anonymous
  • anonymous
help!! how do i solve for n in: (.9985atm)(.02L)= (0.082 atm*L/mole*K)(296K) sooo stumped!!!
Chemistry
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
aaronq
  • aaronq
(.9985atm)(.02L)= n (0.082 atm*L/mole*K)(296K) n=[(0.082 atm*L/mole*K)(296K)]/[(.9985atm)(.02L)]
anonymous
  • anonymous
i cant get a definite number/figure for n tho? i dont understand how to caclulate 0.082 atm*L/mole*K or what it means
chmvijay
  • chmvijay
its PV =nRT n=PV/RT = .9985atm *02L / 0.082 atm*L/mole*K *296K = find it atm*L/mole*K its just unit of gas constant don't worry about it

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chmvijay
  • chmvijay
@missy101 got it ???:)
chmvijay
  • chmvijay
@aaaronq is wrong !!!
aaronq
  • aaronq
my baddd i copied and pasted backwards
chmvijay
  • chmvijay
its all right it happens some times !!!:)
anonymous
  • anonymous
when you go out on a limb, you sometimes fall off the tree. it's really minor anyway.
toxicsugar22
  • toxicsugar22
CAN YOU PLEAE HELP ME
anonymous
  • anonymous
hmmm ok so theres no specific answer? i was told i needed to find one. . . dont see how its posible :/
anonymous
  • anonymous
if you want the numbers, give me a sec.
chmvijay
  • chmvijay
8.2 10^-4 is answer LOL:)
anonymous
  • anonymous
you want the number of gas particles in moles, or n. using the ideal gas law: n = P V / (R T) putting everything in atm, L, and K: n = (.9985 * .02) / (0.082 * 296) mol it comes out to be 8.2 x 10(-4) mol, just like chmvijay posted

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