Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Frederick95 on March 11, 2008, 11:41:19 AM
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Please calculate the number of mols of oxygen gas produced per second
Remember PV=nRT and that you must account for the partial pressure of water vapour.
The reactions conditions are:
Atmospheric pressue:102.6 kPa;
Temperature:290 K
Partial pressure of water vapour:1.94 kPa.
The reaction is:
Volume of 6% H2O2 (mL):10.0
Volume of distilled water (mL):0.0
Volume of 1.0 M NaI (mL):10.0
Volume of distilled water (mL):0.0
O2 production in 60 s (mL):151.0
:)
What I have done thus far is rearranged the Ideal Gas Law in order to get the number of mols
n = PV/RT
Is this the correct first step ???
Also, I have substituted the following values into the equation:
P = 102.6 kPa
V = ?
R = 8.3145 J/mol K
T = 290 K
???
Thank you for any help it is greatly appreciated.
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O2 production in 60 s (mL):151.0
Did you misunderstand this parameter, for a start?
P = 102.6 kPa
Yeah that's what they give you, but maybe they've modified it in some way?
Let's see a balanced equation for the reaction first, the answer to a balanced equation is the moles, after all.
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The balanced equation would be
2H2O2 + NaI = 2H2O + O2 + Na + I
Am I correct, Arkcon?
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The balanced equation would be
2H2O2 + NaI = 2H2O + O2 + Na + I
Does that make chemical sense? Sodium metal in the vicinity of water? Monoatomic iodine?
S
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It make sense to me since both sides have equal quantities of elements/compounds.
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It make sense to me since both sides have equal quantities of elements/compounds.
Just because it seems to be balanced doesn't mean it makes sense. Reread sjb post. What happens when you throw piece of sodium into water? You have sodium and water on the RHS of the equation - is it possible?
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Hmmmm....Sodium is a highly reactive metal, which breaks down water into hydrogen and oxygen when it comes in contact with it. Hence, the balanced equation should be:
2H202 + NaI = NaOH + I +H20?
I am not sure...
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Elemental iodine exists as a diatomic molecule I2. Also, isnt this reaction, as per the original question, supposed to form oxygen gas?
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I beleive I have it figured out:
NaI + 2H2O2 --> NaI + 2H2O + O2
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I beleive I have it figured out:
NaI + 2H2O2 --> NaI + 2H2O + O2
Great. A little Googling on my part confirms that NaI is just a catalyst. (i hope that's true, Google could be wrong you know)
Now, how much O2 do you get?
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It appears we get 1 molecule of O2
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What do I need to know the number of O2 molecules for? :-\
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O2 production in 60 s (mL):151.0
Did you misunderstand this parameter, for a start?
Did you miss this, or did you not realize that mLs is a volume measurement, that you can plug into PV=nRT? 1ml=1cm3
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Sorry Arkcon, I forgot ::)
Since 1ml=1cm3
151 mL = 151 cm^3
I beleive I now rearrange the Ideal Gas Law to find # of mols?
n = PV / RT
=(102.6 kPa)(151 cm^3) / (8.3145 J/mol K)(290 K)
= 15492.6 / 2411.205
= 6.4 mols
???
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Partial pressure of water vapour:1.94 kPa.
Now, what's this part for?
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6.4 mols will tell us how many mols of O2 there are? Since there is one molecule of O2...
6.4 mols x 1 = 6.4 mols O2
Am I correct so far?
Now, since the oxygen production takes place in a time period of 60 seconds, I divide 6.4 mols/ 60 s
which gives 0.11 mols O2/sec :-\
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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2)
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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2)
Augh. So close. And yet so far.
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Since this is a pressure of the substance in the gas phase which is established at a particular temperature,
The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2), at 290 K?
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Since this is a pressure of the substance in the gas phase which is established at a particular temperature,
OK
The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water ,
OK
which is oxygen (O2)
What, you say water is oxygen? I mean, you've typed it, so in a sense, you've said it, but, would you ever say that? To a large group of people?
at 290 K?
