A. The molarity of solution containing 5.2 g of sodium sulphate is 0.0732 M
B. The number of mole of sodium in the solution is 0.0732 mole
<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
- Mass of Na₂SO₄ = 5.2 g
- Molar mass of Na₂SO₄ = (23×2) + 32 + (16×4) = 142 g/mol
Mole = mass / molar mass
Mole of Na₂SO₄ = 5.2 / 142
Mole of Na₂SO₄ = 0.0366 mole
<h3>A. How to determine the molarity </h3>
- Mole of Na₂SO₄ = 0.0366 mole
- Volume = 500 mL = 500 / 1000 = 0.5 L
Molarity = mole / Volume
Molarity = 0.0366 / 0.5
Molarity = 0.0732 M
<h3>B. How to determine the mole of sodium (Na) </h3>
1 mole of Na₂SO₄ contains 2 moles of Na.
Therefore,
0.0366 mole of Na₂SO₄ will contain = 0.0366 × 2 = 0.0732 mole of Na
Thus, 0.0732 mole of Na is present in the solution.
Learn more about molarity:
brainly.com/question/15370276
A pair of electrons(2 electrons) are shared in a single bond.
Hope this helps you.
Gas pressure is measured in mmHg with a barometer
Answer:
The answer to your question is 280 g of Mg(NO₃)₂
Explanation:
Data
Efficiency = 30.80 %
Mg(NO₃)₂ = ?
Magnesium = 147.4 g
Copper (II) nitrate = excess
Balanced Reaction
Mg + Cu(NO₃)₂ ⇒ Mg(NO₃)₂ + Cu
Reactants Elements Products
1 Mg 1
1 Cu 1
2 N 2
6 O 6
Process
1.- Calculate the theoretical yield
Molecular weight Mg = 24
Molecular weight Mg(NO₃)₂ = 24 + (14 x 2) + (16 x 6)
= 24 + 28 + 96
= 148 g
24 g of Mg -------------------- 148 g of Mg(NO₃)₂
147.4 g of Mg ------------------- x
x = (147.4 x 148) / 24
x = 908.96 g of Mg(NO₃)₂
2.- Calculate the Actual yield
yield percent = 
Solve for actual yield
Actual yield = Yield percent x Theoretical yield
Substitution
Actual yield = 
x 908.96
Actual yield = 279.95 ≈ 280g
