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Offline Stephen

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Gasses
« on: June 23, 2009, 07:38:53 AM »
Hi, I've some exercises which I am not sure am I doing them right and some of them I don't even know to start.So here they are:
1.A sample of a smoke stack emission was collected into a 1.25l tank at 752mm Hg and analyzed.The analysis showed 92% CO2 3,6% NO 1.2% SO2 and 4.1% H2O by MASS!!What is the partial pressure exerted by each gas?
This is one of them which I don't know to start.
2.A sample of gas collected over water at 42c occupies a volume of one liter.The wet gas has a pressure of 0.986atm. The gas is dried and the dry gas occupies 1.04l with pressure of 1.0atm at 90c.Using this information calculate the vapor pressure of water at 42c.
n1=n2
pv=nrt
n(gas)=pv/rt=0.035mol I've got this from second half of exercise where is dry gas
And first half is wet gas
Ptot=p(gas)+p(water)
p(gas)=nrt/v=0.9atm
==> p(water)=0.086atm
I think this is it.
3.A sample of oxygen gas is collected over water at 25c(p(water)=23.8mm Hg).The wet gas occupies a volume of 7.28l at a total pressure of 1.25 bar.If all the water is removed what volume will the dry oxygen occupy at a pressure of 1.07atm and the temperature 37c?
Ptot=P(water)+P(oxygen)
P(oxygen)=1.26625atm-0.0352223=1.23atm
pv=nRT
n=0.361mol oxygen
Pv=nRT
V=8.586l
Is it correct?
4.A sample of oxygen is collected over water at 22c and 752mm Hg in 125ml flask.The vapor pressure of water at 22c is 19.8mm Hg.
a)partial pressure of water =?
b)how many moles of dry gas are collected?
c)How many moles of wet gas are in the flask?
d)if 0.0250g N2 are added to the flask at the same temperature what is the partial pressure of nitrogen in the flask?
e)what is the total pressure in the flask after N2 is added?
I've done this too:D
a)ptot=p(water)+p(oxygen)
p(oxygen)=732.2mm Hg=0.963421 atm
b)pv=nrt==>n=p(oxygen)*v/TR=0.00497atm(dry gass that means only oxygen)
e)ptot*v=ntot*R*T
ntot=0.0051045mol(wet gass)
e)wet gass+N2=p'tot
m(N2)=0.025g=0.0008928mol
p'tot*v=ntot*r*t
p'tot=1.1623atm
d)P(N2)=0.1728264atm
5.A mixture of 3.5 mol Kr and 3.9 mol He occupies a 10.00l container at 300K.Which gas has the larger?
a)average transitional energy?
b)partial pressure?
c)mole fraction?
d)effusion rate?

my work:
a)With obvious calculation I found a formula Ek=3/2*RT/Na
If I am right that means that both gasses are going to have the same average transitional energy,no?
b)Ptot=PKr+PHe
PKrV=nKrRT
PKr=8.62atm
and on the same way PHe=9.6atm
c)PHe/Ptot=nHe/ntot=W
WHe=0.53 and on the same way WKr=0.47
d)effusion rte of Kr/effusion rate of He=(MMHe/MMKr)1/2
that means that effusion rate of He=0.22*effusion rte of Kr so Kr has bigger effusion rate!

Please check as soon as you can!
Thank's

Offline Stephen

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Re: Gasses
« Reply #1 on: June 23, 2009, 08:29:42 AM »
One more exercise coming....This is type of exercises which I don't know to do....Finding an empirical formula that's very hard for me so if someone can do this for me..
Glycine is an amino acid made up of carbon, hydrogen, oxygen, nitrogen atoms.Combustion of a 0.2036g sample gives 132.9ml of CO2 at 25oC and 0.122g of water.What are the percentages of C and H in glycine?Another sample of glycine wighing 0.2500g treated in such way that all nitrogen atoms are converted to N2.The gas has a volume of 40.8 ml at 25oC and 1.00 atm.What are the percentages of nitrogen in glycine?What is percentages of oxygen?What is the empirical formula of glycine?

