Answer:
Q = 1267720 J
Explanation:
∴ QH2O = mCpΔT
∴ m H2O = 500 g
∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C
∴ ΔT = 120 - 50 = 70°C
⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ
∴ ΔHv H2O = 40.7 KJ/mol
moles H2O:
∴ mm H2O = 18.015 g/mol
⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O
⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ
⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J
Answer:
Gas giants.
Explanation:
Jupiter, Saturn, Uranus, and Neptune are the gas giants of our solar system.
In a redox reaction electrons are lost and gained in equal numbers. The species that is oxidized gives electrons to the species that is reduced. I hope this helps. Let me know if anything is unclear.
Answer:
The energies of combustion (per gram) for hydrogen and methane are as follows: Methane = 82.5 kJ/g; Hydrogen = 162 kJ/g
<em>Note: The question is incomplete. The complete question is given below:</em>
To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/℃. When a 1.00-g sample of methane gas burned with
<em>excess oxygen in the calorimeter, the temperature increased by 7.3℃. When a 1.00 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Compare the energies of combustion (per gram) for hydrogen and methane.</em>
Explanation:
From the equation of the first law of thermodynamics, ΔU = Q + W
Since there is no expansion work in the bomb calorimeter, ΔU = Q
But Q = CΔT
where C is heat capacity of the bomb calorimeter = 11.3
kJ/ºC; ΔT = temperature change
For combustion of methane gas:
Q per gram = (
11.3
kJ/ºC * 7.3°C)/1.0g
Q = 83 kJ/g
For combustion of hydrogen gas:
Q per gram = (
11.3
kJ/ºC * 14.3°C)/1.0g
Q = 162 kJ/g
I think it’s A not sure though
