An increase in the temperature will speed up the reaction by increasing the frequency and efficiency of the collisions of molecules.
<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.
The metallic pan iis most likely going to be used on a stove.
The stove is heating something, and the conductive metallic pan will, well, conduct that heat throughout the entire body of the pan. Doing this will spread the heat to the handle, burning your hands.
Both wood and plastic are insulators, and they do not conduct heat or electricity. They will insulate your hands and protect them from the heat.
The answer is C. 146g because you add all of the masses of the individual elements and then mulyiply by 1.72 to get your answer.
This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
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