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
The correct option in here is the first one: <span>pentane-1,5-diamine. This is also known as cadaverine. This is a common compund that produces that unpleasant odor from the fish. </span>
The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
b) law of conservation of energy
