19. A sample of neon occupies a volume of 461 mL at STP.
What will be the
volume of the neon when the pressure is reduced to 93.3 kPa?
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Ответ:
500.65mL
Explanation:
The following information were obtained from the question:
V1 (initial volume) = 461 mL
P1 (initial pressure) = stp = 101325Pa
P2 (final pressure) = 93.3 kPa
Recall: 1KPa = 1000Pa
Therefore, 93.3 kPa = 93.3x1000 = 93300Pa
V2 (final volume) =?
Using the Boyle's law equation P1V1 = P2V2, the final volume of the gas can be obtained as follow:
P1V1 = P2V2
461 x 101325 = 93300 x V2
Divide both side by 93300
V2 = (461 x 101325)/93300
V2 = 500.65mL
Therefore, the volume of Neon at 93.3 kPa is 500.65mL
Ответ:
hail Hydra kandksmak Yes