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frez [133]
3 years ago
13

Assuming that the container is completely full, that the temperature is 22.1 ∘C, and that the atmospheric pressure is 1.1 atm ,

calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 g/mL.) Express your answer using two significant figures.
Chemistry
1 answer:
Kisachek [45]3 years ago
5 0

Answer : The percent (by volume) of air displaced if all of the liquid nitrogen evaporated would be 29 %

Explanation :

Let us assume that 1.2 L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m.

First we have to calculate the mass of nitrogen.

Density=\frac{Mass}{Volume}

Given:

Density of nitrogen = 0.807 g/mL

Volume of nitrogen = 1.2 L = 1200 mL    (1 L = 1000 mL)

0.807g/mL=\frac{Mass}{1200mL}

Mass=0.807g/mL\times 1200mL=968.4g

Now we have to calculate the moles of nitrogen.

\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}

Molar mass of nitrogen (N_2) = 28 g/mole

\text{Moles of }N_2=\frac{968.4g}{28g/mole}=34.58mole

Now we have to calculate the volume of nitrogen.

As we know that,

PV = nRT

where,

P = pressure of nitrogen = 1.1 atm

V = volume of nitrogen = ?

n = number of moles of nitrogen = 34.58 mole

R = gas constant = 0.0821 L.atm/mol.K

T = temperature of nitrogen = 22.1^oC=273+22.1=295.1K

Now put all the given values in this formula, we get:

1.1atm\times V=34.58mole\times 0.0821L.atm/mol.K\times 295.1K

V=761.6L

Now we have to calculate the volume of container.

The formula of volume of cuboid, we use the equation:

V=lbh

where,

V = volume of container = ?

l = length of container = 1.0 m

b = breadth of container = 1.3 m

h = height of container = 2.0 m

Putting values in above equation, we get:

V=1.0m\times 1.3m\times 2.0m=2.6m^3=2600L

Conversion used : 1m^3=1000L

Now we have to calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated.

\% \text{ of displaced volume}=\frac{761.6L}{2600L}\times 100=29.29\% \approx 29\%

Therefore, the percent (by volume) of air displaced if all of the liquid nitrogen evaporated would be 29 %

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At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
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Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

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