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
These are both physical changes can I get a brain list please
Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol
KOH -------> K+ OH-
Ba(OH)2 ------> Ba+2. 2OH-
