Posted by liz on Tuesday, April 24, 2012 at 1:50pm.
how many sulfur atoms are in 2.57 g of iron (III) sulfate?
MgCO3.
mols O atoms = 1.58E24/6.02E23 = ?
mols MgCO3 = 1/3 of that.
g MgCO3 = mols MgCO3 x molar mass MgCO3.
mols Fe2(SO4)3 = grams/molar mass = ?
mols S atoms is 3x that.
1 mol S atoms will contain 6.02E23 atoms; therefore, 3 mols will be >>>>>>.
73.7g
1.16 x 10^22 sulfur atoms
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