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dimaraw [331]
3 years ago
5

The exhaust gas from an automobile contains 3% by volume of carbon monoxide (CO). Express this concentration in mg/m3 at 25oC an

d 1 atm.
Chemistry
1 answer:
ser-zykov [4K]3 years ago
4 0

Answer:

24540\frac{mg}{m^3}

Explanation:

Hello,

In this case, since the 3% by volume is represented as:

\frac{3L\ CO}{L\ gas}

By using the ideal gas equation we compute the density of CO:

\rho =\frac{MP}{RT} =\frac{28g/mol*1atm}{0.082\frac{atm*L}{mol*K}*298K}= 0.818g/L

Then we apply the conversion factors as follows:

=\frac{3L\ CO}{100L\ gas}*\frac{0.818g\ CO}{1L\ CO} *\frac{1000mg\ CO}{1g\ CO} *\frac{1000L\ gas}{1m^3\ gas} \\\\=24540\frac{mg}{m^3}

Regards.

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RSB [31]

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Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

PV=nRT

where,

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V = Volume of N_2 gas = 985 mL = 0.982 L    (1 L = 1000 mL)

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(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K

n=5.78\times 10^{-11}mol

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains 6.022\times 10^{23} number of molecules.

As, 1 mole of N_2 gas contains 6.022\times 10^{23} number of molecules

So, 5.78\times 10^{-11} mole of N_2 gas contains (5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13} number of molecules

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3 years ago
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Explanation:

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2 years ago
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astra-53 [7]

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5 0
3 years ago
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hope this helps!




6 0
3 years ago
Heyy pls help anyone?
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7 0
3 years ago
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