Question

cum aflu ms 1 si ms 2 a 2 substante de c 15% respectiv 35% care au fost combinate in ms 3 de 220 g si c 20%

Answer 1

Răspuns:

c,f= (md1+md2)x100/(ms1+ms2)

ms1+ms2=220g

c,final= 20%  

c1= 15%===> md1= c1xms1/100= 0,15ms1

c2= 35%====> md2= c2xms2/100= 0,35ms2

20=(0,15ms1+0,35ms2)X100/220

44= 0,15ms1+0,35ms2

dar ms1= 220-ms2

deci  44= 0,15(220-ms2)+ 0,35ms2

ms2=55g

ms1= 165g

m

Explicație: