Answer:
Moles of KOH in 1000 mL solution = 0.255 moles
Moles of KOH in 1 mL solution = 0.255/1000 = 0.000255 moles
Moles in 95 mL solution = (95 * 255)/1000000 = 24225/1000000
Moles of KOH in 95 mL 0.255M solution = 0.024225 moles
1 atm=7.15/9.25
Volume increase comes from reduced pressure
Maybe
A. 400 ml of 5.0% glucose solution
Answer:
1.85 × 10⁻⁶
Explanation:
0.0003 ÷ 162 = 1.851851852 × 10⁻⁶ ⇒ 1.85 × 10⁻⁶
Hope that helps.
