Answer:
351.43mL
Explanation:
To calculate the original volume of hydrogen gas in this question, the Boyle's law equation will be used. Boyle's law equation is:
P1V1 = P2V2
Where; P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
According to this question, the P1= 1.56atm, V1 = ?, P2 = 0.73atm, V2 = 751mL
Hence;
P1V1 = P2V2
1.56 × V1 = 0.73 × 751
1.56 V1 = 548.23
V1 = 548.23/1.56
V1 = 351.43mL
Therefore, the original volume of hydrogen gas is 351.43 mL.
Answer:
Explanation:
Hello there!
In this case, according to the given data, it is possible to infer that the gas mixture lies on the 15.0 cm-high column of water, so that the total pressure or atmospheric pressure is given by:
Thus, since the atmospheric pressure is 745 mmHg and the vapor pressure of water is 18 mmHg, the pressure of hydrogen turns out to be:
Best regards!
Increasing temperatures in a reaction increases the kinetic energy of the reactant molecules. This causes them to move fast and hence collide with a higher frequency. The higher the rate of collision between the molecules, the faster the reaction.
