Anyone who's really good at chemistry!! Study help.?

Question

I took my chemistry class in order to challenge myself mentally..and now I'm struggling. My question has to do with 4 questions on my study sheet that I really don't know how to do, and I'm going to need it is my guide for the next test. I would really appreciate it if someone could help me out with this if they get it. I don't know the answers, and although I have a kinda-sorta idea of how to get them, I don't have any answers to check it with. Care to try, anyone?

How many moles contain each of the following?
1.) 3.75 x 10^24 molecules CO2
2.) 3.58 x 10^23 formula unites ZNCl2

Determine the number of moles present:
a.) 22.6g AgNO3
b.) 6.50g HCL

thank you!

Answer 1

1. 1 mole of CO2 has 6.02 X10^23 atoms. So now you have two numbers, 6.02 X10^23 and 3.75 X10^24, move the decimal point to the right once so now you have 37.5 X10^23 and now you can divide that by 6.02 X 10^23 and you get 6.23 moles of CO2.

2. Same thing with the first one except that the exponent is the same, so all you do is divide 3.58 X 10^23 (1 mole of whatever is always 6.02 X 10^23 formula units) by 6.02 X 10^23 and you get .595 moles.

a. Find the mass of AgNO3, which is 107.87(Ag) + 14(N) + 16x3(3 moles of O in that compound) and you get 169.87g and divide 22.6 by 169.87 to get .133 moles of AgNO3

b. Same as previous one, HCl is 36.5g, divide 6.50g by 36.5g and you get .178 moles of HCl.

Answer 2

Avogados no. = 6.02252 x 10^23 per mole
this is the no. of atoms or molecules in a mole.
so no moles CO2= 6.02252 x 10^23/ 3.75 x 10^24
simillar calc for ZnCl2

a) you need the molecular weight for each of these Ag =107.87, O=16, N=14 so AgNO3=169.9 so no. moles = 22.6 / 169.9
b) H = 1.008, Cl = 35.45 no moles 6.5 /36.45

Answer 3

1.) 3.75x10^24moleculesCO2 x 1molCO2/6.02x10^23moleculesCO2 = 0.623 x 10^10 = 6.23 moles CO2

3.58x10^23FUZnCl2 x 1molZnCl2/6.02x10^23FUZnCl2 = 0.595 moles ZnCl2

2.) Atomic weights: Ag=108 N=14 O=16 AgNO3=170 H=1 Cl=35.5 HCl=36.5

a.) 22.6gAgNO3 x 1molAgNO3/170gAgNO3 = 0.133 moles AgNO3

b.) 6.50gHCl x 1molHCl/36.5gHCl = 0.178 moles HCl