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At what temperature will 0.505 mole of CO2 occupy a volume of 3.50 x 103 mL at a pressure of 3185 mmHg?

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Answer:

3.5 x 10^3 ml

Step-by-step explanation:

We can use the ideal gas law to solve this problem:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to convert the volume to liters and the pressure to atmospheres:

V = 3.50 x 10^-3 L

P = 3185 mmHg / 760 mmHg/atm = 4.19 atm

Next, we can solve for T:

T = PV / nR

T = (4.19 atm) (0.505 mol) (0.08206 L atm mol^-1 K^-1) / (3.50 x 10^-3 L)

T = 1074 K

Therefore, at a temperature of 1074 K (801°C or 1474°F), 0.505 mole of CO2 will occupy a volume of 3.50 x 10^3 ml at a pressure of 3185 mmHg.

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