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.
because it has ns1 electron configuration like the alkali metals
A metalloid is a metal and a nonmetal
You first need to write the balanced chemical reaction for what is going on.
Ca(OH)₂+2HCl→2H₂O+CaCl₂
After you make the balanced chemical reaction, First you find the moles of HCl used. To do this multiply 0.0375L by 0.124M to get 0.00465mol HCl. Then you multiply 0.00465mol HCl by (1mol Ca(OH)₂)/(2mol HCl) to get 0.002325mol Ca(OH)₂. Finally to find concentration of Ca(OH)₂ used you divide 0.002325mol by 0.020L to get 0.116M Ca(OH)₂.
Therefore the concentration of the unknown solution of Ca(OH)₂ was 0.116M.
I hope this helps. Let me know if anything is unclear.
