Answer:
65.14 g
Explanation:
The change in the entropy of two ,metals on mixing is
![\delta S= R[(n_{cu}+n_{Ni})ln(n_{cu}+n_{Ni})-n_{cu}lnn_{cu}-n_{Ni}lnn_{Ni}]](https://tex.z-dn.net/?f=%5Cdelta%20S%3D%20R%5B%28n_%7Bcu%7D%2Bn_%7BNi%7D%29ln%28n_%7Bcu%7D%2Bn_%7BNi%7D%29-n_%7Bcu%7Dlnn_%7Bcu%7D-n_%7BNi%7Dlnn_%7BNi%7D%5D)
Here, R is real gas constant, n_{cu} is the number of moles of the copper, and n_{Ni} is the number of moles of nickel
Thus the number of moles of Nickel is
![n_{Ni}=\frac{m_{Ni}}{M_{Ni}}](https://tex.z-dn.net/?f=n_%7BNi%7D%3D%5Cfrac%7Bm_%7BNi%7D%7D%7BM_%7BNi%7D%7D)
=1.7 moles
now put n_{Ni} = 1.7 ΔS= 15,
R= 8.314 and solve
![15= R[(n_{cu}+1.7)ln(n_{cu}+1.7)-n_{cu}lnn_{cu}-1.7ln1.7]](https://tex.z-dn.net/?f=15%3D%20R%5B%28n_%7Bcu%7D%2B1.7%29ln%28n_%7Bcu%7D%2B1.7%29-n_%7Bcu%7Dlnn_%7Bcu%7D-1.7ln1.7%5D)
On solving the number of moles of copper in the mixture
n_cu= 1.025
Thus the mass of copper is = n_cu×M_cu = 1.025×63.55 = 65.14 g
Therefore the required mass is = 65.14 g
Answer:
separation of a liquid mixture into fractions differing in boiling point (and hence chemical composition) by means of distillation, typically using a fractionating column.
C. The daughters received a random set of genes from both of their parents
The total effective resistance of several resistors in SERIES is just the sum of all the individual resistances.
So the effective resistance of ten 120-ohm resistors in series is
(120 + 120 + 120 + 120 + 120 + 120 + 120 + 120 + 120 + 120) .
A much easier way to write it is . . . (10) x (120) .
Total effective resistance = <em>1,200 ohms</em> .
Answer:
No, the isothermal process is the not same as an adiabatic process for an ideal gas.
Explanation:
There is no change in the internal energy of an ideal gas undergoing an isothermal process since the internal energy depends only on the temperature.
But in the adiabatic process, there is no heat transfer, but work transfer can take place which causes the change in the internal energy due to which there is a change in temperature of an ideal gas.
Hence, the isothermal process is the not same as an adiabatic process for an ideal gas.