Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction
.

![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
Distance travelled will be equal to displacement when the line drawn is completely straight.
Hopefully, this helps.
If you are in the zone, you are happy or excited because you are doing something very skilfully and easily.
No sorry because i'm only 12 years old so no
There is a missing portion of this question which shows the reaction that needs balancing:
"In a balanced equation, the same number of each kind of atom is shown on each side of the equation. Calculate the number of iron (Fe), oxygen (O), and carbon atoms (C).
Fe2O3+ 3CO --> 2Fe + 3CO<span>2
</span><span>Based on these values, is the equation balanced?</span><span>"
</span>
To check if this equation is balanced we simply compare the number of each element on each side of the equation.
On the reactant side of the equation we have:
2 Fe atoms
6 O atoms
3 C atoms
On the product side of the equation we have:
2 Fe atoms
6 O atoms
3 C atoms
Therefore, both side of the reaction have the correct and equal number of each atom, so the equation is balanced.