<u>Answer:</u> The 
for the reaction is -1406.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical reaction for the formation reaction of 
is:
The intermediate balanced chemical reaction are:
(1) 
( × 6)
(2) 
( × 3)
(3) 
( × 2)
(4) 
The expression for enthalpy of formation of is,
![\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bformation%7D%3D%5B6%5Ctimes%20%5CDelta%20H_1%5D%2B%5B3%5Ctimes%20%5CDelta%20H_2%5D%2B%5B2%5Ctimes%20%5CDelta%20H_3%5D%2B%5B1%5Ctimes%20%5CDelta%20H_4%5D)
Putting values in above equation, we get:
![\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bformation%7D%3D%5B%28-74.8%5Ctimes%206%29%2B%28-185%5Ctimes%203%29%2B%28323%5Ctimes%202%29%2B%28-1049%5Ctimes%201%29%5D%3D-1406.8kJ)
Hence, the for the reaction is -1406.8 kJ.
Remark
The balance numbers in front of Ag and AgNO3 are both 2. That number is in moles.
Rule: if the moles are the same in the equation, then whatever you are given for one, will be the same for the other. So you have 0.854 moles of Ag. You will also have 0.854 moles of AgNO3
Answer: 0.854 <<<<<
Wax is susceptible to heat. Wax is responds to heat addition. The forces in the wax when heat is added are being broken off and are much lesser as its original state. Hope this answers the question. Have a nice day.<span />
Answer:
A beaker
Step-by-step explanation:
Specifically, I would use a 250 mL graduated beaker.
A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.
You don't need precisely 100 mL solution.
If the beaker is graduated, you can easily measure 100 mL of the stock solution.
Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).
