•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol
1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:
(3.9 g)/ (17.03g/mol) = 0.22900763mols
Then convert the moles to molecules by multiplying it with Avogadro’s number:
Avogadro’s number: 6.022 x 10^23
0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
D. Air molecules touch the warm ground, heating them up
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.
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