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Posted: 9/17/2002 5:45:43 PM EDT
[#1]
When you cheat at homework, you're only cheating yourself at test time.
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Posted: 9/17/2002 5:52:37 PM EDT
[#2]
Quoted: When you cheat at homework, you're only cheating yourself at test time. Thanks, ______. if i don't know how to do the friggin homework it doesn't make a pinch of shit difference on the test anyway! [:(!] |
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Posted: 9/17/2002 5:52:59 PM EDT
[#3]
Giving you the answer deprives you of the fun of figuring it out for yourself.
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Posted: 9/17/2002 5:53:47 PM EDT
[#4]
youre a clever one.
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Posted: 9/17/2002 5:56:42 PM EDT
[#5]
Quoted: Giving you the answer deprives you of the fun of figuring it out for yourself. Don't feel bad I don't know either[:D] |
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Posted: 9/17/2002 5:58:30 PM EDT
[#6]
Sorry -
I can't help.... About the only chemistry help I can give is how to instruct a bartender to "ONLY PUT 2 DROPS" of Vermouth in a certain fixed amount of Bombay to yield the correct chemical composition.... |
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Posted: 9/17/2002 6:02:15 PM EDT
[#7]
Quoted: Sorry - I can't help.... About the only chemistry help I can give is how to instruct a bartender to "ONLY PUT 2 DROPS" of Vermouth in a certain fixed amount of Bombay to yield the correct chemical composition.... have some pity and send me some Bombay, then! please?? |
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Posted: 9/17/2002 6:19:06 PM EDT
[#8]
Quoted: this is ridiculously easy but i can not seem to come up with the correct answer: 2.5 mL of glacial (pure) acetic acid is dissolved in 250 mL of water. Sodium Bicarbonate (9.5g) is added to this solution. What mass of carbon dioxide is produced? Enter your answer with 3 significant figures. here is the equation: C2H402 + NaHCO3 --> NaCH3CO2 + H20 + CO2 A cadet will not lie, cheat, steal or tolerate those who do. this little stunt if pulled at my school wouldve had you booted in a nice ceremony at 330 am |
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Posted: 9/17/2002 6:28:34 PM EDT
[#9]
Dude...I just met a girl that I'm talking to now and she's a chemistry teacher. I'll see if she can help at all.
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Posted: 9/17/2002 6:28:49 PM EDT
[#10]
How about help instead of the answer.
First you must calculate the number of mols you have dumped into the water. Since it is a balanced equation already you are lucky. No extra math involved. So you dump in 2.5 mils of pure C2H4O2. How many mols of that did you add to the 250 ml of water. What does that make it now that you are in 250 ml of water? How many mols of NaHCO3 is in 9.5 grams? Now you must find the limiting chemical. Which is used up first. When you find that, it is a 1 mol for 1 mol equivialent for what is used up first to the amount of CO2 created. With that number you can convert into Grams. |
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Posted: 9/17/2002 6:28:56 PM EDT
[#11]
hoplite, you seem to have misunderstood;
I am not cheating, i am not lying, nor am i stealing. I SIMPLY DO NOT UNDERSTAND HOW TO SOLVE THE PROBLEM AND BECAUSE THE HELP ROOM IS CLOSED AND NO ONE ON MY FLOOR SEEMS TO KNOW HOW TO SOLVE IT EITHER, I HAVE TURNED TO AR15.COM FOR ASSISTANCE!! ...So basically, what youre telling me is that YOU never asked for help while in college? RIIIIIIGHHT. hop off your throne and join us serfs... |
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Posted: 9/17/2002 6:29:34 PM EDT
[#12]
So let me get this straight. . .all you guys that don't want to help him musta learned EVERYTHING you know by yourself right? You musta all been born into families where public school was outta the question. OH I KNOW WHAT IT IS! Ya'll are all just perfect omniscient gods with the full understanding of the universe gifted upon you as you were birthed kicking and screaming into this world.
