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Shalnov [3]
3 years ago
13

To calculate the enthalpy change for the reaction, 2CO (g) + O2 (g) Imported Asset 2 CO2 (g), you can use ΔHf0 values for each r

eactant and product. Which setup shows how the enthalpy change should be calculated?
ΔH0reaction = [ΔHf0 CO2(g)] + [ΔHf0 CO(g) - ΔHf0 O2(g)]

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [ΔHf0 CO(g) + ΔHf0 O2(g)] - [ΔHf0 CO2(g)]

ΔH0reaction = [ΔHf0 CO(g) - ΔHf0 O2(g)] + [ΔHf0 CO2(g)]
Chemistry
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

Explanation:

Chemical equation:

CO + O₂   →  CO₂

Balanced chemical equation:

2CO + O₂   →  2CO₂

The standard enthalpy for the formation of CO = -110.5 kj/mol

The standard enthalpy for the formation of O₂  = 0  kj/mol

The standard enthalpy for the formation of CO₂  = -393.5 kj/mol

Now we will put the values in equation:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]

ΔH0reaction = -283 kj/mol

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Answer:

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Explanation:

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V=20*8

=160m/s

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8 0
3 years ago
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4 0
3 years ago
A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

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Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

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                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

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                              = 0.67

Therefore, the mole fraction of Xe is 0.67.

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