Answer:
4552 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 55 mL
Molarity of stock solution (M₁) = 12 M
Molarity of diluted solution (M₂) = 0.145 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
12 × 55 = 0.145 × V₂
660 = 0.145 × V₂
Divide both side by 0.145
V₂ = 660 / 0.145
V₂ ≈ 4552 mL
Thus, the volume of the diluted solution is 4552 mL
They certainly can. However, they have other groups that are used to classify a compound.
Water containing carbonic acid and calcium
beryllium, magnesium, strontium,barium<span>, and </span><span>radium. </span>
Answer: 2 mol
Explanation:
- According to the ideal gas law, One mole of an ideal gas at STP (standard temperature and normal pressure) occupies 22.4 liters.
- Using cross multiplication,
1 mol of (O2) → 22.4 L
? → 43.9 L
Therefore, the number of moles of oxygen in 43.9 L = (43.9 × 1)/ 22.4 = 1.96 mol≈ 2 mol..
