Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
The work:
If it is an ideal gas:
Solving:
Replacing:
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
It is a covalent bond. Whenever a compound uses such suffixes like mono, di, tri, tetra, and so on, it is a covalent compound- thus having covalent bonds.
6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.
0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.
I’m guessing the answer is AX (the first one) sorry if I am wrong tho
Answer:
Hey, I hope this helps. You gave the equation already balanced so there was no need to do so, the next thing we need to do after balancing is to split the strong electrolytes into ions. Once that is completed we cross off any reoccurring ions. That leaves us with our complete net ionic.
I also recommend you check out Wayne Breslyn on Yt. He is so helpful with equations like these.
