Answer: Option (B) is the correct answer.
Explanation:
As the given reaction is as follows.
Equilibrium constant for this reaction will be as follows.
![K_{c} = \frac{[CO_{2}]}{[CO]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCO_%7B2%7D%5D%7D%7B%5BCO%5D%5E%7B2%7D%7D)
According to Le Chatelier's principle, when we increase the temperature then the equilibrium will shift towards the right hand side.
As a result, concentration of carbon dioxide will decrease whereas concentration of carbon monoxide will increase.
Thus, we can conclude that in the given reaction equilibrium constant for this reaction will decrease with increasing temperature.
the semipermeable membrane surrounding the cytoplasm of a cell.
In a redox reaction electrons are lost and gained in equal numbers. The species that is oxidized gives electrons to the species that is reduced. I hope this helps. Let me know if anything is unclear.
Answer: option <span>D. be given a positive charge produced by the movement of electrons to the other end of the ball.
</span>
Explanation:
This phenomenon is called electrostatic induction.
The excess of negative charge on the end of the rod will repel the electrons on the side of the pith ball that have been approached to it.
Then the electrons on the pith ball will move far away from this end with it will be left an excess of positive charge.
In this way the rod has induced that the ball acquires a positive charge on one end and a negative charge on the other end.
<span />
Answer:
0.9359 M
Explanation:
M(CaCO3)=100.1 g/mol
10.305 g * 1 mol/100.1 g = 10.305/100.1 mol CaCO3
(10.305/100.1 mol )/0.1100 L = 0.936 mol/L = 0.9359 M
