Given parameters:
Volume of CuSO₄ = 250mL
Concentration of CuSO₄ = 2.01M
Unknown:
Mass of CuSO₄.5H₂O = ?
To solve this problem, we must write the chemical relationship between both species.;
CuSO₄.5H₂O → CuSO₄ + 5H₂O
Now that we know the expression, it is possible to solve for the unknown mass.
First find the number of moles of CuSO₄;
Number of moles = Concentration x Volume
Take 250mL to L so as to ensure uniformity of units;
Volume = 250 x 10⁻³L
Input the parameters and solve for number of moles;
Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol
From the equation;
1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O
So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O
Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole
Mass of CuSO₄.5H₂O = number of moles x molar mass
= 0.5 x 249.6
= 124.8g
The mass of CuSO₄.5H₂O is 124.8g
Answer:
1-A
2-B
Explanation:
2- mantle is liquid and moves crust
First find the mass
44.1 ml @ 1.55 g/mL = 68.355 grams of Ca
the equation
Ca + O2 → CaO
= 55 grams of O2
Answer:
The partial pressure of chlorine gas in the mixture is 1.55 atm.
Explanation:
Partial pressure of oxygen gas = 
Partial pressure of nitrogen gas = 
Partial pressure of chlorine gas = 
Total pressure of the mixture of gases = P = 3.30 atm
Using Dalton's law of partial pressure:
The partial pressure of chlorine gas in the mixture is 1.55 atm.
