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Hitman42 [59]
3 years ago
15

If 16.4 grams of calcium nitrate is heated as shown in the reaction:

Chemistry
1 answer:
vfiekz [6]3 years ago
4 0

Answer:

7.2

Explanation:

you first have to find the number of moles of nitrogen dioxide by using the number of moles for calcium nitrate and the mole to mole ratios

number of moles of calcium nitrate=mass/mm

=16.4/102

=0.16g/mol

then you use the mole to mole ratios

2 : 4

0.16: x

2x/2=0.64/2

x=0.32g/moles of nitrogen dioxide

then you use the formula for the volume

v=22.4n

=22.4×0.32

=7.2

I hope this helps

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\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

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\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

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\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

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Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

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p_i = partial pressure of gas

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p_{CH_4}=X_{CH_4}\times p_T

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p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

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