Answer:

Explanation:
Hello!
In this case, since the definition of entropy in a random mixture is:
![\Delta S=-n_TR\Sigma[x_i*ln(x_i)]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-n_TR%5CSigma%5Bx_i%2Aln%28x_i%29%5D)
For this silver-gold mixture we write:
![\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-%28n_%7BAu%7D%2Bn_%7BAg%7D%29R%5CSigma%5B%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%2B%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%5D)
By knowing the moles of gold:

It is possible to write the aforementioned formula in terms of the variable
representing the moles of silver:
![20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]](https://tex.z-dn.net/?f=20%5Cfrac%7BJ%7D%7Bmol%7D%3D-%280.508%2Bx%298.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%5CSigma%5B%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%29%2B%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%29%5D)
Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

So the mass is:

Best regards!
This is a double replacement because both Copper and Sodium reacted with each other compounds , hence "double"
Argon, it's a noble gas in the last group on the periodic table.
Explanation:
2-metylbutene should be the profuct
Answer:
See explanation
Explanation:
Calcium is divalent. This means that it donates two electrons during ionic bond formation. Since chlorine atom can only accept one electron during ionic bond formation, two chlorine atoms must accept the two electrons donated by calcium.
For this purpose, each time CaCl2 is formed, there must be two chlorine atoms for each calcium atom.