Ask your own question, for FREE!
Chemistry 8 Online
OpenStudy (toxicsugar22):

How many moles are in 5550 particles of sodium

OpenStudy (aaronq):

\(n=\dfrac{N}{N_A}\) n=moles, N=number of particles, \(N_A\)=avogadros number

OpenStudy (aaronq):

why are you multiplying? just sub your values in

OpenStudy (aaronq):

why are you subtracting?

OpenStudy (aaronq):

\(n=\dfrac{5550}{6.022*10^{23}}\)

OpenStudy (aaronq):

particles go on the top

OpenStudy (aaronq):

thats not right

OpenStudy (aaronq):

divide it 5500 divided by 6.022*10^23

OpenStudy (aaronq):

okay, 5550 divided by 6.022*10^23

OpenStudy (toxicsugar22):

I am doing 5550 over 6.02 times 10^23

OpenStudy (aaronq):

the exponent is -ve as in \(9.21*10^{-25}\)

OpenStudy (toxicsugar22):

Ohh okay 9.21 * 10^-25

OpenStudy (toxicsugar22):

That is ansewer

OpenStudy (aaronq):

yup

OpenStudy (toxicsugar22):

Can you help with 1 pr 2 more please

OpenStudy (toxicsugar22):

The mass of 763 mole sodium chloride formula units NaCl is

OpenStudy (aaronq):

use the same formula, solve for mass this time

OpenStudy (toxicsugar22):

Is is 763 times 6.02 * 10^23

OpenStudy (toxicsugar22):

Divide by mass of nacl

OpenStudy (aaronq):

yeah, no division though

OpenStudy (aaronq):

yeah looks good

OpenStudy (toxicsugar22):

Is it 4.00*10^-4 times by 6.02 *10^23 Divide by 70.906

OpenStudy (aaronq):

yep

OpenStudy (toxicsugar22):

So it is 3.39 * 10^18

OpenStudy (toxicsugar22):

3.3960454*10^18

OpenStudy (aaronq):

i don't know i'm not doing the arithmetic, myself.

OpenStudy (toxicsugar22):

Can you do it and tell me if this is right

OpenStudy (aaronq):

yep it's right

OpenStudy (toxicsugar22):

So is it 3.39*10^18 or is it 3.40 *10^18 pr does not matter

OpenStudy (toxicsugar22):

Does it matter

OpenStudy (aaronq):

it depends if they want a specific number of sig figs

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!