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I cannot figure these two problems out for the life of me, so I'm… - CHS Chem Help Online

About I cannot figure these two problems out for the life of me, so I'm…

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I cannot figure these two problems out for the life of me, so I'm hoping someone here might be able to help.


Compare the amount of cooling experienced by an individual who drinks 400.0 mL of water (0°C) with the amount of cooling experienced by an individual who sweats out 400.0 mL of water. Assume all sweat evaporates. The density of water is 1.00 g/mL, the specific heat of water is 4.18 J/g°C and the heat of vaporization of water is 2.41 kJ/g at 37°C.


9. A 1.80-gram mixture of potassium chlorate, KClO3, and potassium chloride, KCl, was heated until all of the KClO3 had decomposed. The liberated oxygen, after drying, occupied 405 milliliters at 25.0°C when the barometric pressure was 745 torr.

a. How many moles of O2 were produced?
b. What percent of the mixture was KClO3? KCl? (I can do the calculations for these, I just need help with writing out the reaction)
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From:lensovet
Date:July 2nd, 2004 05:10 am (UTC)

help is on the way

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9a will come very soon (after i shower and eat breakfast). 9b and the first ? might take a little more time.
when is this due, and how come you're still in school?
also, if you get answers @ chemistryhelp, post back here so that i can check/not bother doing them.
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From:_sugar_free_
Date:July 2nd, 2004 05:45 pm (UTC)

Re: help is on the way

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Thank you=) This is due on Monday, but I'm pretty sure I have both of them figured out now. I had to go over a lot of stuff that I had forgotten from Chem I. I'm still in school because I couldn't take Chem II last semester as I had other obligations (I was in HS and the college I was taking classes at didn't have any night classes scheduled for Chem II) so I'm taking it in the summer so I can be done with my pre-pharmacy classes next year.
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From:lensovet
Date:July 2nd, 2004 06:10 am (UTC)

9a

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a. You’ve got volume and temperature, and a gas, so you’ll be using the ideal gas equation, PV=nRT (I’m assuming you know what that is). Since I only know the R constant for atmospheres offhand, we will need to convert torr into atms, and keep in mind that temperature must always be absolute in gas law problems. In addition, milliliters have been converted to liters. So:



the last number equals the number of moles of O2
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From:lensovet
Date:July 2nd, 2004 06:37 am (UTC)

strange

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according to the only result on a google search that gave answers on this, it seems that the decomposition of KClO3 occurs as follows:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
Are you sure that you've got the problem down right?
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From:_sugar_free_
Date:July 2nd, 2004 05:26 pm (UTC)

Re: strange

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Yep. That's what made the problem a bit difficult for me; I wasn't sure how to write out the reaction with the KCl in the mixture. It took a while but I think I figured it out. I figured out the percentages with this reaction and they came out pretty close to 100% when added together.

2KClO3(s) + KCl → 3KCl + 2O2</s>
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From:lensovet
Date:July 3rd, 2004 08:32 pm (UTC)

wait, those oxygens

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heh, the coefficient on the oxygen should still be 3, right? because right now you've got 6 on the left and 4 on the right...
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From:_sugar_free_
Date:July 3rd, 2004 09:02 pm (UTC)

Re: wait, those oxygens

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Yeah, I meant for it to be three...heh. I should probably look at my comments more carefully before I post them.
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