how many grams of lead are in .9020 grams of lead 2 acetate
how many grams of lead are in .9020 grams of lead 2 acetate
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The mass xx lead contained xx xxxxxxxxxxxxxxx is:
xxxxx = w(Pb) x m(Pb(CH3COO)2)
xxxxx xxxxx xx xxx xxxx xxxxxxxx xx xxxx in xxxxxxxxxxxxxxxx xx xxx xxxxxxxxx it xx formula:
xxxxx = xxxxx x xxxxxxxxxxxxxxx
xx the xxxxxx of the xxxxx xxxxxx xx lead xxx xxxxxxxxxxxxxxxx xx read those xxxxxx from PSE and we have:
M(Pb) x xxxxx g/mol
xxxxxxxxxxxxxx x xxxxx x xxxxxxxxx x xxxxxx + 3*M(H)) = xxxxx x xxxxxx x 4*M(O) x 6*M(H) x xxxxxx x 4*12.01 + 4*15.99 x xxxxxxx g/mol x xxxxxx g/mol
xxxx we can calculate xxxxxx
w(Pb) = xxxxx x M(Pb(CH3COO)2)
xxxxx = 207.2 g/mol / xxxxxx xxxxx
w(Pb) = xxxxx x 325.29
xxxxx = 0.63697
xxxx xxx xxxx of xxxx is:
xxxxx x w(Pb) * xxxxxxxxxxxxxx
m(Pb) = 0.63697 * xxxxxx x
m(Pb) = xxxxxx g
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