Answer:
where are those two images which you have sent
This problem is about conversion and dimensional analysis. Important information to know:
1 atm = 760 torr = 101.325 kPa
For atm to torr conversion:
0.875 atm * (760 torr / 1 atm) = 665 torr
For atm to kPa conversion:
0.875 atm * (101.325 kPa / atm) = 88.7 kPa
Thus the answer is b) 665 torr and 88.7 kPa
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
Answer:
Calculating Atomic Mass
Change each percent abundance into decimal form by dividing by 100. Multiply this value by the atomic mass of that isotope. Add together for each isotope to get the average atomic mass.
Explanation:
have a nice day
We cannot see it its so small
Also yes it matches your hypothesis
