How many grams are in 2.7 x 10^24 molecules of copper (I) sulfate
Favorites|Homepage
Subscriptions | sitemap
HOME > > How many grams are in 2.7 x 10^24 molecules of copper (I) sulfate

How many grams are in 2.7 x 10^24 molecules of copper (I) sulfate

[From: ] [author: ] [Date: 12-01-05] [Hit: ]
Cu = 63.CuSO4 = 63.5 + 32 + 16*4 = 159.2.7*10^24 molecules * 159.5 g CuSO4/mole * 1 mole/6.......
Chem help needed! Please explain how to do the problemm!

-
A mole is 6.02*10^23 molecules.
The molar weight of a compound is the sum of the atomic weights
Copper sulfate is CuSO4
Atomic weights
Cu = 63.5
S = 32
O = 16
CuSO4 = 63.5 + 32 + 16*4 = 159.5

2.7*10^24 molecules * 159.5 g CuSO4/mole * 1 mole/6.02*10^23 molecules = (2.7 * 159.5 * 10)/6.02 = 715.37 grams
1
keywords: grams,of,are,copper,10,2.7,How,many,in,sulfate,molecules,24,How many grams are in 2.7 x 10^24 molecules of copper (I) sulfate
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .