Answer:
Explanation:
Cheese, Meat, dough, Sauce
I have completed the reaction with both possible products. Below are the products formed both in presence and absence of peroxide. And have connected the proton with corresponding peaks with different color lines.
Answer:
At equal concentration of HBCG and BCG^-, the colour is green. This colour first appears at pH = 3.8
Explanation:
HBCG is an indicator that is prepared by dissolving the solid in ethanol.
Since
Ka=[BCG−][H3O+][HBCG]When [BCG-] = [HBCG], then Ka = [H3O+].
If pH = 3.8
Ka= [H3O+] = -antilog pH = -antilog (3.8)
Ka= 1.58 ×10^-4
Answer:
The mass percentage of calcium carbonated reacted is 2.5%.
Explanation:
The reaction is:

Thus the Kp of the equilibrium will be:
Kp = partial pressure of carbon dioxide [as the other are solid]
Moles of calcium carbonate initially present = 
Let us apply ICE table to the equilibrium given:

Initial 0.2 0 0
Change -x +x +x
Equilibrium 0.2-x x x
Kp = partial pressure of carbon dioxide
Kp = Kc(RT)ⁿ
where n = difference in the number of moles of gaseous products and reactants
for given reaction n = 1
R = gas constant = 8.314 J /mol K
T = temperature = 800 ⁰C = 1073 K
Putting values
Kc =
Kc = ![\frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B2%7D%5D%5BCaO%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D%3D%20%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.2-x%29%7D%3D1.3X10%5E%7B-4%7D)


On calculating
x = 0.005
where x = the moles of calcium carbonate dissociated or reacted.
Percentage of the moles or mass reacted =
%
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.