There are some radioactive nuclides can be used to measure time on an archeological scale. One is the best example of this is radiocarbon dating. This process is based on the ratio of caebon-14 to carbon-12 in the atmosphere which is relatively constant.

The half time of C-14 5730 years

Carbon-14 is a radioactive nucleus. It has a half-life of 5730 years.

All living tissues like plants and animal absorbed carbon-12 along with carbon-14 with same ratio of caebon-14 to carbon-12 in the atmosphere.

Carbon-14 dating is based on the ratio of carbon-14 to carbon-12 in the atmosphere which is relatively constant

3 0

3 years ago

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Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

Answer: The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago

The enthalpies of formation of the compounds in the combustion of methane, , are CH4 (g): Hf = –74.6 kJ/mol; CO2 (g): Hf = –393.

algol [13]

Answer:

The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol

Explanation:

The chemical reaction of the combustion of methane is given as follows;

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor

Where:

CH₄ (g): Hf = -74.6 kJ/mol

CO₂ (g): Hf = -393.5 kJ/mol

H₂O (g): Hf = -241.82 kJ/mol

Therefore, the combustion of 1 mole of methane releases;

-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol

Hence the combustion of 2 moles of methae will rellease;

2 × -802.54 kJ/mol or 1,605.08 kJ/mol.

8 0

3 years ago

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