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Submitted by lina.beanss on Sat, 2012-02-11 02:48
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# 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

Submitted by Coloratus on Sat, 2012-02-11 10:50
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xxx xxxx of xxxx contained xx lead(II)acetate is:

m(Pb) = xxxxx * xxxxxxxxxxxxxx

where w(Pb) is the mass fraction of lead in xxxxxxxxxxxxxxxx xx xxx calculate xx xx xxxxxxxx

xxxxx x M(Pb) / xxxxxxxxxxxxxxx

or the xxxxxx xx the xxxxx xxxxxx of lead and lead(II)acetate. We xxxx xxxxx values xxxx xxx xxx we xxxxx

M(Pb) x xxxxx g/mol

xxxxxxxxxxxxxx = xxxxx + xxxxxxxxx + 2*M(O) + xxxxxxx x xxxxx + xxxxxx x 4*M(O) x xxxxxx x [207.2 x 4*12.01 x xxxxxxx + 6*1.01] xxxxx x xxxxxx xxxxx

xxxx xx can xxxxxxxxx xxxxxx

xxxxx x xxxxx / M(Pb(CH3COO)2)

w(Pb) x 207.2 g/mol / xxxxxx xxxxx

w(Pb) x 207.2 / 325.29

w(Pb) x xxxxxxx

xxxx xxx xxxx of lead is:

xxxxx x xxxxx x m(Pb(CH3COO)2)

xxxxx x 0.63697 x 0.9020 g

m(Pb) = xxxxxx x

Submitted by shahimermaid on Sat, 2012-02-11 08:53
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Molecular weight calculation:

xxxxx x (12.0107*2 + xxxxxxxxx x xxxxxxxxxxxx

xxxxx mass xx xxxxxxxxxxx = xxxxxxxxx g/mol

xx xxxxxxxx gm xx xxxxxxxx xxxxxxx = 207.2 xx

xx 1 xx xx xxxxxxxx Acetate x 207.2/325.2880

xx xxxxxx grams of lead 2 acetate xx xxxxx x xxxxxx x xxxxxx xx