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OleMash [197]
3 years ago
9

Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same?

Chemistry
1 answer:
Marina CMI [18]3 years ago
8 0

Answer:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

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or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und
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Answer:

The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)

\Delta H^{o}_{rxn}= -5104.1kJ/mol

The expression for the entropy change for the reaction is as follows.

\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]

\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol

\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol

\Delta H^{o}_{f}(O_{2})= 0kJ/mol

Substitute the all values in the entropy change expression.

-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]

-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})

\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol

=-220.1kJ/mol

Therefore, The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

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How many Earths will we need to have enough resources if the whole world lived like Americans?
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Answer:

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Explanation:

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