And this addition adds, what, to your previous statement?
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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is H20.
Sorry, oxygen cannot be water since it has no hydrogen to perform hydrogen bonding
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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is water(g)(H2)), at 290 K meaning that at 290 K, the state of water is a gas.
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The partial pressure, of water vapor, is pertinent to the oxygen generated, exactly how.
The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is H20.
Sorry, oxygen cannot be water since it has no hydrogen to perform hydrogen bonding
The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is water(g)(H2)), at 290 K meaning that at 290 K, the state of water is a gas.
When you write semi-logical, technically incorrect, mish-mashes of facts like this, I begin to suspect I'm talking to a bot. :D
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The partial pressure, of water vapor, is related to the oxygen generated because oxygen in water obeys Henry's law rather well; the solubility is roughly proportional to the partial pressure of oxygen in the air:
pO2 = KO2 xO2
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Was my statement correct? :-\
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Arkcon, when you said
NaI + 2H2O2 --> NaI + 2H2O + O2
is correct, my teacher told me you were wrong.
Does anyone know what is the correct balanced equation?
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Arkcon, when you said
NaI + 2H2O2 --> NaI + 2H2O + O2
is correct, my teacher told me you were wrong.
Does anyone know what is the correct balanced equation?
I think it's one correct equation, Googling leads me to believe that NaI is a catalyst for the decomposition of H2O2
Lookie here:
http://www.google.com/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=GFx&q=NaI+catalyst+hydrogen+peroxide&btnG=Search
Second one, I can't get the PDF to download, or I'd attach it for you.
Anyway, we did establish, I believe, that we won't be using the reaction, as volume of O2 was a given.
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I was taught to put catalyst over arrow, not on both reactants and products sides. I believe I have already wrote (in one of the numerous threads about this problem - you see now why posting the same zillion times doesn't make sense?) that NaI should be canceled.
Edit: even if - from the logical point of view - putting catalyst on both sides makes sense :)
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NaI + 2H2O2 --> NaI + 2H2O + O2
Cancelling NaI on both sides ;D
2H202 --> 2H20 + 02
But since the purpose of this experiment is to determine the order of the reaction with respect to hydrogen peroxide and iodide ion and to determine the rate law equation and the rate constant for the decomposition of hydrogen peroxide catalyzed by iodide ion.
I am guessing that there would need to be an iodide ion present in the equation. So I came up with the following:
(1) H2O2 + 2NaI = 2Na+ + I2 + 2OH-
(2) H2O2 = H2O + 1/2 O2
Probably, (if pH is >7) there is also the formation of some IO3-:
3I2 + 6OH- = IO3- + 5I- + 3H2O
I got this information from http://www.thenakedscientists.com/forum/index.php?topic=2067.75;topicseen
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(1) H2O2 + 2NaI = 2Na+ + I2 + 2OH-
Unless you will later reduce I2 to I- NaI is consumed, so it is hardly a catalyst.
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The iodidie ion is the catalyst? :D
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The iodidie ion is the catalyst? :D
Yes - I'm pretty sure I- can be a catalyst - I remember it from one of my lectures last year.
H2O2 + I- --> H20 + IO-
IO- + H202 ----> H20 +O2 +I-
2H202 -----> 2H20 + O2
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hmm.. my chemistry and physics teachers might be wrong then :P
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How would I use the balanced equation to determine the number of mols of hydrogen peroxide decomposed per second?
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How would I use the balanced equation to determine the number of mols of hydrogen peroxide decomposed per second?
Anyway, we did establish, I believe, that we won't be using the reaction, as volume of O2 was a given.
:P
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Quote from: Marcus Soutlo on Yesterday at 07:10:54 PM
How would I use the balanced equation to determine the number of mols of hydrogen peroxide decomposed per second?
Quote from: Arkcon on Yesterday at 11:59:42 AM
Anyway, we did establish, I believe, that we won't be using the reaction, as volume of O2 was a given.
