Answer:
Option D. 0.115 M
Explanation:
The following data were obtained from the question:
Mass of CuSO4 = 36.8 g
Volume of solution = 2 L
Molar mass of CuSO4 = 159.62 g/mol
Molarity of CuSO4 =..?
Next, we shall determine the number of mole in 36.8 g of CuSO4.
This can be obtained as shown below:
Mass of CuSO4 = 36.8 g
Molar mass of CuSO4 = 159.62 g/mol
Mole of CuSO4 =.?
Mole = mass /Molar mass
Mole of CuSO4 = 36.8 / 159.62
Mole of CuSO4 = 0.23 mole
Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:
Mole of CuSO4 = 0.23 mole
Volume of solution = 2 L
Molarity of CuSO4 =..?
Molarity = mole /Volume
Molarity of CuSO4 = 0.23 / 2
Molarity of CuSO4 = 0.115 M
Therefore, the molarity of the CuSO4 solution is 0.115 M.
Answer:
800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
Explanation:
3000 lb of 13% solution is required .
Total adhesive in weight = 3000 x .13 = 390 lb of adhesive
Available = 500 lb of 10% solution = 50 lb of adhesive
Rest = 390 - 50 = 340 lb required .
rest mass of solution = 3000 - 500 = 2500 lb
mass of adhesive required = 340 lb
Let the mass of 20% required be V
mass of adhesive = .20 V
.20 V = 340
V = 1700
rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent
So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
The concentration of hydrogen can be shown as:
[H+ ] = 3 * 10-5 M
pH can be determined as:
pH = - log [H+ ]
= - log (3 * 10-5)
= 4.53
Thus the pH of solution is 4.53
group 1 elements are metals with<u> low</u> density
