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7.) A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To
what temperature must the gas in the syringe be heated/cooled in order to have a volume of 435
mL at 2.50 atm?
A) 139 K
B) 572 K
C) 175 K
D) 466 K
E) 721 K
8.) What mass of NO2 is contained in a 13.0 L tank at 4.58 atm and 385 K?
A) 18.8 g
B) 86.7 g
C) 24.4 g
D) 53.1 g
E) 69.2 g

Respuesta :

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Answer:

[tex]\large \boxed{\text{E) 721 K; B) 86.7 g}}[/tex]

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

[tex]\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}[/tex]

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻Âčmol⁻Âč

T = 385 K

M = 46.01 g/mol

(b) Calculation

[tex]\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}[/tex]

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