Assuming the volume of the gas is measured at standard temperature and pressure, Then one mole of the Gas would occupy 22.4 liters.
Therefore, 1 liter is 1/224 moles
one mole of nitrogen 14 is 14
Therefore 1 liter of the nitrogen weighs 1/224×14
0.0625 grams
The answer is A.
Which is... KCI
Metals are to the left of the zig-zag, nonmetals are to the right, and metalloids lie on/beside the line.
Answer:
Explanation: In my opinion law is stated as fact because it is passes through and approved by many peoples but theory is an idea someone has and is not approved and is set on a base of beliefs.
Answer:
the heat of formation of isopropyl alcohol is -317.82 kJ/mol
Explanation:
The heat of combustion of isopropyl alcohol is given as follows;
C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)
The heat of combustion of CO₂ and H₂O are given as follows
C (s) + O₂ (g) → CO₂(g) = −393.50 kJ
H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ
Therefore we have
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as
3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =
4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol
-1180.5 - 1143.32 +2006 = -317.82 kJ/mol
Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.
