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tresset_1 [31]
3 years ago
10

The half-life of carbon-14 is about 6000 years. assume that a sample of charcoal is formed by burning of living wood 15,000 year

s ago. how much of the original carbon-14 would remain today?
Chemistry
1 answer:
zimovet [89]3 years ago
4 0
The daughter isotope has an atomic number one less than the parent and a mass number two less. : A. : 1 20) The half-life<span> of </span>carbon-14<span> is about </span>6000 years<span>. </span>Assume<span> that a </span>sample<span> of </span>charcoal formed<span> by</span>burning<span> of </span>living wood 15,000 years ago<span>. How </span>much<span> of the </span>original carbon-14 would remain today? A) more than one-<span>half</span>
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Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you
Leni [432]

Answer:

a) \Delta G=2.6kJ

b) \Delta G=-979.57kJ

c) \Delta G=264.21kJ

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol

So we proceed as follows:

a)

\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ

b)

\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ

c)

\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ

Regards.

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