Chemistry
8 Online
OpenStudy (toxicsugar22):
How many moles are in 5550 particles of sodium
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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
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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 \(9.21*10^{-25}\)
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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
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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
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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.
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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