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Ela100
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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!!!



Răspuns :

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