Can you please specify what you mean?
God bless
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Quote from: Marcus Soutlo on Yesterday at 07:10:54 PM
How would I use the balanced equation to determine the number of mols of hydrogen peroxide decomposed per second?
Quote from: Arkcon on Yesterday at 11:59:42 AM
Anyway, we did establish, I believe, that we won't be using the reaction, as volume of O2 was a given.
Can you please specify what you mean?
God bless
OK, review the enhanced portions of the original post:
Please calculate the number of mols of oxygen gas produced per second
Remember PV=nRT and that you must account for the partial pressure of water vapour.
The reactions conditions are:
Atmospheric pressue:102.6 kPa;
Temperature:290 K
Partial pressure of water vapour:1.94 kPa.
The reaction is:
Volume of 6% H2O2 (mL):10.0
Volume of distilled water (mL):0.0
Volume of 1.0 M NaI (mL):10.0
Volume of distilled water (mL):0.0
O2 production in 60 s (mL):151.0
:)
What I have done thus far is rearranged the Ideal Gas Law in order to get the number of mols
n = PV/RT
Is this the correct first step ???
Also, I have substituted the following values into the equation:
P = 102.6 kPa
V = ?
R = 8.3145 J/mol K
T = 290 K
???
Thank you for any help it is greatly appreciated.
Could you quote and highlight your new question again, giving all the given information, regarding the moles of H2O2 decomposed, and the order of the reaction? Have you solved this original one?
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Absolutely,
The question did not ask us to solve for the order of the reaction yet. For the moles of O2 produced, I received an answer of 0.00003846 mol per second, or 0.0023076 mol per minute of O2 gas produced.
The information given regarding the relationship between O2 and H202 is the balanced equation. Thus, from knowing the number of coefficients in the balanced equation, this is how I can solve for the number of moles of hydrogen peroxide decomposed?
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I received an answer of 0.0023076 mol per minute of O2 gas produced.
http://www.chemicalforums.com/index.php?topic=24222.msg91844#msg91844
???
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THe reason the two answers are different is because the lab asks to find out 5 different answers for 5 seperate reactions
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Sorry if I confused anyone
I only need to know how to use the balanced equation to determine the number of mols of hydrogen peroxide decomposed per second?
Of the two reactions I did, I got two answers, according to my calculations, of
0.0023076 mol per minute of O2 gas produced
and
0.0062990 mol per minute of O2 gas produced
Lets say I choose 0.0023076 mol per minute of O2 gas produced. How can I determine the number of mols of hydrogen peroxide decomposed per minute from this? I know that I then need to divide by 60 but I am stuck on the first step.
Again, sorry for any confusion I may have caused :)
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One of my reactions is:
Volume of 6% H2O2 (mL):10.0
Volume of distilled water (mL):0.0
Volume of 1.0 M NaI (mL):10.0
Volume of distilled water (mL):0.0
O2 production in 60 s (mL):151.0
I have collected oxygen gas by the downward displacement of water (results are in the last column). This oxygen was produced in the first 60 s of the reaction.
Im asked to calculate the change in hydrogen peroxide concentration per second. This is the rate. (Hint: What volume of liquid does the hydrogen peroxide now occupy?).
I know that
rate = change in concentration / time
But how do I get change in concentration? ???
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From reaction equation. You know how much oxygen was produced per minute, you can calculate how much hydrogen peroxide was consumed. Simple stoichiometry.
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b) Write a complete balanced equation and use it to determine the number of mols of hydrogen peroxide decomposed per second.
c) Calculate the change in hydrogen peroxide concentration per second. This is the rate. (Hint: What volume of liquid does the hydrogen peroxide now occupy?).
So what is the difference between the number of mols of hydrogen peroxide decomposed per second in part b and the change in hydrogen peroxide concentration per second in part c?
