9. the stopcock between a 6.25l bulb containing hydrogen at 345mmhg and a 7.50l bulb containing nitrogen at…

Explanation:

Step1: Use Boyle's law ($P_1V_1 = P_2V_{total}$) for each gas to find partial - pressures in the combined volume.

For hydrogen: The initial volume of hydrogen $V_{H_2}=6.25L$ and initial pressure $P_{H_2}=345mmHg$. The total volume after opening the stop - cock is $V_{total}=6.25L + 7.50L=13.75L$. Using Boyle's law $P_{1}V_{1}=P_{2}V_{2}$, we have $P_{H_2,final}=\frac{P_{H_2,initial}V_{H_2}}{V_{total}}$. $P_{H_2,final}=\frac{345mmHg\times6.25L}{13.75L}=\frac{345\times6.25}{13.75}mmHg=\frac{2156.25}{13.75}mmHg = 156.82mmHg$.

Step2: Use Boyle's law for nitrogen.

The initial volume of nitrogen $V_{N_2}=7.50L$ and initial pressure $P_{N_2}=678mmHg$. Using Boyle's law $P_{N_2,final}=\frac{P_{N_2,initial}V_{N_2}}{V_{total}}$. $P_{N_2,final}=\frac{678mmHg\times7.50L}{13.75L}=\frac{678\times7.50}{13.75}mmHg=\frac{5085}{13.75}mmHg = 370.55mmHg$.

Step3: Use Dalton's law of partial pressures.

Dalton's law states that $P_{total}=P_{H_2,final}+P_{N_2,final}$. $P_{total}=156.82mmHg + 370.55mmHg=527.37mmHg$.

Answer:

$527mmHg$ (rounded to the nearest whole number)