Question

Is anyone willing to help me with chemistry calculations based on moles and molar mass?

Answer 1

ask your question don't ask to ask

Answer 2

and I can help you with such questions

Answer 3

https://docs.google.com/viewer?a=v&q=cache:qQWzmeJWXUQJ:www.npenn.org/557770125102149/lib/557770125102149/5.0%2520The%2520Mole/Mole%2520Concept%2520WS.doc+&hl=en&gl=us&pid=bl&srcid=ADGEESg-hM-LyfIxkqAOpOk0VlEIEdJbPFo_UxF1W0pptg8LXz4uJBbzEdrenGLQE4cIhU9_TGyiaB9PzG-5T-QubyB42eJ5PDyvd6trIfi5CrmJMcx_K0lAdlR5cHM4MGfSV0AbVslu&sig=AHIEtbS1O-vIflza9B_EGcdrSIj1_sGqrQ

Answer 4

are u able to see that^

Answer 5

I need help on the chart

Answer 6

?

Answer 7

sorry got distracted which problem are you having problems with?

Answer 8

I dont want to do your entire worksheet so first off you need to use the formula

mole = grams/molecular mass

Molecular mass can be found on the periodic table it has units grams/mole


Secondly,

\[1 mol = 6.022*10^{23} atoms\]

avogadro's number = 6.022*10^{23}

Answer 9

to find the molecular mass of a compound such as NaOH

you would add the molecular mass of Na + Oxygen + Hydrogen then sub that into the formula I provided you

Answer 10

that formula being

1mol = grams/molecular mass

Answer 11

to convert something from moles to atoms

multiply by

\[ \frac{6.022*10^{23}atoms}{1mol}\]

to convert something from atoms to moles

multiply by

\[ \frac{1mol}{6.022*10^{23}atoms}\]

Answer 12

hope that helps

Answer 13

\[mole = \frac{grams}{molecular mass}\]

Answer 14

if you have any questions feel free to ask

Answer 15

but I think I covered everything that is found on that worksheet

Answer 16

number of particles = atoms

Answer 17

sorry I should correct this

to find the molecular mass of a compound such as NaOH you would add the molecular mass of sodium + Oxygen + Hydrogen.

so
22.99g/mol + 15.9994g/mol + 1.0079g/mol = molecular mass of NaOH

then sub that into the formula I provided you, if you want to find grams or moles.

Answer 18

also next time post this problem in the chemistry section :)