N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹
n(Mg)=N/Nₐ
m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ
m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g
The question has missing information! But, with the objects and their respectives masses and specific coefficiets and with the following explanation, the question can be aswered!
Answer:
The one which has the highest product of the mass by specific heat.
Explanation:
The heat is a way of "energy in motion", which means that is a way that energy can be exchanged between substances. This exchange occurs when there is a difference in the temperature of the substances that are in contact, and then, to achieve thermal equilibrium, one substance loses heat (the one with higher temperature) and had a decrease in temperature, and the other gain heat and had an increase of temperature.
The heat loss is the same heat gained, and it can be calculated. If the substance doesn't suffer a physical change (which is the case here), then:
Q = m*c*ΔT
Where Q is the heat, c is the specific heat, and ΔT is the temperature change. So, rearranged the expression:
ΔT = Q/(m*c)
Because Q is the same for all of the objects, the one that will experience the smallest temperature change is the one that has the highest value of m*c (because it is indirectly proportional to ΔT).
Answer:
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I hope it helps you although
<u>Answer:</u> The 
for the reaction is 72 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 given chemical reaction follows:
The intermediate balanced chemical reaction are:
(1) 
(2) 
( × 2)
(3) 
( × 2)
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%28%5CDelta%20H_1%29%5D%2B%5B2%5Ctimes%20%28-%5CDelta%20H_2%29%5D%2B%5B2%5Ctimes%20%28%5CDelta%20H_3%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-1184%29%29%2B%282%5Ctimes%20-%28-234%29%29%2B%282%5Ctimes%20%28394%29%29%5D%3D72kJ)
Hence, the for the reaction is 72 kJ.
