# 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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xxxxxxxx xxxxxxx molecular xxxxxx

xxxxxxxxx weight calculation:

xxxxx + xxxxxxxxxx x xxxxxxxxx + xxxxxxxxxxxx

Molar xxxx xx xxxxxxxxxxx x xxxxxxxxx xxxxx

xx xxxxxxxx gm of Lead(II) xxxxxxx x 207.2 xx

xx x xx of xxxxxxxx Acetate = xxxxxxxxxxxxxx

xx 0.9020 xxxxx xx xxxx x acetate xx xxxxx x 0.9020 = 0.5746 xx

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