does anyone think they can help me with chem?

Question

3. Calculate the density, g/cm3, of carbon dioxide at 103.0oC and 818 torr.

4.A mixture of 5.90 g of O2 and 10.3 g of Xe exerts a pressure of 0.480 atm.
What is the partial pressure, in atm, of the O2 gas ?

5.In a mixture of H2 and CH4 gases, the partial pressure of H2 is 0.155 atm. The total pressure is 0.375 atm. The temperature is 34.0oC. If the total volume of the gas mixture is 445 mL, what is the mass of CH4 in grams?

Any help will be much appreciated!!

Answer 1

For #3; to do this, just use the gas law equation; set the volume to 1 ml and solve for n:

n = PV/RT

Get things in the proper units:

103 C = 273 + 103 = 276K

818 torr = 818torr/760torr/atm = 1.08atm

Plug into equation

n = (1.08atm x 0.001L)/(0.08206Latm/MolK x 376K) =

0.00035mole [this # of moles in 1 ml or cm^3]

to get density multiply mole/cm^3 by g/mole for the gas:

0.00035mol/ml x 44 g/mole = 0.00154g/cm^3

For part #4, The partial pressure is proportional to the mole fraction of the gas:

5.9g O2 = 5.9g/32g/mol = 0.184 mole O2

10.3g Xe = 10.3g/131.3g/mol = 0.0784mole Xe

partial pressure of O2 is [0.184/(0.184 + 0.0784)] x 0.480atm

= 0.336atm from O2

For part # 5; you could use the ideal gas law to calculate moles and then get to grams but here is another method:

First, determine volume of the CH4 portion of the mixture at STP:

Partial pressure of CH4 = 0.375 - 0.155 = 0.220atm

Now use P1V1/Ti = P2V2/T2 to solve for V at STP:

[(0.220atm x 0.455L)/307K] x 273K/1atm = 0.0871liter

Now, divide by 22.4 l/mol for gases at STP to get moles:

0.0871L/22.4L/mol = 0.00388mole

Now multiply by molecular weight to get grams of CH4

0.00388 x 16g/mole = 0.0622g CH4

Answer 2

i would convert my torr to megapascals of pressure and convert my degrees c to kelvin then i would grab an r from http://en.wikipedia.org/wiki/Gas_constant
if you convert pressure from torr to MPa then you can use the fourth r from the top in the table at the right. then you can choose a volume of one cubic centimeter or one cm^3 for convenience. then use pv=nrt and solve for n, which is mols of co2. t is given as 103c, convert to kelvins. p is given as 818torr, convert to megapascals, v is chosen by you, choose one cubic cm, r is got from the web page or from your chem book so solve for n. n=pv/rt. now that you have n go to the periodic table and get the molecular wt of co2. c plus o plus o again to get the weight. now get grams. mols x wt = grams.
now you have density because those grams are the grams for one cubic centimeter.

for number 4 go back to the periodic table and get the weight of O2 and of Xe and find mols of O2 and mols of Xe.
mols= grams/atomic weight. get mols of each one. add the mols together to get total mols. get a mol fraction for Xe by dividing mols of Xe/(total mols). now multiply that mol fraction times 0.48. thas the partial pressure of xe. the partial pressue of xe and the partial pressuer of o2 add to 0.48

Answer 3

http://www.kentche
mistry.com/aplinks/cha
pters/5Gases.htm