1 ) A sample of vapor occupies a volume of 125 ml at 100 celcius and 748 torr. What is its volume under STP conditions?

2) A sample of vapor weighing 0.745 g filled a 250 ml flask when placed in a water bath at 98 degree celcius. The barometric presure that day was 753 torr. Calculate the volume of the vapor at STP, and the weight of one mol of the violate liquid?

Please explain, will greatly appreciate!

write a balanced chemicl equation for the reversible reaction of carbon monoxide and chlorine gases to form carbonyl choride gas, COCl2.

1) Use PV = nRT. Note the correct spelling of Celsius.

1) You can use (P1V1/T1) = (P2V2/T2) to solve for volume at STP.

2)Use the same equation from #1 above and solve for volume at STP. For the molar mass, use PV = nRT, solve for n = number of moles of the gas, then use n = grams/molar mass and solve for molar mass.

For the following reaction, 6.80 grams of carbon monoxide are mixed with excess sulfur. The reaction yields 5.90 grams of sulfur dioxide.

sulfur (s) + carbon monoxide (g) sulfur dioxide (g) + carbon (s)

What is the theoretical yield of sulfur dioxide ? grams
What is the percent yield of sulfur dioxide ? %

To solve these questions, we need to use the ideal gas law and the concept of STP (Standard Temperature and Pressure).

1) To find the volume of the vapor under STP conditions, we can use the ideal gas law, which states:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin

To convert the temperature from Celsius to Kelvin, we add 273.15:

T = 100°C + 273.15 = 373.15 K

Now we can rearrange the ideal gas law to solve for the volume under STP conditions:

V₁ = (P₁ * V₁ * T₂) / (P₂ * T₁)

Where:
V₁ = initial volume (125 mL)
P₁ = initial pressure (748 torr)
T₁ = initial temperature (373.15 K)
P₂ = final pressure (STP = 1 atm)
T₂ = final temperature (STP = 273.15 K)

Substituting the given values:

V₁ = (748 torr * 125 mL * 273.15 K) / (1 atm * 373.15 K)
V₁ ≈ 206.2 mL

Therefore, the volume of the vapor under STP conditions is approximately 206.2 mL.

2) To calculate the volume of the vapor at STP, we can use the same approach as in the first question. However, in this case, we need to find the number of moles first.

We can use the ideal gas law again, but this time rearrange it to solve for the number of moles:

n = (PV) / (RT)

Where:
P = pressure (753 torr)
V = volume (250 mL)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (98°C + 273.15 K = 371.15 K)

Now we can calculate the number of moles:

n = (753 torr * 250 mL) / (0.0821 L•atm/mol•K * 371.15 K)
n ≈ 0.0447 moles

Next, to find the weight of one mole of the volatile liquid, we need to know the molecular weight of the liquid. Let's assume it is M grams.

The weight of one mole (molar mass) can be calculated using the equation:

Weight = Moles * Molar mass

Given that the weight of the vapor is 0.745 g and the number of moles is 0.0447, we can rearrange the equation and solve for the molar mass:

M = (Weight) / (Moles)

M = 0.745 g / 0.0447 moles
M ≈ 16.66 g/mol

Therefore, the weight of one mole of the volatile liquid is approximately 16.66 grams.