2.03 kg = 2030 grams
2030g C6H8O7 (1 mol/ 192.124g) = 10.6 mol C6H8O7
1) moles = mass/mR
CaCO3 Mr = 40 + 12 + (16×3)
= 52 + 48
= 100
mass = 15
so the moles would be 15 ÷ 100
which is 0.15 moles of CaCO3
2) moles = mass ÷ Mr
Mr of Al2O3 = 27 + (16×3)
= 27 + 48
= 75
mass = 204
so the moles would be 204/75 which is 2.72 moles of Al2O3
Answer is: f<span>ormula for the hydrated compound is CuSO</span>₄·3H₂O.
ω(H₂O) = 25,3% = 0,253.
ω(CuSO₄) = 100% - 25,3%.
ω(CuSO₄) = 74,7% = 0,747.
ω(H₂O) : M(H₂O) = ω(CuSO₄) : M(CuSO₄).
0,253 : M(H₂O) = 0,747 : 159,6 g/mol.
M(H₂O) = (0,253 · 159,6 g/mol) ÷ 0,747.
M(H₂O) = 54 g/mol.
N(H₂O) = 54 g/mol ÷ 18 g/mol.
N(H₂O) = 3.
Answer:
So if we need to react with 88 gm. of copper 2 then at least 3-4 liters.
Explanation:
Electrons carry a negative charge, so when you add on to the electron, the ion gets more negative.
hope this helps!