A) The image is a drawing of the sun
B) Energy (light) from the sun
C) carbon dioxide
D) water
E) glucose
F) oxygen
Remember that the equation for photosynthesis is:
carbon dioxide + water -> (light energy) -> glucose + oxygen
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
There are 1,000m is 1k. So just move the decimal one position right. 127.56m
There are 10,000cm in 1k. Move the decimal two positions right. 1275.6cm
I think that would be the third one
Answer:
6. IDK, google only says glacial erratics
7. Deposition
8. Building landfills, I think
Explanation:
