Answer:
Explanation:
The balanced equation is
I₂(g) + Br₂(g) ⇌ 2IBr(g)
Data:
Kc = 8.50 × 10⁻³
n(IBr) = 0.0600 mol
V = 1.0 L
1. Calculate [IBr]
2. Set up an ICE table.
3. Calculate [I₂]
4. Convert the temperature to kelvins
T = (150 + 273.15) K = 423.15 K
5. Calculate p(I₂)
Answer:
The heat needed to boil 1 gallon of water is 81,490.62 Joules.
Explanation:
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
We have :
Volume of water = V = 1 gal = 4546.09 mL
Density of water , d= 1 g/mL
mass of water = m = d × V = 1g/mL × 4546.09 mL = 4546.09 g
Specific heat of water = c = 1 Cal/g°C
ΔT = 100°C - 25°C = 75 °C
9 (boiling pint of water is 100°C)
Heat absorbed by the water to make it boil:
1 calorie = 4.184 J
The heat needed to boil 1 gallon of water is 81,490.62 Joules.