Answer:
You must remove
.
Explanation:
There are three heat transfers in this process:
Total heat = cool the vapour + condense the vapour + cool the liquid
q = q₁ + q₂ + q₃
q = nC₁ΔT₁ + nΔHcond + nC₂ΔT₂
Let's calculate these heat transfers separately.
Data:
You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.
C₁ = specific heat capacity of vapour = 90 J·K⁻¹mol⁻¹
C₂ = specific heat capacity of liquid = 115 J·K⁻¹mol⁻¹
ΔHcond = -38.56 kJ·mol⁻¹
Tmax = 300 °C
b.p. = 78.4 °C
Tmin = 25.0 °C
n = 0.782 mol
Calculations:
ΔT₁ = 78.4 - 300 = -221.6 K
q₁ = 0.782 × 90 × (-221.6) = -15 600 J = -15.60 kJ
q₂ = 0.782 × (-38.56) = -30.15 kJ
ΔT = 25.0 - 78.4 = -53.4 K
q₃ = 0.782 × 115 × (-53.4) = -4802 J = 4.802 kJ
q = -15.60 - 53.4 - 4.802 = -50.6 kJ
You must remove of heat to convert the vapour to a gas.
3. Cl₂ + 2KI --> 2KCl + I₂
Cl = 2 Cl = 2
K = 2 K = 2
I = 2 I = 2
4. 2NaCl --> 2Na + Cl₂
Na = 2 Na = 2
Cl = 2 Cl = 2
5. 4Na + O₂ --> 2Na₂O
Na = 4 Na = 4
O = 2 O = 2
6. 2Na + 2HCl --> H₂ + 2NaCl
Na = 2 Na = 2
H = 2 H = 2
Cl = 2 Cl = 2
7. 2K + Cl₂ --> 2KCl
K = 2 K = 2
Cl = 2 Cl = 2
It is going through the process of decomposition. The solution has weaker bonds than would be necessary to keep it together given the outside conditions, causing the bonds to break and the bubbles are oxygen bubbles escaping the solution.
Li (3)
N (7)
F (9)
Depends on if you are putting them in smallest to greatest.