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How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?

Respuesta :

hey...
Use the molarity formula
M=moles/L and then convert to grams

0.07268*0.15=moles
0.010902 mol
Pb(NO3)2
1 mole=331.22g
0.010902 moles=

3.61 g


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