The steam releases 39.9 kJ when it condenses..
The steam condenses and transfers its energy to the skin.
<em>q = m</em>Δ<em>H</em>_cond = 17.7 g × (-2257 J/1 g) = -39 900 kJ = -39.9 kJ
The negative sign shows that the steam is releasing energy
1s2 2s2 2p6 3s2 3p6 4s1
s orbital can hold 2 electron
p orbitals can hold 6 electron
Answer:
81.5 L
Explanation:
We can use the combined gas law equation that gives the relationship among pressure, temperature and volume of gases for a fixed amount of gas.
P1V1 / T1 = P2V2 / T2
where P1 - pressure, V1 - volume and T1 - temperature at the first instance
P2 - pressure, V2 - volume and T2 - temperature at the second instance
substituting the values in the equation
1240 Torr x 47.2 L / 298 K = 730 Torr x V2 / 303 K
V2 = 81.5 L
the new volume the gas would occupy when the conditions have changed is 81.5 L
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.
