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3 years ago

I need a answer ASAP !!! Need help !!

Chemistry

1 answer:

vova2212 [387]3 years ago

7 0

1. Inertia

2. Energy

3. Heat

4. Reflection

5. Refraction

6. Compression

7. Rarefaction

8. Friction

Sorry it took a min, hope this helps!

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A mixture of CO2 and Kr weighs 31.7 g and exerts a pressure of 0.665 atm in its container. Since Kr is expensive, you wish to re

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Answer:

11.94 grams of carbon dioxide were originally present.

19.94 grams of krypton can you recover.

Explanation:

Mass of carbon dioxide gas = x

Mass of krypton gas = y

x + y = 31.7 g

Moles of carbon dioxide gas = $n_1=\frac{x}{44 g/mol}=$

Moles of krypton gas = $n_2=\frac{y}{84 g/mol}=$

Mole fraction of krpton = $\chi '$

Total pressure of the mixture = P = 0.665 atm

Partial pressure of carbon dioxide gas = p

Partial pressure of krypton gas before removal of carbon dioxide gas = p'

Partial pressure of krypton gas after removal of carbon dioxide gas = p'' = 0.309 atm

p' = p'' = 0.309 atm

0.665 atm = p + 0.309 atm

p = 0.665 atm - 0.306 atm = 0.359 atm

Partial pressure of krypton can also be given by :

$p'=P\times \chi '$

$0.309 atm=0.665 atm\times \frac{n_2}{n_1+n_2}$

$0.309 atm=0.665 atm\times \frac{\frac{y}{84}}{\frac{x}{44}+\frac{y}{84}}$

$0.4645=\frac{\frac{y}{84}}{\frac{x}{44}+\frac{y}{84}}$ ..[2]

Solving [1] and [2]:

x = 11.94 g

y = 19.76 g

11.94 grams of carbon dioxide were originally present.

19.94 grams of krypton can you recover.

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Answer:

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Explanation:

Equation of reaction

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