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DiKsa [7]
3 years ago
10

I have a 5 liter container of oxygen (O 2 ) gas at 2.0 atm and 25 0 C, what mass of gas is inside this

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

m=13.1gO_2

Explanation:

Hello,

In this case, by using the ideal gas equation, we first compute the moles of oxygen at the given volume, pressure and temperature:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{2.0atm*5L}{0.082\frac{atm*L}{mol*K}(25+273.15)K}  =0.41mol

Then, since molar mass of gaseous oxygen is 32 g/mol, we compute the contained mass in grams as shown below:

m=0.41mol*\frac{32g}{1mol}\\\\m=13.1gO_2

Best regards.

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2) Infere the molar ratios:

1 mol H2 : 2 mol of water

3) Make the calculus as the direct proportion relation:

[2 mol H2O] / [1 mol H2] * 7 mol H2 = 14 mol H2

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6 0
3 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

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where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

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T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

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5 0
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6 0
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If arranged in order of increasing atomic mass, which element would come after chlorine?
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Ionic equation:

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5 0
2 years ago
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