<span>The root mean square speed is given by V_rms = âšRT/M where r, t, and m are the rate constant, temperature and molar mass the gas
Average molar kinetic energy of the gas
E = 1/2 M * (V_rms)^2 = 8750 ms/1
So (V_rms)^2 = (2 * 8750) / M
Molar mass of 2 chlorine atoms in kg is 2 * 35 * 10^(-3)
Hence we have (V_rms)^2 = (2 * 8750)/ (2 * 35 * 10^(-3))
(V_rms)^2 = 8750/0.035 = 250000
So V_rms = âš 250000 = 500</span>
Answer:
13.85 kJ/°C
-14.89 kJ/g
Explanation:
<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>
<em />
The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qcomb + Qcal = 0
Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ
The heat capacity (C) of the calorimeter can be calculated using the following expression.
Qcal = C . ΔT
where,
ΔT is the change in the temperature
Qcal = C . ΔT
50.72 kJ = C . 3.661 °C
C = 13.85 kJ/°C
<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>
Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ
The heat of combustion per gram of B is:

Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.
Answer: 43 L
Answer:
2.24 Liters are in 4.4 grams of CO2 at STP
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