<h3>
Answer:</h3>
0.5 moles/L or 0.5 M
<h3>
Explanation:</h3>
- Molarity is the concentration of a solution in moles per liter.
- It is calculated by;
Molarity = Number of moles ÷ Volume
In this case;
Volume = 125 mL
Mass of sodium nitrate = 5.31 g
We are required to calculate the molarity of sodium nitrate
<h3>Step 1: Calculate the number of moles of sodium nitrate </h3>
Moles = Mass ÷ Molar mass
Molar mass of sodium nitrate = 84.9947 g/mol
Therefore,
Moles of NaNO₃ = 5.31 g ÷ 84.9947 g/mol
= 0.0625 moles
<h3>Step 2: Molarity of sodium nitrate </h3>
Remember;
Molarity = Moles ÷ Volume
Volume of NaNO₃ = 0.125 L
Therefore;
Molarity = 0.0625 moles ÷ 0.125 L
= 0.5 Moles/L or 0.5 M
Thus; the concentration of sodium nitrate is 0.5M
Missing question: What is the vapor pressure of the solution at 25°<span>C?
n(NaCl) = 100 g </span>÷ 58,4 g/mol.
n(NaCl) = 1,71 mol.
NaCl → Na⁺ + Cl⁻, amount of ions are 2 · 1,71 mol = 3,42 mol.
n(CaCl₂) = 100 g ÷ 111 g/mol = 0,9 mol.
CaCl₂ → Ca²⁺ + 2Cl⁻, amount of ions 3 · 0,9 mol = 2,7 mol.
m(solution) = 1000 ml (1,00 L) · 1,15 g/ml = 1150 g.
m(H₂O) = 1150 g - 100 g - 100 g = 950 g.
n(H₂O) = 950 g ÷ 18 g/mol = 118,75 mol.
<span>water's mole fraction = 118,75 mol </span>÷ (118,75 mol + 2,7 mol + 3,42 mol).
water's mole fraction = 0,95.
p(solution) = 0,95 · 23 mmHg = 21,85 mmHg.
Explanation:
oxidation is adding of oxygen and removal of hydrogen.
Reduction means removal of oxygen and addition of hydrogenm
Answer:
4 mol of Al₂O₃ are produced from 8 mole of Al
Explanation:
The reaction:
4 Al + 3O₂ → 2Al₂O₃
Ratio is 4:2
So, If we have 8 mol of Al, we will produce 4 mol of Al₂O₃