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.
Answer: has properties similar to other elements in group 18, does not react readily with other elements, is part of the noble gas group
Explanation: I’ve done on edg before
Answer : The balanced chemical equation is,
Explanation :
Rules for the balanced chemical equation in basic solution are :
- First we have to write into the two half-reactions.
- Now balance the main atoms in the reaction.
- Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
- If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
- If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion
at that side where the less number of hydrogen are present. - Now balance the charge.
The half reactions in the basic solution are :
Reduction : 
......(1)
Oxidation : 
.......(2)
Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.
The balanced chemical equation in a basic solution will be,
Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol
Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.
W are given a chemical reaction:
To calculate the enthalpy change, we use the formula:
This is the amount of energy released when 0.1326 grams of sample was burned.
So, energy released when 1 gram of sample was burned is = 
Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:
