For physics the answer is 47500
The balanced equation for the reaction is as follows
2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂
stoichiometry of Al to H₂SO₄ is 2:3
number of Al moles reacted - 15.0 mol
if 2 mol of Al react with 3 mol of H₂SO₄
then 15.0 mol of Al reacts with - 3/2 x 15.0 mol = 22.5 mol
22.5 mol of H₂SO₄ is required
Answer:
the compound formed between nitrogen and magnesium is Magnesium Nitride
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 .
Answer:
See explanation.
Explanation:
Hello,
In this case, we could have two possible solutions:
A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

That is the mass of copper (II) sulfate contained in 1 mol of substance.
B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

So you can solve for the moles of the solute:

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

But this is just a supposition.
Regards.