Answer:
in this file you will find it https://xlbrands.com/wp-content/uploads/2014/06/DEW-POINT-CALCULATION-CHARTenglish.pdf
Explanation:
Answer:
Mass of the salt: 105.6g of KCl.
Mass water: 958.9g of water.
Molality: 1.478m.
Explanation:
<em>Mass of the salt:</em>
In 1L, there are 1.417 moles. In grams:
1.417 moles KCl * (74.54g / mol) = 105.6g of KCl
<em>Mass of the water:</em>
We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:
1L * (1064.5g / L) = 1064.5g - Mass solution.
Mass water = 1064.5g - 105.6g = 958.9g of water
<em>Molality:</em>
Moles KCl = 1.417 moles KCl.
kg Water = 958.9g = 0.9589kg.
Molality = 1.417mol / 0.9589kg = 1.478m
Answer:
The heat of combustion is -25 kJ/g = -2700 kJ/mol.
Explanation:
According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.
Qcomb + Qcal = 0
Qcomb = - Qcal
The heat absorbed by the calorimeter can be calculated with the following expression.
Qcal = C × ΔT
where,
C is the heat capacity of the calorimeter
ΔT is the change in temperature
Then,
Qcomb = - Qcal
Qcomb = - C × ΔT
Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ
Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:

The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:
