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Thread: Chemistry Problems

  1. #1 Chemistry Problems 
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    1) A certain gas has a volume of 31.9-L at a temperature of -10.2C. What would be the volume of this gas at a temperature of 17.8 C assuming the pressure remains constant?

    2) A certain gas has a volume of 4.64-L at a pressure of 355 mmHg. What would be the volume of this gas at pressure of 679 mmHg assuming that the temperature remained constant?

    3) What mass of hydrogen, H2, is contained in a 76.1-L tank at a pressure of 899 mmHg and a temperature of 16.2C?

    4) Determine the molar mass of a gas if 3.75-g occupis 727mL at 766mmHg and 35C.

    5) Calculate the density of H2S gas at 1.45 atm and 34C.

    6) The metabolic oxidation of glucose C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas:

    C6H12O6(aq) + 6O2(g) ->6CO2(g) + 6H2O(l)

    Calculate the volume of CO2 produced at a body temperature (37C) and 976 mmHg when 38.6g of glucose is consumed in this reaction?

    7) Calcium hydride, CaH2, reacts with water to form hydrogen gas:

    CaH2(s) + 2H2O(l) -> Ca(OH)2(aq) + 2H2(g)

    This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2 is desired. How many grams of CaH2 are needed to generate 53.5L of H2 gas if the pressure is 917mmHg at 18C?

    8 ) A certain gas has a volume of 31.5L at a temperature of 37C and a pressure of .954 atm. What would be the volume of this gas at STP?


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  3. #2  
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    It looks like most of those are simple PV=nRT problems. Isn't school out for the summer?


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  4. #3  
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    Which ones are PV=nRT and which ones aren't?
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  5. #4  
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    Actually, it's the forum's policy that we don't do students' homework. We'll help you out if you get stuck, but you have to show a good faith effort at doing it yourself.
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  6. #5  
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    Honestly, how do I work on the problems if I don't know what formulas to use? You tell me they are so easy. If that is true, then lend some help!
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  7. #6  
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    Okay, in the first problem you are going to use P1V1/T1=P2V2/T2. This is a form of the ideal gas law. Did you learn that?

    Pressure is constant so P1=P2, then V1/T1=V2/T2. Remember to convert the temperature to absolute. Work that one out, then we'll move on to the others.
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  8. #7  
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    Ok, I think I'm somewhere. I ask that you have pity on me and be specific with your corrections. Here is what I have so far:

    1)

    Vf=(Vi)(Tf)/Ti = (31.9L)(290.95K) / 262.95K

    35.3L

    2)

    (P1)(V1)/Pf = (335mmHg)(4.64L)/679mmHg = 2.28L

    3)

    PV = nRT

    899mmHg/760mmHg = 1.183 atm

    (1.183atm x 76.1 L) / (.0821 x 289.35K) = 3.789molH2 x 2.02g H2/1 mol = 7.655g

    4) Mm = mRT/PV

    766mmHg/760mmHg = 1.01atm

    ((3.75g)(.0821)(308.15K))/((1.01 atm)(.727L)) = 129g/mol

    5) d=(PMm)/RT

    (1.45 atm x 34.08g/mol) / (.0821 (L x atm / mol x K) x 307.15K)=

    1.96 g/L

    6) V = nRT/P

    976mmHg /760 mmHg = 1.28atm

    38.6g C6H12O6/180.15 C6H12O6=.2142mol C6H12O6

    (.2142 mol C6H12O6)x6 = 1.29 mol CO2

    V = (310.15 x 1.29 x .0821)/1.28atm = 25.46L

    7) V=nRT/P

    917mmHg/760= 1.20atm

    VP/RT =n

    (53.5L x 1.20 atm) / (.0821 x 291.15) = 2.68mol CaH2

    25g/mol CaH2 x 2.68 mol = 67g CaH2

    8 )

    V=nRT/P

    VP/RT = n

    (31.5L x .954 atm) / (.0821 x 310.15K) = 1.18mol

    V = (1.18mol x .0821 x 273.15K)/1atm = 26.462L
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  9. #8  
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    I didn't check all your math but it looks like you are on the right track, except I didn't follow what you did in problem 4. It looks like you converted a volume of 727 milliliters to a pressure and where did you get the 15.5 grams when the mass was 3.76 grams?

    In problem 5 google says 35C=308.15K not 307.15.
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  10. #9  
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    I have corrected my answer for number four and have corrected the question for number five which was supposed to be 34 degrees C.
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  11. #10  
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    Quote Originally Posted by donumabdeo
    I have corrected my answer for number four and have corrected the question for number five which was supposed to be 34 degrees C.
    Okay. Good job.
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  12. #11  
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    I appreciate your help.
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