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lesya692 [45]
1 year ago
14

An unknown quantity of methane gas, CHA, is held in a 2.00-liter container at 77°C. The pressure inside the container is 3.00 am

. How many moles of methane must be in the container?
Chemistry
1 answer:
kolbaska11 [484]1 year ago
6 0

Answer:

0.21\text{ mol}

Explanation:

Here, we want to calculate the number of moles of methane in the container

From the ideal gas law:

\begin{gathered} PV\text{ = nRT} \\ n\text{ = }\frac{PV}{RT} \end{gathered}

where:

P is the pressure inside the container which is 3 atm

V is the volume of the container which is 2 L

R is the molar gas constant which is 0.0821 Latm/mol.k

T is the temperature in Kelvin (we convert the temperature in Celsius by adding 273 : 273 + 77 = 350 K)

n is the number of moles that we want to calculate

Substituting the values, we have it that:

n=\text{ }\frac{3\times2}{0.0821\text{ }\times350}\text{ = 0.21 mol}

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A 365 L balloon is blown up inside at 283 K. It is then taken outside in the hot sun and Heated to 300K. What is the new volume?
asambeis [7]

V₁ = initial Volume of the balloon after it is blown up = 365 L

V₂ = new Volume of the balloon after it is taken outside = ?

T₁ = initial temperature of the balloon = 283 K

T₂ = new temperature of the balloon = 300 K

using the equation

V₁/V₂ = T₁/T₂

365/V₂ = 283/300

V₂ = 387 L

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MrRa [10]
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How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three
Dima020 [189]

Answer:

a. =9.1x10^3gC_3H_8

b. 2.1x10^2molC_3H_8

c. Q=-4.6x10^5kJ

Explanation:

Hello,

a. By applying the given information, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8

c. Here, the propane's combustion chemical reaction is stated:

C_3H_8+5O_2-->3CO_2+4H_2O

This enthalpy of reaction is computed via:

ΔrH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol

Finally, since it is done for 20 lb of propane (2.1x10^2molC_3H_8), the obtainable energy is:

Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ

Best regards.

4 0
3 years ago
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