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| Mad Scientists Science and Mathematics discussion-- theories, arguments, citations, proofs and pudding. |
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# 1

2008-01-24, 06:17
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Chemistry Help
Yes I realize that there is Mad Scientist but I figured that I would get quicker replies here and hopefully 5% of those replies will be useful.
Anyway, My first question is:
In a titration experiment, a student neutralized 25.0 mL of sulphuric acid solution using 36.8 mL of 0.125 mol / L sodium hydroxide solution. Calculate the concentration of the sulphuric acid solution. Now I got a final answer of 0.093mol/L which seems weird. My Calculations Are:
n=cv
n(NaOH)=0.0368L*0.125mol/L=4.6E-3
Since balanced part of equation is H2SO4 +2NaOH that means that we calculated for 2 NaOH, so its actually 2.3E-3 mol for H2SO4. Then use c=n/v:
=2.3E-3/0.025L
and get 0.092 mol/L which seems odd to me.
Now,
What mass of lithium was reacted with an excess of 0.250 mol/L HCl(aq) to form 1.35 L of hydrogen gas collected by the downward displacement of water measured at 97.4 kPa and 26.0 °C?
This is the one I'm having most trouble with, Now I know the balanced equation does not need any coefficients, and that since you are in water you need to subtract water vapour pressure to get like a total of 94.04kPa. But what next? Would you just use PV=nRT?, plus I dont get why they tell you the concentration of HCl.
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# 2

2008-01-24, 06:20
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Regular
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█ █ █
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Re: Chemistry Help
If nuclearrabbit was still here he could have helped you.
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# 3

2008-01-24, 06:29
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Re: Chemistry Help
Fuck your homework.
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# 4

2008-01-24, 07:10
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Regular
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Kingston, Canada
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Re: Chemistry Help
Oh yeah? Well fuck your couch.
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# 5

2008-01-24, 07:27
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Re: Chemistry Help
I don't have chem anymore because I graduated High school.
hahhahaahahahahahhahaahah
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# 6

2008-01-24, 12:17
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Re: Chemistry Help
Quote:
Originally Posted by magixx
Yes I realize that there is Mad Scientist but I figured that I would get quicker replies here and hopefully 5% of those replies will be useful.
Anyway, My first question is:
In a titration experiment, a student neutralized 25.0 mL of sulphuric acid solution using 36.8 mL of 0.125 mol / L sodium hydroxide solution. Calculate the concentration of the sulphuric acid solution. Now I got a final answer of 0.093mol/L which seems weird. My Calculations Are:
n=cv
n(NaOH)=0.0368L*0.125mol/L=4.6E-3
Since balanced part of equation is H2SO4 +2NaOH that means that we calculated for 2 NaOH, so its actually 2.3E-3 mol for H2SO4. Then use c=n/v:
=2.3E-3/0.025L
and get 0.092 mol/L which seems odd to me.
Now,
What mass of lithium was reacted with an excess of 0.250 mol/L HCl(aq) to form 1.35 L of hydrogen gas collected by the downward displacement of water measured at 97.4 kPa and 26.0 °C?
This is the one I'm having most trouble with, Now I know the balanced equation does not need any coefficients, and that since you are in water you need to subtract water vapour pressure to get like a total of 94.04kPa. But what next? Would you just use PV=nRT?, plus I dont get why they tell you the concentration of HCl.
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PV=nrt the hydrogen gas to get the moles of hydrogen. Number of moles of hydrogen = same number of moles of HCl. I assume it's a lithium salt you're using so moles of lithium is also the same as the moles of hydrogen. Check the balanced equation of Li + HCl to make sure it's a 1:1 reaction. Thus, just multiply the moles of hydrogen with the molecular mass of lithium. Too tired to look for my periodic table lol
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# 7

2008-01-24, 12:18
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Re: Chemistry Help
Quote:
Originally Posted by DonJabronyo
PV=nrt the hydrogen gas to get the moles of hydrogen. Number of moles of hydrogen = same number of moles of HCl. I assume it's a lithium salt you're using so moles of lithium is also the same as the moles of hydrogen. Thus, just multiply the moles of hydrogen with the molecular mass of lithium. Too tired to look for my periodic table lol
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This.
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# 8

2008-01-24, 12:26
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Re: Chemistry Help
Quote:
Originally Posted by magixx
Yes I realize that there is Mad Scientist but I figured that I would get quicker replies here and hopefully 5% of those replies will be useful.
Anyway, My first question is:
In a titration experiment, a student neutralized 25.0 mL of sulphuric acid solution using 36.8 mL of 0.125 mol / L sodium hydroxide solution. Calculate the concentration of the sulphuric acid solution. Now I got a final answer of 0.093mol/L which seems weird. My Calculations Are:
n=cv
n(NaOH)=0.0368L*0.125mol/L=4.6E-3
Since balanced part of equation is H2SO4 +2NaOH that means that we calculated for 2 NaOH, so its actually 2.3E-3 mol for H2SO4. Then use c=n/v:
=2.3E-3/0.025L
and get 0.092 mol/L which seems odd to me.
Now,
What mass of lithium was reacted with an excess of 0.250 mol/L HCl(aq) to form 1.35 L of hydrogen gas collected by the downward displacement of water measured at 97.4 kPa and 26.0 °C?
This is the one I'm having most trouble with, Now I know the balanced equation does not need any coefficients, and that since you are in water you need to subtract water vapour pressure to get like a total of 94.04kPa. But what next? Would you just use PV=nRT?, plus I dont get why they tell you the concentration of HCl.
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WTF is 4.6E-3?
Anyway you need two moles of NaOH for each H2SO4 right? Just get the number of moles of NaOH by multiplying 0.0368 to 0.125 and then multiply by two to get the moles of H2SO4 that you oxidized. Divide that by 25mL but first convert it to litres to get the initial concentration of the sulfuric acid solution.
Last edited by DonJabronyo : 2008-01-24 at 12:39.
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# 9

2008-01-24, 13:27
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Re: Chemistry Help
i cant provide you with any help here.
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# 10

2008-01-24, 13:32
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Regular
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A cave, because I'm a hermit.
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Re: Chemistry Help
Murder/suicide.
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This thread continued for 2 pages in the real archive, 15 posts total - only page 1 survived here.
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