Ile gramów chlorku srebra I powstanie w wyniku reakcji 0.5 mola chlorku wapnia z azotanem V srebra I ?
mAgCl=143,5u
CaCl2 + 2AgNO3---->2AgCl + Ca(NO3)2
1mol CaCl2-------287g AgCl
0,5mola CaCl2------xg AgCl
x = 143,5g AgCl
CaCl2 + 2AgNO3 -----> 2AgCl + Ca(NO3)2
1 mol CaCl2 ---- 287g AgCl
0,5 mola CaCl2 ---- X
X=143,5g AgCl
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mAgCl=143,5u
CaCl2 + 2AgNO3---->2AgCl + Ca(NO3)2
1mol CaCl2-------287g AgCl
0,5mola CaCl2------xg AgCl
x = 143,5g AgCl
CaCl2 + 2AgNO3 -----> 2AgCl + Ca(NO3)2
1 mol CaCl2 ---- 287g AgCl
0,5 mola CaCl2 ---- X
X=143,5g AgCl