A 5267ml sample of h2 is collected at 772 mmhg and 21C. What would the volume be if it is cooled to oC and 1 atm?

To find the volume of the hydrogen gas at 0°C and 1 atm, we will use the combined gas law.

The combined gas law can be expressed as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 and P2 are the initial and final pressures,
V1 and V2 are the initial and final volumes,
T1 and T2 are the initial and final temperatures.

Let's plug in the given values:
P1 = 772 mmHg
V1 = 5267 mL
T1 = 21°C = 21 + 273.15 K (convert to Kelvin)

P2 = 1 atm
V2 = Unknown (what we need to find)
T2 = 0°C = 0 + 273.15 K (convert to Kelvin)

Now we can rewrite the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2

Let's plug in the values and solve for V2:
(772 mmHg * 5267 mL) / (21 + 273.15 K) = (1 atm * V2) / (0 + 273.15 K)

To calculate the volume at the new conditions, we rearrange the equation and solve for V2:
V2 = (772 mmHg * 5267 mL * (0 + 273.15 K)) / (1 atm * (21 + 273.15 K))

Let's do the calculations:
V2 = (772 mmHg * 5267 mL * 273.15 K) / (1 atm * 294.15 K)

Now we can solve the equation:
V2 = (3399740146.55 mmHg mL K) / (294.15 atm K)

To convert from mmHg mL to liters, we divide by 101325 (1 atm in mmHg):
V2 = (3399740146.55 mmHg mL K) / (294.15 atm K * 101325 mmHg atm/L)

V2 ≈ 113.9 L

Therefore, if the hydrogen gas is cooled to 0°C and 1 atm, the volume would be approximately 113.9 liters.