Question

Peste 400cm³ solutie de hidoxid de sodiu de c=40%(densitatea=1,437g/cm³) se adauga 80cm³ solutie de hidroxid de c=20%(densitatea=1,225g/cm³).Concentratiafinala a solutiei va fiȘ
a)27,39%
b)45%
c)32,5%
Datele problemei densitatea= m/v 
v=400cm³
c₁=40%
densitatea=1,437g/cm³

v=80cm³
c₂20%
densitatea=1,224g/cm³

c=?

Va rog faceti-mi aceasta problema,repede!!!

Answer 1

c1=40%
V1=400$cm^{3}$
ρ1=1,437$\frac{g}{ cm^{3} }$
ρ=$\frac{m}{V}$=1,437$\frac{g}{ cm^{3} }$
ms1=1,437×400=574,8g
c2=20%
V2=80$cm^{3}$
ρ2=1,225$\frac{g}{ cm^{3} }$
ρ=$\frac{m}{V}$=1,225$\frac{g}{ cm^{3} }$⇒ms2=1,225×80=98g
c=$\frac{md}{ms}$×100⇒40=$\frac{md1}{574,8g}$×100=$\frac{md}{5,748g}$⇒md1=5,748×40=229,92g
c$\frac{md}{ms}$×100⇒20=$\frac{md2}{98g}$×100=$\frac{md2}{49g}$×50⇒50md=980g⇒md2=980:50=19,6g
$c_{finala}$=$\frac{ md_{1}+ md_{2} }{ ms_{1}+ ms_{2} }$×100=$\frac{229,92g+19,6g}{574,8g+98g}$×100=$\frac{249,52g}{672,8g}$×100=0,37×100=37%