OpenStudy (anonymous):

How many grams are there in 9.4 X 1025 molecules of hydrogen gas, H2? How do I solve this I need step ppl! @Frostbite anychance u can help me?

4 years ago
OpenStudy (frostbite):

You wanna give it a try or totally clueless?

4 years ago
OpenStudy (anonymous):

totally clueless

4 years ago
OpenStudy (anonymous):

I don't even remember learning this :(

4 years ago
OpenStudy (frostbite):

All we are going to use is good old \[\Large m=M \times n\]And I'm going to introduce you to a new equation: \[\Large N=N_{A} \times n\] Where \(N\) is the amount of molecules and \(N_{A}\) is the Avogadro constant. Think you can do it now? :)

4 years ago
OpenStudy (anonymous):

moles=mass *n? whats n? Avogadros number is 6.02*10^23 I believe

4 years ago
OpenStudy (frostbite):

"moles=mass *n" not sure I follow. n is the amount of moles.

4 years ago
OpenStudy (anonymous):

ooooo so moles=mass*(6.02*10^23)

4 years ago
OpenStudy (anonymous):

i am confused

4 years ago
OpenStudy (frostbite):

Not mass. Amount of particles = amount of moles * Avogadro's constant. :)

4 years ago
OpenStudy (anonymous):

eh still not really clear

4 years ago
OpenStudy (anonymous):

amount of particles? I thought I needed grams

4 years ago
OpenStudy (frostbite):

I know, but we need to find \(n\) the amount of substance in mol first, and we do so by using the equation I wrote.

4 years ago
OpenStudy (anonymous):

ooo wait so would it be amount of particles=9.4*10^25

4 years ago
OpenStudy (frostbite):

Exactly! :)

4 years ago
OpenStudy (anonymous):

so how do I get grams from amount of particles?

4 years ago
OpenStudy (anonymous):

uh can u show me this one, im lagging a bit

4 years ago
OpenStudy (frostbite):

Sure :) We start by finding the amount of substance: \[\Large N=n \times N _{A} \to n=\frac{ N }{ N _{A} }\] Now that we know the amount of substance we put in into the equation: \[\Large n=\frac{ m }{ M } \to m=n \times M=\frac{ N }{ N _{A} }\times M\] So: \[\Large m=\frac{ 9.4 \times 10^{25} }{ 6.02214×10^{23} ~ \frac{ 1 }{ mol } } \times 2.01588 \frac{ g }{ mol }\]

4 years ago
OpenStudy (anonymous):

312?

4 years ago
OpenStudy (frostbite):

I haven't calculated it, but it should be correct what I have written.

4 years ago
OpenStudy (anonymous):

I know but I had to round it :)

4 years ago
OpenStudy (anonymous):

so im guessing it's probably right, right?

4 years ago
OpenStudy (frostbite):

312 is what you got when you typed in what I had written and round it?

4 years ago
OpenStudy (frostbite):

I get 314.68 g

4 years ago
OpenStudy (anonymous):

156 * 2=312

4 years ago
OpenStudy (frostbite):
4 years ago

OpenStudy (anonymous):

o what about sig digits?

4 years ago
OpenStudy (frostbite):

lol they only give us two. So if we should be good students we should only use two as well and say: \[\large m=3.1 \times 10^{2} ~ \sf g\]

4 years ago
OpenStudy (frostbite):

But does my calculations makes sense to you?

4 years ago
OpenStudy (anonymous):

yes very much, but I have trouble remembering the steps sometimes

4 years ago
OpenStudy (anonymous):

ty so much!!!!!!!!!

4 years ago
OpenStudy (frostbite):

You are more than welcome anytime.

4 years ago