Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is


= 0.0385 mol
Mass of PbO needed is

After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is

After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.
Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
Answer:
The three laws of Chemical Reaction are .The law of constant proportions. The law of multiple proportions. The law of reciprocal proportions.
A chemical compound is always found to be made up of the same elements combined together in the same fixed proportion by mass.
potassium and chlorine gas ---> chloride.
Hope this helps, have a good day.✌
B and D is out. It cant be A because heat of combustion is substance not compound. So the answer is D.