Answer:
1038.96 kPa
Explanation:
We’ll use the ideal gas law; P1V1/T1 = P2V2/T2
P1*14.8/75.5 = 101.3*16.5/70.2
P1 = (101.3 * 16.5 * 75.5) / (70.2 *14.8)
P1 = 1038.96
Answer:
The higher the vapor pressure of a liquid at a given temperature, the lower the normal boiling point of the liquid.
Explanation:
Answer:
it will be classical as gas
Answer:
The answer to your question is V2 = 23.52 l
Explanation:
Data
Volume 1 = V1 = 22.5 l
Pressure 1 = P1 = 734 mmHg
Volume 2 = V2 = ?
Pressure 2 = 702 mmHg
Process
To solve this problem use Boyle's law.
P1V1 = P2V2
-Solve for V2
V2 = P1V1 / P2
-Substitution
V2 = (734 x 22.5) / 702
-Simplification
V2 = 16515 / 702
-Result
V2 = 23.52 l
-Conclusion
If we diminish the pressure, the volume will be higher.
Answer : The volume of NaOH stock solution used should be, 250 ml
Solution :
According to the neutralization law,
where,
= molarity of NaOH solution = 0.50 M
= volume of NaOH solution = 3.0 L
= molarity of NaOH stock solution = 6.0 M
= volume of NaOH stock solution = ?
Now put all the given values in the above law, we get the volume of NaOH stock solution.
(1 L = 1000 ml)
Therefore, the volume of NaOH stock solution used should be, 250 ml