Answer:
V₁ = 96.2 mL
Explanation:
Given data:
Initial volume of NH₄OH required = ?
Initial molarity = 15.6 M
Final molarity = 3.00 M
Final volume = 500.0 mL
Solution:
Formula:
M₁V₁ = M₂V₂
M₁ = Initial molarity
V₁ = Initial volume of NH₄OH
M₂ =Final molarity
V₂ = Final volume
Now we will put the values.
15.6 M ×V₁ = 3.00 M×500.0 mL
15.6 M ×V₁ = 1500 M.mL
V₁ = 1500 M.mL /15.6 M
V₁ = 96.2 mL
Answer:
I think carbon and hydrogen
Answer:
The answer to your question is: 58.4 g of NaCl
Explanation:
Data
Volume = 200 ml = 0.2 l
Concentration = 5M
MW = 58.4 g
mass NaCl = ?
Formula
Molarity = (# of moles ) / volume
# of moles = Molarity x volume
# of moles = 5 x 0.2
# of moles = 1
58.4 g ---------------------- 1 mol
x --------------------- 1 mol
x = (1 x 58.4) / 1
x = 58.4 g of NaCl
