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.
<h3>
<u>Answer</u></h3>
4. loses 2 electrons
<h3>
<u>EXPLANATION</u></h3>
Mg²+ has 10 electrons while Mg has 12 electrons.
Actually Rb or Rubidium in zero state has the following
electron configuration:
<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>
However we can see that the ion has a 1 positive charge,
which means that it lacks 1 electron, therefore the answer from the choices is:
<span>d.
rb+: 1s22s22p63s23p64s23d104p6</span>