Thank you again ;)
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H2O2 NaI
Reaction Volume of 6% H2O2 (mL) Volume of distilled water (mL) Volume of 1.0 M NaI (mL) Volume of distilled water (mL) O2 production in 60 s (mL)
1 10.0 0.0 10.0 0.0 151.0
2 8.0 2.0 10.0 0.0 93.0
3 6.0 4.0 10.0 0.0 61.0
4 4.0 6.0 10.0 0.0 41.0
5 10.0 0.0 8.0 2.0 99.0
6 10.0 0.0 6.0 4.0 73.0
7 10.0 0.0 4.0 6.0 44.0
The above table is the information I am given. Say I choose reaction 2.
what is the difference between the number of mols of hydrogen peroxide decomposed per second in part b and the change in hydrogen peroxide concentration per second in part c?
My guess is it has something to do with the distilled water volume?
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I would start converting all these volume numbers into hydrogen peroxide and sodium iodide concentrations.
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So since
n = m / M
is the volume (in ml) a mass?
For example, I am assuming that 8.0 mL of 6% H2O2 is actually 8.0 grams?
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Wouldnt I use the formula
molarity = moles of solute / liter of solution
in order to convert all these volume numbers into hydrogen peroxide and sodium iodide concentrations?
If this is what I must do, how would I find the moles of solute? For reaction 2, for instance should I assume that 8.0 mL of 6% H2O2 is actually 8.0 grams, and from this mass I would find the moles of solute by using the formula n=m/M? Would the 2.0 mL of distilled water be 0.002 L for litre of solution? ??? Please help for this assignment was due yesterday and I need to finish it :'(
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This is regarding reaction 2
Reaction 2
H2O2 NaI
Reaction Volume of 6% H2O2 (mL) Volume of distilled water (mL) Volume of 1.0 M NaI (mL) Volume of distilled water (mL) O2 production in 60 s
(mL)
2 8.0 2.0 10.0 0.0 93.0
My teacher told me If I find the number of moles of peroxide decomposed per second(0.00006470 mol O2/second), it is only a single calculation to convert that to change in concentration by using volume. What is this single calculation?
Ive tried n = c/v (mols = concentration / volume) but I do not think this is it. Can anyone tell me this calculation and I will sub in my values to see if it is correct.
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Wouldnt I use the formula
molarity = moles of solute / liter of solution
Yes.
should I assume that 8.0 mL of 6% H2O2 is actually 8.0 grams
Yes.
and from this mass I would find the moles of solute by using the formula n=m/M?
No. You should use definition of percent concentration, solved for mass of the solute.
Would the 2.0 mL of distilled water be 0.002 L
Yes.
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My teacher told me If I find the number of moles of peroxide decomposed per second(0.00006470 mol O2/second), it is only a single calculation to convert that to change in concentration by using volume. What is this single calculation?
You have to find out how many moles of hydrogen peroxide were decomposed to produce given amount of oxygen. Concentration is amount of solute per volume of solution - assume that volume doesn't change, calculate how much solute was left (you know initial amount, you know how much was decomoposed) - and you have everything needed.
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C%w/w = msubstance / msubstance + msolvent x 100%
= 8.0 / 8.0 + 2.0 x 100%
= 8.0 / 10 x 100%
= 80%
Therefore reaction 2 is made of 80% hydrogen peroxide?
But how do I get mass from this?
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How would I get this mass and is this mass the change in hydrogen peroxide?
Thanks again
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C%w/w = msubstance / msubstance + msolvent x 100%
Solve for msubstance, simple algebra. Note, it is even easier than it looks, as you don't have to treat mass of solution as (msubstance + msolvent) - you already know total volume and you can assume density of the solution to be 1 g/mL. That's not exactly true, but will do.
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2 moles were decomposed to produce 1 mol of oxygen.
Concentration = amount of solute per volume of solution
= 8 g / 10 g
Therefore mass of hydrogen peroxide must be 8 grams ???
If this is right, then would I use 8 grams to find mols and then plug the number of mols into the value
moles of solute / liter of solution to get M?
