Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, 
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
![[Cl^{-}{diluted}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bdiluted%7D%5D)
= 0.0222 M
![[Cl^{-}{concentrated}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D)
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V
Answer:
- 7.48
Explanation:
Given:
Concentration of the sugar solution, C = 0.3 M
Temperature, T = 27° C = 273 + 27 = 300 K
Now,
The solute potential is given as:
solute potential = - iCRT
where,
i is the number of particles the particular molecule will make in water
i = 1 for sugar
R is the universal gas constant = 0.0831 liter bar/mole-K
on substituting the respective values, we get
solute potential = - 1 × 0.3 × 0.0831 × 300
or
The solute potential = - 7.479 ≈ - 7.48
Answer:
feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6
Explanation:
Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.
Overall balance: 
Water: 
Using substitution method, we have:
= 220.77 kg/s
= 4.16 kg/s
The maximum production rate of solid crystal is 
= 10*4.16 = 416 kg/s
Around evaporator:
kg/s
Around the mixing point:
Solid crystal: 
Using the last two equations, we can obtain:
kg/s
The rate of supplying fresh feed to obtain the production rate is:
= 352.5/220.77 = 1.6
Answer:
premium: 91 octane rating
Explanation:
Octane number refers to the percentage or volume fraction of isooctane in a fuel.
The octane number gives a picture of how safe a fuel is for an engine. The higher the octane rating the lesser the tendency of the fuel to cause knocking of the engine.
The type of gasoline with the highest percentage of octane among the options is premium.
So each 19.3g of gold is equivalent to one ml
So if we divide the mass of the gold nugget by the density we can find the volume: 93.5/19.3 = 4.79ml
Hope that helps
