N=3.5 mol
c=3.5 mol/L
n=cv
v=n/c
v=3.5/3.5=1.0 L
A) 1.0 liter of solution
The solution for this problem is:
1. number of mole / kg of vitamin K = (4.43 °C) / (40.0 °C/m) = 0.11075 m *molality freezing point depression constant for camphor is 37.7 C Kg/mo
2. (0.11075 mol/kg) x (0.0100 kg) = 0.0011075 mol
3. (0.500 g) / (0.0011075 mol) = 451 g/mol
The noble gas notation is the short or abbreviated form of the electron configuration.
It means that you use the symbol of the previous noble gas as part of the electron configuration of an element.
The gas noble previous to antimony is Kr, so you do not use Xe to write the electron configuration of Sb.
The gas noble previous to radium is Rn, so you do not use Xe to wirte the electron configuration of Ra.
The gas noble previous to uranium is Rn, so you do not use Xe to write the electron configuration of U.
The gas noble previous to cesium is Xe, so you use Xe to write the noble notation for Sb. This is it: Cs: [Xe] 6s.
Answer: cesium
The ga
<u>Answer:</u> The concentration of other (diluted) 
solution is 0.0026 M
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So, 
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}_{(diluted)}]}{[Fe^{2+}_{(concentrated)}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D%7D%7B%5BFe%5E%7B2%2B%7D_%7B%28concentrated%29%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= 0.047 V
![[Fe^{2+}_{(diluted)}]](https://tex.z-dn.net/?f=%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D)
= ? M
![[Fe^{2+}_{(concentrated)}]](https://tex.z-dn.net/?f=%5BFe%5E%7B2%2B%7D_%7B%28concentrated%29%7D%5D)
= 0.10 M
Putting values in above equation, we get:
![0.047=0-\frac{0.0592}{2}\log \frac{[Fe^{2+}_{(diluted)}]}{0.10M}](https://tex.z-dn.net/?f=0.047%3D0-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%5Cfrac%7B%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D%7D%7B0.10M%7D)
![[Fe^{2+}_{(diluted)}]=0.0026M](https://tex.z-dn.net/?f=%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D%3D0.0026M)
Hence, the concentration of other (diluted) solution is 0.0026 M
