Answer:
<h3>Right answer is: ( a) chalk powder remains suspended in water.</h3>
Explanation:
Filtration is the technique used to separate suspended solute particles from a solution . The chalk powder remains suspended in the solution and can easily be filtered through a filter paper , the chalk powder can be collected on the filter paper and clear solvent is collected as the filtrate.
<u>Answer:</u> The equilibrium concentration of
is 1.285 M.
<u>Explanation:</u>
The chemical equation for the decomposition of phosphorus pentachloride follows:

The expression for equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
We are given:

![[PCl_3]=0.18M](https://tex.z-dn.net/?f=%5BPCl_3%5D%3D0.18M)
![[Cl_2]=0.30M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D0.30M)
The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.
Putting values in above equation, we get:
![0.042=\frac{0.18\times 0.30}{[PCl_5]}](https://tex.z-dn.net/?f=0.042%3D%5Cfrac%7B0.18%5Ctimes%200.30%7D%7B%5BPCl_5%5D%7D)
![[PCl_5]=1.285](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D1.285)
Hence, the equilibrium concentration of
is 1.285 M.
It's the balance of chemical equation.
Answer:
B.
Explanation:
the SI unit for density is the kilogram per Cubic meter
Explanation:
It is given that the total volume is (10 mL + 60 mL) = 70 mL.
Also, it is known that
= 
Where,
= total volume
= initial volume
Therefore, new concentration of
= 
= 
= 0.43 M
New concentration of NaOH = 
= 
= 0.14 M
So, the given reaction will be as follows.

Initial: 0.43 0.14 0
Change: -0.14 -0.14 0.14
Equilibrium: 0.29 0 0.14
As it is known that value of
= 4.74
Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.
pH = ![pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 
= 4.74 + (-0.316)
= 4.42
Therefore, we can conclude that the pH of given reaction is 4.42.