Oblicz ile gramów kwasu siarkowego (VI) znajduje się w 0,5 roztworu o stężeniu 59,7%. Gęstość wody wynosi 1,5.
V = 0,5dm³ = 500cm³
d = 1,5g/cm³
m = dV
m = 1,5g/cm³ x 500cm³ = 750g
cp = 59,7%
ms = mr x cp/100%
ms = 750g x 59,7%/100% = 447,75g
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V = 0,5dm³ = 500cm³
d = 1,5g/cm³
m = dV
m = 1,5g/cm³ x 500cm³ = 750g
cp = 59,7%
ms = mr x cp/100%
ms = 750g x 59,7%/100% = 447,75g
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)