Offline Phlogiston

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Re: Gasses
« Reply #2 on: June 29, 2009, 03:05:04 PM »
Hi, I've some exercises which I am not sure am I doing them right and some of them I don't even know to start.So here they are:
1.A sample of a smoke stack emission was collected into a 1.25l tank at 752mm Hg and analyzed.The analysis showed 92% CO2 3,6% NO 1.2% SO2 and 4.1% H2O by MASS!!What is the partial pressure exerted by each gas?
This is one of them which I don't know to start.

It's probably assumed that you should use the ideal gas law here, so that's a good start.  Dalton's law of partial pressures will then come in handy, and finally, you could relate the percent by mass to mole fraction by choosing an arbitrary total mass such as 100. grams.


2.A sample of gas collected over water at 42c occupies a volume of one liter.The wet gas has a pressure of 0.986atm. The gas is dried and the dry gas occupies 1.04l with pressure of 1.0atm at 90c.Using this information calculate the vapor pressure of water at 42c.
n1=n2
pv=nrt
n(gas)=pv/rt=0.035mol I've got this from second half of exercise where is dry gas
And first half is wet gas
Ptot=p(gas)+p(water)
p(gas)=nrt/v=0.9atm
==> p(water)=0.086atm
I think this is it.

Your method is correct.  I haven't checked the numbers.

3.A sample of oxygen gas is collected over water at 25c(p(water)=23.8mm Hg).The wet gas occupies a volume of 7.28l at a total pressure of 1.25 bar.If all the water is removed what volume will the dry oxygen occupy at a pressure of 1.07atm and the temperature 37c?
Ptot=P(water)+P(oxygen)
P(oxygen)=1.26625atm-0.0352223=1.23atm
pv=nRT
n=0.361mol oxygen
Pv=nRT
V=8.586l
Is it correct?

Correct reasoning.  I haven't double checked the calculations.

4.A sample of oxygen is collected over water at 22c and 752mm Hg in 125ml flask.The vapor pressure of water at 22c is 19.8mm Hg.
a)partial pressure of water =?
b)how many moles of dry gas are collected?
c)How many moles of wet gas are in the flask?
d)if 0.0250g N2 are added to the flask at the same temperature what is the partial pressure of nitrogen in the flask?
e)what is the total pressure in the flask after N2 is added?
I've done this too:D
a)ptot=p(water)+p(oxygen)
p(oxygen)=732.2mm Hg=0.963421 atm
b)pv=nrt==>n=p(oxygen)*v/TR=0.00497atm(dry gass that means only oxygen)
e)ptot*v=ntot*R*T
ntot=0.0051045mol(wet gass)
e)wet gass+N2=p'tot
m(N2)=0.025g=0.0008928mol
p'tot*v=ntot*r*t
p'tot=1.1623atm
d)P(N2)=0.1728264atm

Seems plausible enough.

5.A mixture of 3.5 mol Kr and 3.9 mol He occupies a 10.00l container at 300K.Which gas has the larger?
a)average transitional energy?
b)partial pressure?
c)mole fraction?
d)effusion rate?

my work:
a)With obvious calculation I found a formula Ek=3/2*RT/Na
If I am right that means that both gasses are going to have the same average transitional energy,no?
b)Ptot=PKr+PHe
PKrV=nKrRT
PKr=8.62atm
and on the same way PHe=9.6atm
c)PHe/Ptot=nHe/ntot=W
WHe=0.53 and on the same way WKr=0.47
d)effusion rte of Kr/effusion rate of He=(MMHe/MMKr)1/2
that means that effusion rate of He=0.22*effusion rte of Kr so Kr has bigger effusion rate!

Please check as soon as you can!
Thank's

a-c look fine.  You should double check your formula for d.

Offline Phlogiston

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Re: Gasses
« Reply #3 on: June 29, 2009, 03:08:07 PM »
One more exercise coming....This is type of exercises which I don't know to do....Finding an empirical formula that's very hard for me so if someone can do this for me..
Glycine is an amino acid made up of carbon, hydrogen, oxygen, nitrogen atoms.Combustion of a 0.2036g sample gives 132.9ml of CO2 at 25oC and 0.122g of water.What are the percentages of C and H in glycine?Another sample of glycine wighing 0.2500g treated in such way that all nitrogen atoms are converted to N2.The gas has a volume of 40.8 ml at 25oC and 1.00 atm.What are the percentages of nitrogen in glycine?What is percentages of oxygen?What is the empirical formula of glycine?

