Answer:
D
Explanation:I alredy know this i am in 7th gread
Mass of water produced : 0.146 g
<h3>Further explanation</h3>
Given
33.2 mL of 0.245 M lithium hydroxide
Required
mass of water
Solution
Reaction
HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)
mol LiOH :
= M x V
= 0.245 x 33.2 ml
= 8.134 mmol
From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol
mass H₂O :
= mol x MW
= 8.134 x 10⁻³ mol x 18 g/mol
= 0.146 g
Answer:
45.02 L.
Explanation:
- Firstly, we need to calculate the no. of moles of water vapor.
- n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.
- We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.
<em><u>Using cross multiplication:</u></em>
1.0 mole of CO occupies → 22.4 L.
2.01 mole of CO occupies → ??? L.
∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.
