Question

Va rog e foarte urgent rămân corigent!!!!!

Answer 1

a).

m.apa = 180 g , md=? g

ms = md+m.apa

c= mdx100/ms => 18 = mdx100/(189+md)

=> 3402 + 18md = 100md

=> md = 41,49 g sare

facem proba:

c= 41,49x100/(189+41,49) = 18% , deci verifica calculul

b).

ms2 = ms1 - m.apa.evap (a)

=> a = ms2-ms1

ms2 = mdx100/c2 = 41,49x100/12 = 345,75

=> a = ms2-ms1 = 345,75-230,49 = 115,26 g apa

c).

c3 = 23% , md.final = ?g

md.final = 41,49 + b

ms.fin = 345,75 + b

=> 23 = (41,49+b)100/(345,75+b)

=> 4149 + 100b = 7952,25 + 23b

=> b = 49,40 g sar adaugata...

facem proba:

c = 90,89x100/395,15 = 23% , deci verifica

d).

regula dreptunghiului:

66% -------------------------> 6 p. sol. 66%

l                                    l

l                16%              l

l                                     l

10% --------------------------> 50 p. sol 10%

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

                     total = 56 p. sol. 16%

56 p. sol total ........ 6 p. sol 66% ......... 50 p. sol 10%

n ................................. m ........................... 500 g

m = 500x6/50 = 60 g sol. 66%

e).

ms.f = ms1 + a

c1 = 30% => md = ms1xc1/100 = 400x30/100

= 120 g

c2 = 24% => ms,f = mdx100/c2 = 120x100/24

= 500 g

m.apa adaugata = ms2-ms1 = 500-400 = 100 g apa

f).

ms1 = 80 g , c1 = 4%

=> md1 = 80x4/100 = 3,2 g

ms2 = 160 g , c2 = 40%

md2 = 160x40/100 = 64 g

a = X = ? , c3 = 33,538%

ms.f = ms1+ms2+a = 80x+160+a

md.f = md1+md2+a = 3,2x64+a

33,538 = (67,2+a)100/(240+a) =

=> 8049,12 + 33,538a = 6720 + 100a

=> a = X = 20 g