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wel
3 years ago
9

If hydrogen were used as a fuel, it could be burned according to the following reaction: H2(g)+1/2O2(g)→H2O(g). Use average bond

energies to calculate ΔHrxn for this reaction Use average bond energies to calculate ΔHrxn for the combustion of methane (CH4). Which fuel yields more energy per mole and per gram?
Chemistry
1 answer:
serious [3.7K]3 years ago
7 0
The deltaHrxn = -243 kJ/mol the deltaHrxn of CH4(methane) = -802 kJ/mol 

The fuel that yields more energy per mole is METHANE. The negative sign merely signifies the release of energy. Thus, 802 kJ/mol is greater than 243 kJ/mol.

The fuel that yields more energy per gram is HYDROGEN. Here is the computation:
deltaHrxn = (-243 kJ/mol)(1 mol/2.016 g H2)  <span>= -120.535714286 kJ/g or -121 kJ/g

</span>deltaHrxn of CH4(methane) = (-802 kJ/mol)(1 mol/16.04 g) 
<span>= -50 kJ/g 
</span>
As discussed the negative sign serves as the symbol of released energy. Thus, 121 is greater than 50.
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4 0
3 years ago
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Irina-Kira [14]

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}

Where:

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%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:  

1 = \%_{16} + \%_{17} + \%_{18}

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Hence, the percent abundance of O-18 is:  

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x = 0.980614 \times 100 = 98.0614 \%                                                              

Hence, the percent abundance of oxygen-18 is:

\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%                      

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!                                                      

8 0
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7 0
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