Give me a fucking break. He's asking for some help on a chemistry problem. Have a problem with giving him the answer? Help him along with the basic concepts so he can find the answer himself. I'd try to mess with it, but I hate chemistry and I have 2 more test this week and no more time. I was just totally discusted at the attitudes of those that posted. All I can say is balance the equation then work the molar formula. But make sure there isn't a limited reagent. - Matt |
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Posted: 9/17/2002 6:32:38 PM EDT
[#13]
Quoted: How about help instead of the answer. First you must calculate the number of mols you have dumped into the water. Since it is a balanced equation already you are lucky. No extra math involved. So you dump in 2.5 mils of pure C2H4O2. How many mols of that did you add to the 250 ml of water. What does that make it now that you are in 250 ml of water? How many mols of NaHCO3 is in 9.5 grams? Now you must find the limiting chemical. Which is used up first. When you find that, it is a 1 mol for 1 mol equivialent for what is used up first to the amount of CO2 created. With that number you can convert into Grams. yes, i understand that as well as determining which reactant limits the production of CO2. however, i'm not sure how to convert 2.5 mL of glacial acetic acid to moles... is it 1:1 ml to grams? i did try that previously but still ended up with the wrong answer. and really, what i was looking for was HOW to solve the problem rather than just the answer because an answer in and of itself doesn't help me on tomorrow night's exam. |
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Posted: 9/17/2002 6:33:02 PM EDT
[#14]
That is correct it is a moler formula, but I think the equation is already balanced.
Check me on it, but all the atoms are equal on both sides. |
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Posted: 9/17/2002 6:33:24 PM EDT
[#15]
which i covered in my second post on this thread...
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Posted: 9/17/2002 6:35:19 PM EDT
[#16]
correct, andrew. i entered the equation as a balanced equation.
edited to add: i have .113 moles of baking soda. (check my math but i think its right) but how do i determine the moles of acetic acid? |
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Posted: 9/17/2002 6:35:21 PM EDT
[#17]
My guess is since they say it is pure, it is 1 molar acid, but you disolved it into 250 ml. You must convert. Basically 1 mol says you have 1 mol in a liter of water. You have 2.5 ml of 1 mol fluid you are pouring into 250 ml of water.
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Posted: 9/17/2002 6:35:28 PM EDT
[#18]
"Glacial acetic acid boils at 118°C, and has a density of 1.049 g/mL at 25°C. "
M = D/V - Matt |
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Posted: 9/17/2002 6:42:53 PM EDT
[#19]
I think (and you can correct me if I am wrong because it has been a while) it may be redundant to calculate mass. If I am correct it will work out to be .625 mols of acid. I don't have a periodic chart to find the mols of base.
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Posted: 9/17/2002 6:46:27 PM EDT
[#20]
Quoted: "Glacial acetic acid boils at 118°C, and has a density of 1.049 g/mL at 25°C. " M = D/V - Matt does that M stand for mass? i think it is M=dv then...?? |
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Posted: 9/17/2002 6:48:06 PM EDT
[#21]
Yes it should be mass. If you check the units it comes out in g.
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Posted: 9/17/2002 6:49:33 PM EDT
[#22]
Quoted: Quoted: "Glacial acetic acid boils at 118°C, and has a density of 1.049 g/mL at 25°C. " M = D/V - Matt does that M stand for mass? i think it is M=dv then...?? Yes. Anyone who took the time to think about the units of density could determine that Density=Mass/Volume. |
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Posted: 9/17/2002 6:52:07 PM EDT
[#23]
Ack, I'm sorry. I told yall I don't like chemistry :) But I digress, I must study. . .
Sorry again about that slip up, but then again at least I don't profess to be all knowing. . . - Matt |
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Posted: 9/17/2002 6:52:17 PM EDT
[#24]
yes, IMHO, but i didn't have the mass of the acetic acid
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Posted: 9/17/2002 6:52:55 PM EDT
[#25]
don't listen to these other fellas...they refuse to answer because they can't.
