Saturday
May 25, 2013

# Homework Help: Chemistry

Posted by jecce. on Wednesday, July 30, 2008 at 8:41pm.

im curious if i post my work on here will anyone check it for me to see if it is right?

• Chemistry - DrBob222, Wednesday, July 30, 2008 at 8:44pm

Sure. One question per post. If you give us the problem and your answer we can give you a yes or no. If you show your work we can tell you where you are going wrong if the answer is wrong.

• Chemistry - jecce., Wednesday, July 30, 2008 at 8:56pm

ok well i had to find the molar mass of these substances.

A) NH3 = 17.00 grams
B) N2H4 = 32.00 grams
C) (NH4)2 Cr2O7 = 776 grams

For the second part i had to find how many moles of compounds are present in 1.00 grams of each substance i found above.

A) i did 1.00g x (1 mole)/(17.00g) and got 5.9x10^-2

B) i did 1.00g x (1 mole)/(32.00g) and got 3.1x10^/2

C) i did 1.00g x (1 mole)/(776g) and got 12.9x10^-4

The second part finding the moles of the compounds confused me. So if you could please check my work and answer and explain why its wrong that would help me a lot.

Thanks!

• Chemistry - jecce., Wednesday, July 30, 2008 at 8:57pm

this is what its suppose to be.

B) i did 1.00g x (1 mole)/(32.00g) and got 3.1x10^-2

• Chemistry - bobpursley, Wednesday, July 30, 2008 at 9:02pm

C and C are wrong. The molar mass of ammonium dichromate is about a third of what you used. Recheck that.

• Chemistry - jecce., Wednesday, July 30, 2008 at 9:11pm

can you tell me how to add that?
because what im doing is

(NH4)2 Cr2)7

N= 14.00
H= 4x 1.00 = 4

so..
18.00 x 2 = 36.00

and
Cr= 2x 52 = 104
O= 7X 16 = 112

which equals to 216 and then i add 36 and get 776.

• Chemistry - jecce., Wednesday, July 30, 2008 at 9:11pm

can you tell me how to add that?
because what im doing is

(NH4)2 Cr2O7

• Chemistry - bobpursley, Wednesday, July 30, 2008 at 9:16pm

Hmmmm. Here in Texas adding 216 plus 36 is considerably less than 776.

• Chemistry - jecce., Wednesday, July 30, 2008 at 9:19pm

whoa.

i must have put that wrong in the calculator!

thanksss!

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