------------------------------------------------------------------------------------------------------------------------------------------
OR
How do I rearrange
C%w/w = msubstance / msubstance + msolvent x 100% to find m substance?
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http://en.wikipedia.org/wiki/Elementary_algebra
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I think I understand
C%w/w = msubstance / msubstance + msolvent x 100%
rearranging gives
msubstance = C%w/w divided by msolution x 100% ???
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Cp = msubst/msol x 100%
How do you "move" msolution to the LHS?
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To move msolution to the left hand side, I would rearrange the formula to get
msol = Cp / msubst x 100%
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You are doing it wrong. Slow down. Try to multiply BOTH sides of the equation by msolution. If you do some operation to BOTH sides of the equation, it still holds.
Geez, you really were not taught how to do such things?
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Cp(msol) = (msol)(msubst)(100%)
Sir, algebra is not my strong point.
Is there any other method or formula I could use?
Here is what I have in mind:
The only thing I know about the rate is the oxygen production: 93.0 mL / 60 s
The reaction that gives the oxygen is:
2 H2O2 --> 2 H2O + O2
So, I need to calculate the amount of moles of oxygen produced. Then I know the amount of moles of peroxide that are reacting away... which is the answer I need.
n = 93 g / 32g per mol
= 2.90635 mol
= 2.90635 mol / 60 seconds
= 0.0484375 mol / sec of O2 are created in reaction 2
Since there is twice the amount of H2O2, multiply 0.0484375 mol / sec by 2 to get
0.096875 mol/L s of H2o2
Therefore the rate is 0.096875 mol/L s ?
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Cp(msol) = (msol)(msubst)(100%)
No.
Cp = msubst/msol x 100%
If you multiply BOTH sides by msol you get
msol x Cp = msol x msubst/msol x 100%
Can you cancel something?
Sir, algebra is not my strong point.
Don't sir me. And I already know you struggle.
Is there any other method or formula I could use?
Any other approach needs separate approach to each question, which doesn't make sense. Once you know how to rearrange formula and solve for particular unknown you have a tool to solve each question. Well, almost each question.
The only thing I know about the rate is the oxygen production: 93.0 mL / 60 s
The reaction that gives the oxygen is:
2 H2O2 --> 2 H2O + O2
So, I need to calculate the amount of moles of oxygen produced. Then I know the amount of moles of peroxide that are reacting away... which is the answer I need.
n = 93 g / 32g per mol
= 2.90635 mol
= 2.90635 mol / 60 seconds
= 0.0484375 mol / sec of O2 are created in reaction 2
Since there is twice the amount of H2O2, multiply 0.0484375 mol / sec by 2 to get
0.096875 mol/L s of H2o2
Therefore the rate is 0.096875 mol/L s ?
Could be you are on the right track, but you have started with 93 mL of gaseous oxygen and then it miraculously became 93 g of oxygen.
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Regarding your quote
msol x Cp = msol x msubst/msol x 100%
Can you cancel something?
I can cancel msol on both sides to get Cp = msubst x 100%
Regarding your quote
Could be you are on the right track, but you have started with 93 mL of gaseous oxygen and then it miraculously became 93 g of oxygen.
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hmm.. oxygen gas is not water so 1 g cannot equal 1 mL.
Could 93 mL be converted to 0.093 L? Could volume be accepted as mass?
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Sorry I misread
Cp = msubst/msol x 100%
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msol x Cp = msol x msubst/msol x 100%
Can you cancel something?
I can cancel msol on both sides to get Cp = msubst x 100%
No. Grouping msl on the right:
msol x Cp = msol/msol x msubst x 100%
as
msol/msol = 1
so
msol x Cp = msubst x 100%
and msol was mobed to the LHS. Now, try to do the same to 100% (hint: divide both sides).
hmm.. oxygen gas is not water so 1 g cannot equal 1 mL.
Could 93 mL be converted to 0.093 L? Could volume be accepted as mass?
