Answer:
Let the mixture is X% by mass of CuSO
4
.5H
2
O and 100 - X % by mass of MgSO
4
.7H
2
O. 5.0 g of mixture will contain 0.05X g CuSO
4
.5H
2
O and 5.0 - 0.05X g MgSO
4
.7H
2
O
The molar masses of CuSO
4
.5H
2
O and MgSO
4
.7H
2
O are 249.7 g/mol and 246.5 g/mol respectively.
The number of moles of CuSO
4
.5H
2
O=
249.7
0.05X
=2.00×10
−4
X moles.
Explanation:
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Answer:
More information so I can answer please.
Explanation:
Answer:
b molartiary will decrease
Explanation
Answer:Is this from the stochiometry unit?
Explanation:
Answer:
4.50 L
Explanation:
First we <u>calculate how many moles are there in 3.84 L of a 8.50 mol/L solution</u>:
- 3.84 L * 8.50 mol/L = 32.64 mol
Now, keeping in mind that
- Concentration = Mol / Volume
we can calculate the volume of a 7.25 mol/L solution that would contain 32.64 moles:
- Volume = Mol / Concentration
- Volume = 32.64 mol ÷ 7.25 mol/L
So we could take 4.50 L of the 7.25 mol/L solution and evaporate the solvent until only 3.84 L remain.
