The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
Learn more: brainly.com/question/13164491
Explanation:
A) particles are close together in random positions with about equal kinetic energy and intermolecular forces.
These points are about liquid state.
B) particles are close together in fixed positions with low kinetic energy
These points satisfy the qualities of Solid state
C)particles are far apart with greater kinetic energy and low intermolecular forces.
The above qualities are for Gaseous state of matter
A) Liquid
B)Solid
C)Gas
Answer:
14.048 moles I believe.
Explanation:
(8.46 * 10^24) / ( 6.022 * 10^23) = 14.048
0.001 would be the smallest.
Good Luck! :)
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