No, it is gas, so PV = nRT. Marcus already got it right.
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Regarding Marcus's post
n = PV / RT
Thus,
=(102.6 kPa)(0.93 L) / (8.3145)(290 K)
= 95.418 / 2411. 205
= 0.0395727447 mol of O2
Since there exists twice the amount of hydrogen peroxide,
0.0791 mol of H2O2
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msol x Cp = msubst x 100%
Dividing both sides:
msol x cP / msol = msubst x 100% / msol
msol cancels out on the LHS.
Thus,
cP = msubst x 100%
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Regarding Marcus's post
n = PV / RT
Thus,
=(102.6 kPa)(0.93 L) / (8.3145)(290 K)
Almost OK. Almost, as you have forgot to subtract water vapor pressure. Marcus did.
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msol x Cp = msubst x 100%
Dividing both sides:
msol x cP / msol = msubst x 100% / msol
msol cancels out on the LHS.
Thus,
cP = msubst x 100%
Please, pay attention to what you are doing, reading and writing. I told you to move 100% to LHS, why do you divide by msol? But even then you have lost it on the RHS for no reason.
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102.6 kPa - 1.94 kPa = 100.7 kPa
n = PV / RT
Thus,
=(100.7 kPa)(0.93 L) / (8.3145)(290 K)
= 93.65 / 2411. 205
= 0.03884 mol of O2
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Please, pay attention to what you are doing, reading and writing. I told you to move 100% to LHS, why do you divide by msol? But even then you have lost it on the RHS for no reason.
I divided 100% by msol by reading your hint (divide both sides). I lost msol on the RHS because one is the denominator, the other is numerator. They are both the same so I cancelled them out :-\
On a second attempt,
100% x msol x Cp = msubstance
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I divided 100% by msol by reading your hint (divide both sides). I lost msol on the RHS because one is the denominator, the other is numerator. They are both the same so I cancelled them out :-\
On a second attempt,
100% x msol x Cp = msubstance
It is hopeless. You have to move 100%, so you should divide by 100%.
msol x Cp = msubst x 100%
Both sides divided by 100%:
msol x Cp / 100% = msubst x 100%/100%
canceling:
msol x Cp / 100% = msubst
Solved for msubst.
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:o honestly, I can't beleive I didn't realize this
msubstance = msolution x cP / 100%
= 10 g x 80% / 100%
= 8 g
Is this correct?
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:o honestly, I can't beleive I didn't realize this
msubstance = msolution x cP / 100%
= 10 g x 80% / 100%
= 8 g
Is this correct?
As written - yes. If it fits the problem - I don't know, you have long lost me.
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Let me rewrite the question in a neat order :P
The question was:
Calculate the change in hydrogen peroxide concentration per second. This is the rate. (Hint: What volume of liquid does the hydrogen peroxide now occupy?).
The reaction was:
Reaction Volume of 6% H2O2 (mL) 8.0
Volume of distilled water (mL) 2.0
Volume of 1.0 M NaI (mL) 10.0
Volume of distilled water (mL) 0.0
O2 production in 60 s (mL) 93.0
2 H2O2 --> 2 H2O + O2
Since you have helped me solve mass for H2O2, I can now find number of moles for H2O2 by using finding molarity.
Molarity = moles of solute (H2O2) / litres of solution (0.01 L)
gives me value of M
but what is next?
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To Frederick, please i need some help with this rates lab
Thank you
I appreciate it
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The results of this discussion are incorrect
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please someone, i need some help with the rates lab
thank you very much
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Sorry Arkcon, I forgot ::)
Since 1ml=1cm3
151 mL = 151 cm^3
I beleive I now rearrange the Ideal Gas Law to find # of mols?
n = PV / RT
=(102.6 kPa)(151 cm^3) / (8.3145 J/mol K)(290 K)
= 15492.6 / 2411.205
= 6.4 mols
???
where did u get 8.3145 j/mol k from???