i'll try and walk you through this (i was there once myself). first eliminate non-important information (250 mL of water). since your only concerned with the amount of acetic acid (the moles of the acid) the volume of water is of little importance. your equasion is balance, so that is already taken care of (one mole of acetic acid-strong acid with one mole of sodium bicarb-weak base will give you one mole of corbon dioxide). no figure out your limiting reagent. the MW of acetic acid is 60 and that of sodium bicarb is 84. so when you calculate out your molar amounts you will end up with 0.0000417 moles of acetic acid and 0.1131 moles of sodium bicarb (THIS IS ASSUMING THAT THE DENSITY OF ACETIC ACID IS 1.0 G/mL...which I think it is). so you've determined that the limiting reagent (acetic acid) is present in 0.0000417 moles. this will give you 0.0000417 moles of carbon dioxide (mw 44) which should give you 0.00183 G (1.83 mg's) of carbon dioxide. now can you tell me, what volume this carbon diozide will occupy at standard atmospheric conditions (gas laws). i hope this is right...i haven't done it in quite some time. sloth |
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Posted: 9/17/2002 6:53:34 PM EDT
[#26]
And I don't want you to get sidetracked here, so I'll suggest that you turn to the section in your book, or day of lecture notes, that deals with stoichiometrics, and specifically stoichiometrically equivalent molar ratios from balanced equations:
Write a balanced equation Convert the given mass of the first substance to the amount (mol) Use the appropriate molar ratio from the balanced equation to determine the amount (mol) of the second substance Convert the amount of the second substance to the desired mass |
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Posted: 9/17/2002 6:55:23 PM EDT
[#27]
Right, but you could have gotten it by the mols if in fact I did the calcs right and made the assumptions correctly. Let me know if my way would have short cut it for you, as I am curious now.
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Posted: 9/17/2002 6:55:52 PM EDT
[#28]
shoot...you've have to account for the density (1.05 g's/mL) so that will give you a mass of 0.00238 g's acetic acid...sorry.
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Posted: 9/17/2002 6:56:22 PM EDT
[#29]
God, apparently I just read EVERYTHING wrong. Sorry nm_man I read it as you have 250mL of acetic acid, which is why I said use density. Skoool make meh brain BAD!! ;)
But yeah, I hope you have better luck in chemistry than I did. - Matt |
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Posted: 9/17/2002 6:57:18 PM EDT
[#30]
Quoted: yes, IMHO, but i didn't have the mass of the acetic acid Well, yes you did. . . haven't you got a periodic table for reference? You already know that acetic acid is C2H4O2. . . add it up. |
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Posted: 9/17/2002 6:59:15 PM EDT
[#31]
IMHO-i had the MOLAR mass of the acid but not the mass itself. sorry bout that.
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Posted: 9/17/2002 7:04:10 PM EDT
[#32]
Quoted: IMHO-i had the MOLAR mass of the acid but not the mass itself. sorry bout that. Am I missing something here? If you know the molar mass, and the number of moles, then the mass ought not be so hard to calculate. . . |
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Posted: 9/17/2002 7:07:41 PM EDT
[#33]
Quoted: now can you tell me, what volume this carbon diozide will occupy at standard atmospheric conditions (gas laws). sloth Yes, the ideal gas laws, fondly remembered in my mind as the "Pervert" law. . . PV=nRT (just make sure you use the right units) |
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Posted: 9/17/2002 7:08:13 PM EDT
[#34]
how did you determine the number of moles of acetic acid?
here's how i did it: m=dv 1.049*2.5=2.6225g of acetic acid. divide that by 60.05 g/mol and you have .04367 moles of acetic acid. since you have .113 moles of baking soda, and it is used on a 1:1 basis, the acid is the limiting reactant, right? and there fore, because everything is on a one to one basis on both sides of the equation, you will also have .04367 moles of CO2 which comes out to be 1.92 g am i correct? or where am i erring if not? |
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Posted: 9/17/2002 7:09:38 PM EDT
[#35]
Quoted: Quoted: IMHO-i had the MOLAR mass of the acid but not the mass itself. sorry bout that. Am I missing something here? If you know the molar mass, and the number of moles, then the mass ought not be so hard to calculate. . . EXCEPT that i didn't have the number or moles nor the mass... |
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Posted: 9/17/2002 7:59:07 PM EDT
[#36]
I come up with 1.92g CO2 so nm man you got it correct assuming a density of 1.049g/ml which I cannot seem to verify.
