Asked by aayushi
x gm og Ag was dissolved in HNO3 and the solution was treated with excess of NaCl when 2.87 g of AgCl was precipitated. find the value of x?
Answers
Answered by
Ayush
2.16 gm
Answered by
shabaz ahmed dar (vkv baragolai)
143.5 g AgCl contains Ag = 108
2.87 g Agcl contains Ag = (108×2.87)/143.5
Answer= 2.16 g
Thanks :) with regards from SHABAZ AHMED DAR (VIVEKANANDA KENDRA VIDYALAYA, BARAGOLAI, ASSAM)
2.87 g Agcl contains Ag = (108×2.87)/143.5
Answer= 2.16 g
Thanks :) with regards from SHABAZ AHMED DAR (VIVEKANANDA KENDRA VIDYALAYA, BARAGOLAI, ASSAM)
Answered by
LAD PRATHAMKUMAR RAKESHBHAI
The molecular mass of AgCl is 143.37......
143.37g AgCl contains 108 Ag.
So , 2.87 g AgCl contains (?)
then,
(2.87×108)/143.37
= 2.16
THANKS ................
143.37g AgCl contains 108 Ag.
So , 2.87 g AgCl contains (?)
then,
(2.87×108)/143.37
= 2.16
THANKS ................
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