1. Nickel (II) Bromide
2. Iron (II) Oxide
3. Iron (III) Oxide
4. Tin (IV) Chloride
5. Lead (IV) tetrachloride
6. Tin (II) Bromide
7. Chromium (III) Phosphide
8. Iron (II) Fluoride
9. Gold (III) Chloride
I hope this helps. I'm more than 100% sure that all the answers except for number 7 are correct. I knew all of them off the top of my head except for this one. I hope the other answer has the correct answer for that one. Good luck and have a great day.
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
Answer:
if 12 moles are produced then it came 24 moles of Al(OH)3
Explanation:
