T=(36 + 273.15)\text{K}=309.15\text{K}$

Step2: Use the ideal - gas - density formula $M=\frac{dRT}{P}$

Given $d = 7.71\text{ g/L}$, $R=0.0821\text{ L}\cdot\text{atm/mol}\cdot\text{K}$, $T = 309.15\text{K}$, $P = 2.88\text{ atm}$
$M=\frac{7.71\times0.0821\times309.15}{2.88}$
First, calculate the numerator: $7.71\times0.0821\times309.15=7.71\times25.381215 = 195.689177$
Then, divide by the denominator: $M=\frac{195.689177}{2.88}\approx67.9\text{ g/mol}$

Step1: Use Boyle's law $P_1V_1 = P_2V_2$

Given $P_1 = 760\text{ mmHg}$, $V_1=1.40\times10^{3}\text{ mL}$, $P_2 = 380\text{ mmHg}$
We can solve for $V_2$: $V_2=\frac{P_1V_1}{P_2}$
Substitute the values: $V_2=\frac{760\times1.40\times 10^{3}}{380}$
Since $\frac{760}{380}=2$, then $V_2 = 2\times1.40\times10^{3}=2.80\times10^{3}\text{ mL}$

Step1: Calculate the mass of Al

The molar mass of $Al$ is $M = 26.98\text{ g/mol}$. For $n = 10.5\text{ mol}$, $m=n\times M=10.5\times26.98 = 283.29\text{ g}$

Step2: Use the heat formula $q = mc\Delta T$

$c = 0.900\text{ J/g}\cdot^{\circ}\text{C}$, $\Delta T=(225 - 30.5)^{\circ}\text{C}=194.5^{\circ}\text{C}$
$q=283.29\times0.900\times194.5$
First, $283.29\times0.900 = 254.961$
Then, $254.961\times194.5=254.961\times(200 - 5.5)=254.961\times200-254.961\times5.5$
$=50992.2-1402.2855 = 49589.9145\text{ J}$
Convert to kJ: $q = 49.6\text{ kJ}$

Answer:

$67.9$

Question 2