I'm in middle school too and I could help u but I have absolutely no idea what you are asking.
Li2O
Fe(NO3)3
Al2O3
CuCl2
ZnSO4
All you have to do here is make sure your charges are balanced when you write the compound. For example, Iron (III) has a +3 charge, and nitrate has a -1 charge. You need 3 nitrates to match that charge, hence Fe(NO3)3.
Answer:
q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.
Explanation:
The Specific Heat capacity of Lead is 0.128 
This means, increase in temperature of 1 gm of lead by
will require 0.128 J of heat.
Formula Used :

q = amount of heat added / removed
m = mass of substance in grams = 85.0 g
c = specific heat of the substance = 0.128
= Change in temperature
= final temperature - Initial temperature
= 10 - 200
= -
put value in formula
q = - 
On calculation,
q = - 2067.2 J
- sign indicates that the heat is released in the process
Answer: I don’t know lol
Explanation: I am so sorry I thought this was easy