Answer:
Imma just say acids have more acidity and bases have lower acidity...
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
Answer:
The answer to your question is P2 = 84.16 kPa
Explanation:
Data
Volume 1 = V1 = 4.52 L Volume 2 = V2 = 4.83 l
Pressure 1 = P1 = 102 kPa Pressure 2 = P2 = ?
Temperature 1 = T1 = 23°C Temperature 2 = T2 = -12°C
Process
1.- Convert the temperature to °K
Temperature 1 = 23 + 273 = 296°K
Temperature 2 = -12 + 273 = 261°K
2.- Use the Combined Gas law to solve this problem
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (102)(4.52)(261) / (296)(4.83)
-Simplification
P2 = 120331.44 / 1429.68
-Result
P2 = 84.16 kPa
