I have a bad chemistry teacher, and these type of questions are generally easy but im not sure how to do them.
1. You have to prepare some 2M solutions, with 10g of solute in each. What volume of solution will you preparem for each solute below?
a) Lithium sulfate
b) Magnesium sulfate
c ammonium nitrate
2. Phosphorus forms two oxides, which have the empirical formulae P2O3 and P2O5
a) what mass of phosphorus will combine with 1 mole of oxygen molecules (O2) to form P2O3?
b) What is the molecular formula of the oxide that has a formula mass of 284
NEED URGENT CHEMISTRY HELP!!!!!?
2016-05-17 2:09 pm
回答 (2)
2016-05-17 3:12 pm
1.
(a)
Molar mass of Li₂SO₄ = (6.94×2 + 32.06 + 16.00×4) g/mol = 109.94 g/mol
No. of moles of 10 g of Li₂SO₄ = (10 g) / (109.94 g/mol) = 0.0910 mol
Volume of the solution = (0.0910 mol) / (2/1000 mol/mL) = 45.5 mL
(b)
Molar mass of MgSO₄ = (24.31 + 32.06 + 16.00×4) g/mol = 120.37 g/mol
No. of moles of 10 g of MgSO₄ = (10 g) / (120.37 g/mol) = 0.08308 mol
Volume of the solution = (0.08308 mol) / (2/1000 mol/mL) = 41.5 mL
(c)
Molar mass of NH₄NO₃ = (14.01×2 + 1.01×4 + 16.00×3) g/mol = 80.06 g/mol
No. of moles of 10 g of NH₄NO₃ = (10 g) / (80.06 g/mol) = 0.125 mol
Volume of the solution = (0.125 mol) / (2/1000 mol/mL) = 6.25 mL
====
2
(a)
4P + 3O₂ → 2P₂O₃
According to the equation, mole ratio P : O₂ = 4 : 3
Number of moles of O₂ = 1 mol
Number of moles of P = (1 mol) × (4/3) = 4/3 mol
Molar mass of P = 31 g/mol
Mass of P needed = (31 g/mol) × (4/3 mol) = 41.3 g
(b)
If the molecular formula will be (P₂O₃)ᵤ:
The formula mass = (31×2 + 16×3)u = 110u
If the molecular formula will be (P₂O₅)ᵥ:
The formula mass = (31×2 + 16×5)v = 142v
Obviously, 142 is a factor of 284, and a = 284/142 = 2
The molecular formula of the oxide = P₄O₁₀
(a)
Molar mass of Li₂SO₄ = (6.94×2 + 32.06 + 16.00×4) g/mol = 109.94 g/mol
No. of moles of 10 g of Li₂SO₄ = (10 g) / (109.94 g/mol) = 0.0910 mol
Volume of the solution = (0.0910 mol) / (2/1000 mol/mL) = 45.5 mL
(b)
Molar mass of MgSO₄ = (24.31 + 32.06 + 16.00×4) g/mol = 120.37 g/mol
No. of moles of 10 g of MgSO₄ = (10 g) / (120.37 g/mol) = 0.08308 mol
Volume of the solution = (0.08308 mol) / (2/1000 mol/mL) = 41.5 mL
(c)
Molar mass of NH₄NO₃ = (14.01×2 + 1.01×4 + 16.00×3) g/mol = 80.06 g/mol
No. of moles of 10 g of NH₄NO₃ = (10 g) / (80.06 g/mol) = 0.125 mol
Volume of the solution = (0.125 mol) / (2/1000 mol/mL) = 6.25 mL
====
2
(a)
4P + 3O₂ → 2P₂O₃
According to the equation, mole ratio P : O₂ = 4 : 3
Number of moles of O₂ = 1 mol
Number of moles of P = (1 mol) × (4/3) = 4/3 mol
Molar mass of P = 31 g/mol
Mass of P needed = (31 g/mol) × (4/3 mol) = 41.3 g
(b)
If the molecular formula will be (P₂O₃)ᵤ:
The formula mass = (31×2 + 16×3)u = 110u
If the molecular formula will be (P₂O₅)ᵥ:
The formula mass = (31×2 + 16×5)v = 142v
Obviously, 142 is a factor of 284, and a = 284/142 = 2
The molecular formula of the oxide = P₄O₁₀
2016-05-17 2:29 pm
Vol = moles / Concentration and moles = mass / Molar mass
convert the 10 g into moles of each substance, then use the Volume formula above.
convert the 10 g into moles of each substance, then use the Volume formula above.
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