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
= ?
= 0.0222 M
= 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 : The correct option is, (a) paramagnetic with two unpaired electrons.
Explanation :
According to the molecular orbital theory, the general molecular orbital configuration will be,
![(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%2C%28%5Csigma_%7B1s%7D%5E%2A%29%2C%28%5Csigma_%7B2s%7D%29%2C%28%5Csigma_%7B2s%7D%5E%2A%29%2C%28%5Csigma_%7B2p_z%7D%29%2C%5B%28%5Cpi_%7B2p_x%7D%29%3D%28%5Cpi_%7B2p_y%7D%29%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29)
As there are 14 electrons present in the given configuration.
The molecular orbital configuration of molecule will be,
![(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%5E2%2C%28%5Csigma_%7B1s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2s%7D%29%5E2%2C%28%5Csigma_%7B2s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2p_z%7D%29%5E2%2C%5B%28%5Cpi_%7B2p_x%7D%29%5E1%3D%28%5Cpi_%7B2p_y%7D%29%5E1%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%5E0%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5E0%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29%5E0)
The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.
Hence, the correct option is, (a) paramagnetic with two unpaired electrons.
<u>Answer:</u> The value of
for the reaction at 690 K is 0.05
<u>Explanation:</u>
We are given:
Initial pressure of
= 1.0 atm
Total pressure at equilibrium = 1.2 atm
The chemical equation for the decomposition of phosgene follows:

Initial: 1 - -
At eqllm: 1-x x x
We are given:
Total pressure at equilibrium = [(1 - x) + x+ x]
So, the equation becomes:
![[(1 - x) + x+ x]=1.2\\\\x=0.2atm](https://tex.z-dn.net/?f=%5B%281%20-%20x%29%20%2B%20x%2B%20x%5D%3D1.2%5C%5C%5C%5Cx%3D0.2atm)
The expression for
for above equation follows:


Putting values in above equation, we get:

Hence, the value of
for the reaction at 690 K is 0.05
Answer:
<em><u></u></em>
Explanation:
<u />
<u>1. Dissociation equation</u>
<u />
Assuming 100% dissociation, the equation is:
- Ba(C₂H₃O₂)₂ → Ba²⁺ + 2C₂H₃O₂⁻
↑
acetate ions
<u>2. Molarity</u>
<u />
Calculate the molarity, M, of the solution:
- n = mass in grams / molar mass
- n = 69.g / 255.415g/mol = 0.27015 mol
- M = 0.27015mol / 0.970liter = 0.27850 mol/liter ≈ 0.28M
<u>3. Acetate ions</u>
From the chemical equation, 1 mol of dissolved Ba(C₂H₃O₂)₂ produces 2 acetate ions in solution.
Thus, 0.28 mol/liter × 2 = 0.56 mol/liter = 0.56M ← answer
Answer:
The new pressure of the gas in the chamber is 1,093.75 mmHg
Explanation:
The Gay-Lussac Law is a gas law that relates pressure and temperature to constant volume. This law says that the pressure of the gas is directly proportional to its temperature.
That is, if the temperature increases, the pressure increases, while if the temperature decreases, the pressure decreases. So the Gay-Lussac law can be expressed mathematically as follows:

Having an initial and an end state of a gas, the following expression can be used:

In this case:
- P1= 1560 mmHg
- T1= 445 K
- P2=?
- T2= 312 K
Replacing:

Solving:

P2=1,093.75 mmHg
<u><em>The new pressure of the gas in the chamber is 1,093.75 mmHg</em></u>