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
7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.
Explanation:
Data given:
volume of phosphoric acid, Vacid =60 ml
volume of barium hydroxide, Vbase = 225 ml
molarity of barium hydroxide, Mbase = 2M
Molarity of phosphoric acid, Macid =?
the formula for titration is used as:
Macid x Vacid = Mbase x Vbase
rearranging the equation to get Macid
Macid = 
Macid =
Macid = 7.5 M
the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.