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Alexeev081 [22]
2 years ago
7

What volume of carbon dioxide is produced when 6.40 g of methane gas, CH4 (g), reacts with excess oxygen? All gases are at 35.0C

and 100.0 kPa.
Chemistry
1 answer:
Mandarinka [93]2 years ago
3 0

Answer:

V = 10.3 L

Explanation:

Given data:

Mass of methane = 6.40 g

Volume of CO₂ produced = ?

Temperature = 35°C (35+273 = 308 K)

Pressure = 100.0 KPa (100.0/101 = 0.98 atm)

Solution:

Chemical equation:

CH₄ +  2O₂        →       CO₂ + 2H₂O

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 6.40 g/ 16 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of CO₂ with CH₄.

         CH₄          :         CO₂

           1             :           1

        0.4            :         0.4

Volume of CO₂:

Formula:

PV = nRT

0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K ×  308 K

0.98 atm ×V = 10.11 atm.L

V = 10.11 atm.L /0.98 atm

V = 10.3 L

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ozzi
Hello!

First, we need to determine the pKa of the base. It can be found applying the following equation:

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Explain how you can use boyle's law to determine the new volume of gas when its pressure is increased from 270 kPa to 540 kPa? T
LUCKY_DIMON [66]
There are several information's already given in the question. Based on the information's the answer can be easily deduced. 

We know the formula

<span>P1*V1/T1 = P2*V2/T2
</span>
As the temperature is constant, so T1 and T2 can be negated. The formula changes to 
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What is the density of air at 75 F and latm pressure? Express in lbm/ft3 and kg/m3
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Answer : The density of air in lbm/ft^3 and kg/m^3 is, 0.0743lbm/ft^3 and 1.19kg/m^3 respectively.

Explanation :

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

where,

P = pressure of air = 1 atm

V = volume of air

T = temperature of air = 297 K

The conversion used for the temperature from Fahrenheit to degree Celsius is:

^oC=(^oF-32)\times \frac{5}{9}

^oC=(75-32)\times \frac{5}{9}=24^oC

The conversion used for the temperature from degree Celsius to Kelvin is:

K=273+^oC

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m = mass of air

M = average molar mass of air = 28.97 g/mole

\rho = density of air = ?

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

\rho=\frac{PM}{RT}

\rho=\frac{(1atm)\times (28.97g/mole)}{(0.0821L.atm/mol.K)\times (297K)}

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Now we have to calculate density in kg/m^3.

Conversion used :

1g/L=1kg/m^3

So,

1.19g/L=1.19kg/m^3

The density of air in kg/m^3 is, 1.19kg/m^3

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