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:
The answer for the following question is mentioned below.
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
Explanation:
Given:
Pressure of gas (P) = 1.2 atm
Volume of a gas (V) = 50.0 liters
Temperature (T) =650 K
To calculate:
no of moles present in the gas (n)
We know;
According to the ideal gas equation;
We know;
<u>P × V = n × R × T
</u>
where,
P represents pressure of the gas
V represents volume of the gas
n represents no of the moles of a gas
R represents the universal gas constant
where the value of R is 0.0821 L atm mole^{-1} K^-1
T represents the temperature of the gas
As we have to calculate the no of moles of the gas;
n = 
n = \frac{1.2*50.0}{0.0821*650}
n = \frac{60}{53.365}
n = 1.12 moles
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.
The balanced chemical equation for the reaction can be represented as,
Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL
Heat of the reaction, q = 
Δ
m is mass of the solution = 151.8 mL * 
C is the specific heat of solution = 4.18 
ΔT is the temperature change = 
q = 
Moles of NaOH = 
NaOH
Moles of 
= 
Enthalpy of the reaction = 
Answer:
V₂ = 15.00 atm
Explanation:
Given data:
Initial pressure = 5.00 atm
Initial volume = 3.00 L
Final pressure = 760 mmHg ( 760/760 = 1 atm)
Final volume = ?
Solution:
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = 5.00 atm × 3.00 L / 1 atm
V₂ = 15.00 atm
I think the answer is D no change. Though you add more CO2, but the pressure is not mentioned. If the pressure is constant and the reaction is already balanced, the H2O is also saturation and can not absorb more CO2.