edited because its too f'ing late for this shit. |
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Posted: 9/17/2002 8:18:00 PM EDT
[#37]
Quoted: hoplite, you seem to have misunderstood; I am not cheating, i am not lying, nor am i stealing. I SIMPLY DO NOT UNDERSTAND HOW TO SOLVE THE PROBLEM AND BECAUSE THE HELP ROOM IS CLOSED AND NO ONE ON MY FLOOR SEEMS TO KNOW HOW TO SOLVE IT EITHER, I HAVE TURNED TO AR15.COM FOR ASSISTANCE!! ...So basically, what youre telling me is that YOU never asked for help while in college? RIIIIIIGHHT. hop off your throne and join us serfs... Theres a thin line between help and cheating. You have crossed the line. As far as me on a high horse, are you and the environment around you so corrupted that is it impossible for you to percieve one person (me) or many who will not lie,cheat, or steal because it is the right thing to do? Honor above ones self? If that is the case then this homework assignmenment is the least of your problems |
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Posted: 9/17/2002 8:24:06 PM EDT
[#38]
Quoted: Quoted: Quoted: IMHO-i had the MOLAR mass of the acid but not the mass itself. sorry bout that. Am I missing something here? If you know the molar mass, and the number of moles, then the mass ought not be so hard to calculate. . . EXCEPT that i didn't have the number or moles nor the mass... ...so you used the density(a known constant) and multiplied by the volume to determine mass in beaker, therefore moles, etc, etc. Good job. |
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Posted: 9/17/2002 8:31:58 PM EDT
[#39]
Hoplite, I wouldn't be so hard on him. He used to pieces of advice given to figure out and solve the problem. I don't think anyone solved it for him. Many acted as teachers here. I think it was great.
I think its also safe to assume that he's just starting chem this semester. The key, he has learned to most stochiometric problems, is to find a way to break your reactants, etc into moles. |
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Posted: 9/18/2002 2:28:45 PM EDT
[#40]
where do you get the idea that asking for help is cheating? you seem to be implying that you never asked for help throughout college....? and i have a hard time believing that. to clarify, i think you need to realize that even if you had given me the correct answer and only the correct answer, i still would have worked the answer back through the problem to learn how to do it....like i stated before, an answer doesn't help me on the exam that's in 45 minutes and counting unless i know how to get it, as i outlined in the second post.... dumbass. you DO need to get off your high horse and stop being so anal retentive.
btw, the CAPA system accepted 1.92g of CO2. thanks all for your help! |
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Posted: 9/18/2002 2:34:10 PM EDT
[#41]
Quoted: this is ridiculously easy but i can not seem to come up with the correct answer: 2.5 mL of glacial (pure) acetic acid is dissolved in 250 mL of water. Sodium Bicarbonate (9.5g) is added to this solution. What mass of carbon dioxide is produced? Enter your answer with 3 significant figures. here is the equation: C2H402 + NaHCO3 --> NaCH3CO2 + H20 + CO2 Mmmm... about a hundred dollars. Yeah, 22 minutes till Wapner. Yeah. |
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Posted: 9/18/2002 3:37:23 PM EDT
[#42]
nm,
2.5 mL of glacial (pure) acetic acid is dissolved in 250 mL of water. Sodium Bicarbonate (9.5g) is added to this solution. What mass of carbon dioxide is produced? Enter your answer with 3 significant figures. here is the equation: C2H402 + NaHCO3 --> NaCH3CO2 + H20 + CO2 NaHC03 = 9.5g NACH3Co2 = 2.5ml H20 = 100ml C2H4O2 = (x)g or (x)ml CO2 = (x)g It's been a long time since I've taken chemistry, but it seems you are trying to solve an equation with two unknowns. |
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Posted: 9/18/2002 4:16:32 PM EDT
[#43]
Quoted: how did you determine the number of moles of acetic acid? here's how i did it: m=dv 1.049*2.5=2.6225g of acetic acid. divide that by 60.05 g/mol and you have .04367 moles of acetic acid. since you have .113 moles of baking soda, and it is used on a 1:1 basis, the acid is the limiting reactant, right? and there fore, because everything is on a one to one basis on both sides of the equation, you will also have .04367 moles of CO2 which comes out to be 1.92 g am i correct? or where am i erring if not? ak_ar_man, here is the answer per nm_man. You have to know the density of acetic acid to calculate mass in grams and thereby the moles of acetic acid. |
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