Start with some balanced reactions that relate moles of CO2 and H2O to moles of C and H in glycine.  Then, use the ideal gas law to find the specific moles for this example, then finally convert to mass percent.  Then repeat for N2 for the second part.

Offline 18uos

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Re: Gasses
« Reply #4 on: July 01, 2009, 12:56:49 PM »
 ::)
Hi, I've some exercises which I am not sure am I doing them right and some of them I don't even know to start.So here they are:
1.A sample of a smoke stack emission was collected into a 1.25l tank at 752mm Hg and analyzed.The analysis showed 92% CO2 3,6% NO 1.2% SO2 and 4.1% H2O by MASS!!What is the partial pressure exerted by each gas?
This is one of them which I don't know to start.
2.A sample of gas collected over water at 42c occupies a volume of one liter.The wet gas has a pressure of 0.986atm. The gas is dried and the dry gas occupies 1.04l with pressure of 1.0atm at 90c.Using this information calculate the vapor pressure of water at 42c.
n1=n2
pv=nrt
n(gas)=pv/rt=0.035mol I've got this from second half of exercise where is dry gas
And first half is wet gas
Ptot=p(gas)+p(water)
p(gas)=nrt/v=0.9atm
==> p(water)=0.086atm
I think this is it.
3.A sample of oxygen gas is collected over water at 25c(p(water)=23.8mm Hg).The wet gas occupies a volume of 7.28l at a total pressure of 1.25 bar.If all the water is removed what volume will the dry oxygen occupy at a pressure of 1.07atm and the temperature 37c?
Ptot=P(water)+P(oxygen)
P(oxygen)=1.26625atm-0.0352223=1.23atm
pv=nRT
n=0.361mol oxygen
Pv=nRT
V=8.586l
Is it correct?
4.A sample of oxygen is collected over water at 22c and 752mm Hg in 125ml flask.The vapor pressure of water at 22c is 19.8mm Hg.
a)partial pressure of water =?
b)how many moles of dry gas are collected?
c)How many moles of wet gas are in the flask?
d)if 0.0250g N2 are added to the flask at the same temperature what is the partial pressure of nitrogen in the flask?
e)what is the total pressure in the flask after N2 is added?
I've done this too:D
a)ptot=p(water)+p(oxygen)
p(oxygen)=732.2mm Hg=0.963421 atm
b)pv=nrt==>n=p(oxygen)*v/TR=0.00497atm(dry gass that means only oxygen)
e)ptot*v=ntot*R*T
ntot=0.0051045mol(wet gass)
e)wet gass+N2=p'tot
m(N2)=0.025g=0.0008928mol
p'tot*v=ntot*r*t
p'tot=1.1623atm
d)P(N2)=0.1728264atm
5.A mixture of 3.5 mol Kr and 3.9 mol He occupies a 10.00l container at 300K.Which gas has the larger?
a)average transitional energy?
b)partial pressure?
c)mole fraction?
d)effusion rate?

my work:
a)With obvious calculation I found a formula Ek=3/2*RT/Na
If I am right that means that both gasses are going to have the same average transitional energy,no?
b)Ptot=PKr+PHe
PKrV=nKrRT
PKr=8.62atm
and on the same way PHe=9.6atm
c)PHe/Ptot=nHe/ntot=W
WHe=0.53 and on the same way WKr=0.47
d)effusion rte of Kr/effusion rate of He=(MMHe/MMKr)1/2
that means that effusion rate of He=0.22*effusion rte of Kr so Kr has bigger effusion rate!

Please check as soon as you can!
Thank's

Offline Stephen

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Re: Gasses
« Reply #5 on: July 01, 2009, 02:58:41 PM »
So It's good I suppose:D
OK, thank you guys a lot, you helped me very much! ;D
Here comes a new set of questions soon about structures of atoms!Prepare tourselves!:D

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