77 902
Assignments Done
Successfully Done
In August 2019

Answer to Question #17306 in Inorganic Chemistry for Manolo Javier

Question #17306
Solid Iodine is added to a strongly basic solution of mercury (II) nitrate. What are the products and how do you solve it?
Expert's answer
If ahighly alkaline solution add-crystal iodine, the following reactions:

3I2 + 6KOH = 5KI + KIO3+ 3H2O

Next KI reacts with nitrate of mercury to form a yellow precipitate:
Hg(NO3)2 + 2KI = HgI2 + 2 KNO3

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be first!

Leave a comment

Ask Your question

Privacy policy Terms and Conditions