How would you prepare 500 mL of 0.360 M solution of CaCl2 from solid CaCl2?
1 answer:
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
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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: