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    killasnake's Avatar
    killasnake Posts: 5, Reputation: 1
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    #1

    Jan 26, 2006, 08:47 PM
    3 more chem question; I don't know how to start
    Hi again, I need help with these three problems I don' t know how to start it.

    The heat of combustion of benzene, C6H6, is –41,900 J/g. What is the heat of
    combustion expressed in kilojoules per kilogram?
    a. -4.19 x 101 kJ/kg b. -4.19 x 103 kJ/kg c. -4.19 x 104 kJ/kg
    d. -4.19 x 106 kJ/kg e. -4.19 x 109 kJ/kg

    All I know with this question is that kilo is 10^3.

    -------------------------------------------

    One gram of washing soda, Na2CO3 • 10H2O, has 2.1 x 1021 carbon atoms. How many
    oxygen atoms are present in 1.0g of washing soda?
    a. 2.1 x 1021 b. 6.3 x 1021 c. 8.4 x 1021
    d. 2.7 x 1022 e. 3.2 x 1022

    I'm just confused here
    --------------------------------------------


    One gram of TNT, C7H5N3O6, has 7.95 x 1021 nitrogen atoms. How many hydrogen
    atoms are there in 1.00 g of TNT?
    a. 7.95 x 1021 b. 9.26 x 1021 c. 1.33 x 1022
    d. 1.59 x 1022 e. 1.86 x 1022

    With this question Do I use 6.02 x 10^23 atoms?

    And thanks again for helping me.
    kp2171's Avatar
    kp2171 Posts: 5,318, Reputation: 1612
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    #2

    Jan 26, 2006, 10:00 PM
    I don't answer homework problems for people, but I'll guide you a long way this time. This might be long winded, but I want to show you how to solve the problem, not give you that answer outright.

    These are all fundamental problems that you'll NEED to be able to do yourself if you're going to pass chem. Its ok if you're not sure or need help... but don't just get an answer and then walk away from the problem type. Try a few more like it to make sure you understand. Most students that failed in my general chem classes never got the fundamentals at the beginning and it just got uglier from there. Really, really ugly.


    the first problem has 2 parts to understand. How to write in scientific notation and also how to convert units to other SI units.

    so... you know a kilo is 10^3. that means a kilo is 1000 whatevers... grams, meters, blah blah blah. We happen to be talking about joules here but so what. A kilojoule is a thousand joules. Ten kilojoules is ten thousand joules.

    so if a kilojoule is joule*1000... and therefore a kilogram is gram*1000...
    what happens when you write 1000*joule/1000*gram. Simplify the equation. Basically if you're changing the units on the top and bottom by the same number, it has no effect on the number out in front.

    if a reaction gave off 1 calorie/ounce... it also gives off 1 kilocalorie/kiloounce because it's the same as saying 1000 calories/ 1000 ounces. (stupid units but it shows a point)

    so all you have to do is figure out how to convert -41900 to sci notation. I won't do this for you.

    follow up is then what if I asked you to convert 10 joules/gram to joules/kilogram... that would mean 10 joules/1000g = 0.01joules/kg, without writing in sci notation. Not all problems will be like your example where the top and bottom cancel out by multiplying by the same number.

    Problem two can be answered a few ways... the fastest is to figure out the ratio of carbon atoms to oxygen atoms. So in the formula it says C one time with no subscripts, and the 2 in front is for the Na... so there is one C in this formula. Now how many O's are there?. you need to figure this out. The first O has a 3 behind it. What does that mean? The second O you see is from water, H2O. Every water has one O. the formula says there are 10 waters. How many O's is that? Add all the O's together. Now you know for every 1C there are?? O's.

    Most cars have four tires. One car = 4 tires. Ten cars = 4*10 tires. Same thing here. 2.1*10^21 C atoms is what you know... use the ratio to figure out O.

    The third problem is just like the 2nd. Yes, the instructor may have wanted you to convert to moles but the fastest way to do this is just like the above problem. How many nitrogens are there in the formula? How many H's are there?

    lets say the ratio was 9 N to 5 H (which it is NOT, this is an example) and you had 10 N atoms... to convert to H atoms you would take 10N * (5H/9N)... because 10N divided by 9N maked the N cancel, and * H leaves H on top.

    Threre you go. That's all I can give you. If you try the problems and post your answers I'll look at them.

    The early stuff in gen chem can be B O R I N G... but unfortunately if you don't know sulfate from sulfite or nitrate from nitrite, and you cannot manipulate units and work with sci notation then nothing after will make sense. So put in the time on this stuff. Get as much help as you need as early as possible. When you get it the early stuff the later stuff makes much more sense.
    killasnake's Avatar
    killasnake Posts: 5, Reputation: 1
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    #3

    Jan 27, 2006, 11:35 AM
    Well, I figured out problems 1 and 2 but I'm still having trouble with problem 3

    Problem 1:

    I take -41,900 J of heat. Divide that by 1000. Then I got -41.9 kJ of heat Then I take the 1 gram of mass. Divide that by 1000. This gives me .001 kg Then I took -41.9 kJ and divide it by .001 kg
    -41.9kJ/.001kg = -41900 kJ/kg or -4.19 x 10^4 kJ/kg. The answer is 'c'

    Problem 2:
    I have 13 oxygen atoms.

    Then I just multiply 13 by 2.1 x 10^21 atoms

    Which gives me; 2.7 x 10^22 oxygen atoms

    Problem 3:
    This is where I get stuck; you said that it was similar to problem 2

    I have 5 H and 3 N

    lets say the ratio was 9 N to 5 H (which it is NOT, this is an example) and you had 10 N atoms... to convert to H atoms you would take 10N * (5H/9N)... because 10N divided by 9N maked the N cancel, and * H leaves H on top.
    Then you talked about this I got confused, I know this was just an example but where did the 10 N atoms come from?
    kp2171's Avatar
    kp2171 Posts: 5,318, Reputation: 1612
    Uber Member
     
    #4

    Jan 28, 2006, 10:39 PM
    > Problem 3:
    > This is where I get stuck; you said that it was similar to problem 2

    > I have 5 H and 3 N


    >Quote:
    > lets say the ratio was 9 N to 5 H (which it is NOT, this is an example) and
    > you had 10 N atoms... to convert to H atoms you would take
    > 10N * (5H/9N)... because 10N divided by 9N maked the N cancel,
    > and * H leaves H on top.


    > Then you talked about this I got confused, I know this was just an example
    > but where did the 10 N atoms come from?


    it was as if I gave you a new problem and said you had 10 N atoms... I could've said you have 73946 N atoms. Point is you take the # N atoms, whatever the problem gives you, and you multiply by the ratio that you set up by looking at the formula.

    known # N atoms * ratio H atoms/N atoms = # H atoms

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