Response:
65.14 g
Clarification:
The entropy change resulting from mixing two metals 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 the real gas constant, n_{cu} denotes the moles of copper, while n_{Ni} signifies the moles of nickel.
Therefore, determining the moles of Nickel yields

=1.7 moles
Then substituting n_{Ni} = 1.7 into the equation produces ΔS= 15,
with R= 8.314, and solving gives
![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)
Upon resolving, the quantity of moles of copper in the mixture equals
n_cu= 1.025
Thus, the copper mass is calculated as n_cu×M_cu = 1.025×63.55 = 65.14 g
Consequently, the required mass is found to be = 65.14 g
Answer:
The resistance of the skin is 98 kΩ
Explanation:
Given:
Resistivity
Ωm
Thickness
m
Resistivity of the skin:

With assumed radius for the worker's palm,
m
Area of the worker's palm,



Thus the resistance of palm is,

Ω
Consequently, the resistance of the skin is 98 kΩ
Answer:

Explanation:
It's provided that
Atmospheric pressure=0.67 atm
We need to determine the partial pressure of ozone at 441 ppb.
Partial pressure of ozone=
Inserting the given atmospheric pressure
Thus, we find
Partial pressure of ozone=
Consequently, the partial pressure of ozone is=
a.) 10 Hz b.) 0.1 s c.) 187.4 m/s d.) -412.6 m/s²
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