Answer : The entropy change of reaction for 1.62 moles of 
reacts at standard condition is 217.68 J/K
Explanation :
The given balanced reaction is,
The expression used for entropy change of reaction 
is:
![\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BBr_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28Br_2%29%7D%2Bn_%7BF_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28F_2%29%7D%5D-%5Bn_%7BBrF_3%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28BrF_3%29%7D%5D)
where,
= entropy change of reaction = ?
n = number of moles
= standard entropy of formation
= 245.463 J/mol.K
= 202.78 J/mol.K
= 292.53 J/mol.K
Now put all the given values in this expression, we get:
![\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B1mole%5Ctimes%20%28245.463J%2FK.mole%29%2B3mole%5Ctimes%20%28202.78J%2FK.mole%29%7D%5D-%5B2mole%5Ctimes%20%28292.53J%2FK.mole%29%5D)
Now we have to calculate the entropy change of reaction for 1.62 moles of reacts at standard condition.
From the reaction we conclude that,
As, 2 moles of has entropy change = 268.74 J/K
So, 1.62 moles of has entropy change = 
Therefore, the entropy change of reaction for 1.62 moles of reacts at standard condition is 217.68 J/K
Other group because they weren’t in the right group
Answer:
Density changes with temperature because volume changes with temperature. Density is mass divided by volume. As you heat something up, the volume usually increases because the faster moving molecules are further apart. Since volume is in the denominator, increasing the volume decreases the density.
Explanation:
A. True.
Very true. The quicker or slower the reactants are used up the faster or slower the rate of reaction, and the faster or slower the products are formed, the faster or slower the rate of reaction.
0.01 cubic meters
Hope this helps
