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MArishka [77]
3 years ago
9

Consider the reaction:

Chemistry
1 answer:
mestny [16]3 years ago
5 0

Answer:

ΔHrxn = [(1) x (-1675.7 kJ/mole) + (2) x (0 kJ/mole)] - [(1) x (-824.3 kJ/mole) + (2) x (0 kJ/mole)] = -1675.7 kJ/mole + 824.3 kJ/mole = -851.4 kJ/mole.

Explanation:

  • To calculate ΔH of the reaction, we use the mathematical relation:
  • ΔHrxn = ∑ΔHproducts - ∑ΔHreactants.
  • So, ΔHrxn = [(n x ΔHf for Al₂O₃) + (n x ΔHf for Fe)] - [(n x ΔHf for Fe₂O₃) + (n x ΔHf for Al)]
  • where n is the number of moles of each reactant.
  • ΔHrxn = [(1) x (-1675.7 kJ/mole) + (2) x (0 kJ/mole)] - [(1) x (-824.3 kJ/mole) + (2) x (0 kJ/mole)] = -1675.7 kJ/mole + 824.3 kJ/mole = -851.4 kJ/mole.
  • <em>ΔHf for Al and Fe = 0.0 kJ/mole</em>; since the heat for formation of elements that found in nature is equal 0.0 kJ/mole
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Calculate the Kelvin temperature to which 10.0 L of a gas at 27 °C would have to be heated to change the volume to 12.0 L. Units
KatRina [158]

Hello!

For this problem, we will be applying <em>Charles' Law</em>:

V1/T1 = V2/T2

Now that we have the formula, let's convert the temperature to Kelvin.

27 + 273 = 300K

Let's plug everything in now!

10/300 = 12.0/x

Simplified:

1/30 = 12.0/x

Cross-multiply:

1x = 30*12.0

<u>x = 360</u>

<em>Check!</em>

10/300 = 12/360

300*12 = 360*10

3600 = 3600

Therefore, you would have to heat the gas at a temperature of 360K in order to raise the volume to 12.0L.

 

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don't forget to mulitiply by coefficients

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