Question

Am si eu nevoie de ajutor . Va rog.​

Answer 1

1)

Datele problemei

c=20%

msol = 400g

md = ? (y)

mH2O = ? (x)

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c=md*100/msol => md = c*msol/100 = 20*400/100 = 80g => y = 80g

msol = x + y => x = msol - y = 400 - 80 = 320g H2O

2)

Datele problemei:

msol1 = 200g

c1 = 6%

c2 = 4%

mH2O adaugata = ?

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md ramane constant!

c1 = md*100/msol1 => md = c1*msol1/100 = 6*200/100 = 12g

c2=md*100/msol2 => msol2 = md*100/c = 12*100/4 = 300g

mH2O adaugat = msol2 - msol1 = 300 - 200 = 100g

3)

Datele problemei:

msol1 = 400g

c1=15%

mH2O evaporat = 100g

c2=?

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

md ramane constant!

c1=md*100/msol1 => md = c1*msol1/100 = 15*400/100 = 60g

msol2 = msol1 - mH2O evaporat = 400 - 100 = 300g

c2 = md*100/msol2 = 60*100/300 = 20%

4)

Datele problemei:

mzahar adaugat = 30g

msol1 = 200g

c1 = 10%

c2 = ?

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c1=md1*100/msol1 => md1 = c1*msol1/100 = 10*200/100 = 20g

md2 = md1 + mzahar adaugat = 20 + 30 = 50g

msol2 = msol1 + mzahar adaugat = 200+30 = 230g

c2 = md2*100/msol2 = 50*100/230 = 21,74%

Ma folosesc de formula concentratiei procentuale, c = md*100/msol la toate!