Answer:a) 0.1 mole. b) 4g. c) 2% d) 196 mL
Explanation: in 200mL , 0.1mole
mw NaOH = 40g/mol —> 4g in 0.1 mole
4g in 200mL so 2g in 100mL
density NaOH = 1g/mL so if 4g in 200 mL, 4mL , 196 mL water
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
Answer:
DO NOT TASTE IT, but all the rest are correct
Explanation: