Mieszanina chlorku srebra i bromku srebra zawiera 71.3% Ag . Oblicz procentową zawartość AgCl i AgBr w mieszaninie
AgCl + AgBr
mAg = 108g
mCl = 35,5g
mBr = 80g
m AgCl+AgBr=331,5g
216g - 100%
x g - 71,3%
x= 154,008g
mCl + Br= 105,5g
105,5g - 100%
z g- 28,7%
z=30,2785g
***
35,5 - 100%
x - (28,7% - y)
80g - 100%
30,2785g - x = y%
UKLAD ROWNAN:
35,5(28,7% - y)=100x
80y=100(30,2785g - x)
y=19,04%
x=15,232g
Ag - 71,3%
Br = 19,04% mBr=15,232g
Cl = 9,66% mCl=3,4293g
gdy mamy 35,5g przypada na 108 g Ag
a 3,4293 na x
x=10,4328
mAgCl=13,8621
gdy mamy 80g przypada na 108 g Ag
a 15,232 na x
x=20,5632
20,5632 + 13,8621 = 34,4253
I z tego liczymy procenty
20,5632 -x
34,4253- 100%
x=59,73%
100%- 59,73% = 40,27%
Odp: 40,27% AgCl oraz 59,73% AgBr
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AgCl + AgBr
mAg = 108g
mCl = 35,5g
mBr = 80g
m AgCl+AgBr=331,5g
216g - 100%
x g - 71,3%
x= 154,008g
mCl + Br= 105,5g
105,5g - 100%
z g- 28,7%
z=30,2785g
***
35,5 - 100%
x - (28,7% - y)
80g - 100%
30,2785g - x = y%
UKLAD ROWNAN:
35,5(28,7% - y)=100x
80y=100(30,2785g - x)
y=19,04%
x=15,232g
Ag - 71,3%
Br = 19,04% mBr=15,232g
Cl = 9,66% mCl=3,4293g
gdy mamy 35,5g przypada na 108 g Ag
a 3,4293 na x
x=10,4328
mAgCl=13,8621
gdy mamy 80g przypada na 108 g Ag
a 15,232 na x
x=20,5632
20,5632 + 13,8621 = 34,4253
I z tego liczymy procenty
20,5632 -x
34,4253- 100%
x=59,73%
100%- 59,73% = 40,27%
Odp: 40,27% AgCl oraz 59,73% AgBr