Supstitucija

tame_1's picture

Koliko je atoma broma uvedeno supstitucijom u molekulu toulena ako je procenat broma 46,78% i ako je atomska masa za brom 80? Odgovor je 1

zlaja99's picture

W(Br)=46,78%=0.4678

x(Br)=?

Ar(Br)=80 g/mol

Mr(toluena)=92 g/mol

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x/(x+ m(toluena))=w(Br)

x/(x+92)=0.4678

x=0.4678(x+92)

x=0.4678x +43.037

x-0.4678=43.037

0.5322x=43.037

x≈80 g/mol-to je ekvivalentno Ar(Br)=80 g/mol,znaci 1 Br je supstituisan sa jednim H-atomom

Poz

zlaja99's picture

W(Br)=46,78%=0.4678

x(Br)=?

Ar(Br)=80 g/mol

Mr(toluena)=92 g/mol

----------------------------------

x/(x+ m(toluena))=w(Br)

x/(x+92)=0.4678

x=0.4678(x+92)

x=0.4678x +43.037

x-0.4678=43.037

0.5322x=43.037

x≈80 g/mol-to je ekvivalentno Ar(Br)=80 g/mol,znaci 1 Br je supstituisan sa jednim H-atomom

Poz