Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Carbon dioxide is a carbon atom and two oxygen atoms, therefore CO2
For the others, they are hydrocarbons.
The first part of the name is determined by how many carbon atoms there are. The second part, is by the type. Alcohol, Alkane, Alkene, alkynes, acid, esters, amides.
21 kg x [(3 x 35.45)/(12.01 + 19.00 + (3 x 35.45))] =
21 kg x (106.35/137.36) = 16.3 kg of chlorine
You just multiply the weight of the material by the fraction of chlorine (by weight). The others are done the same way
Answer:
Mass = 0.000176 gram
Steps:
m = V × ρ
= 20 milliliter × 8.8 gram/cubic meter
= 2.0E-5 cubic meter × 8.8 gram/cubic meter
= 0.000176 gram
Explanation: