Sodium Chloride because its still a liquid at the 773 temperature mark<span />
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
Red #40 is soluble in water while zinc oxide is not.
So the easiest way to separate them is as follows:
1- add water to the mixture until red #40 is dissolved in the water
2- filter to separate the zinc oxide
4- heat the solution of red #40 and water until water evaporates and red#40 remains.
Answer:
H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.
Explanation:
The Lewis structure of PF3 shows that the central phosphorus atom has one non bonding and three bonding electron pairs. In this compound each Phosphorus atom contributes 5 valence electrons while each fluoride contributes 7 valence electrons making a total of 26 valence electrons. The central Phosphorus atom forms single bonds with each of the fluoride atoms. Phosphorus, therefore ends up with a non-bonding pair since the fluoride atoms already have 8 electrons around them. <span />
