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ivolga24 [154]
4 years ago
6

Why do real gases not behave exactly like ideal gases?

Chemistry
2 answers:
svp [43]4 years ago
6 0

An ideal gas is a hypothetical gas consisting of identical particles of negligible volume, with no intermolecular forces. Also the atoms or molecules of the gas undergo perfectly elastic collisions with the walls of the container and each other. Real gases do not exhibit these exact properties.

jok3333 [9.3K]4 years ago
6 0

because the particals will not move the same as real gases

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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
beks73 [17]

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.

4 0
3 years ago
How many grams of iron metal do you expect to be produced when 275 grams of an 65.5 percent by mass iron (II) nitrate solution r
Jlenok [28]

Answer:

75.5

Explanation:

4 0
3 years ago
Read 2 more answers
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