<h2>
=
![\dfrac{[H^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
</h2>
Explanation:
- When an aqueous solution of a certain acid is prepared it is dissociated is as follows-
⇄ 
Here HA is a protonic acid such as acetic acid, 
- The double arrow signifies that it is an equilibrium process, which means the dissociation and recombination of the acid occur simultaneously.
- The acid dissociation constant can be given by -
=
- The reaction is can also be represented by Bronsted and lowry -
⇄ ![[H_3O^+] [A^-]](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%5BA%5E-%5D)
- Then the dissociation constant will be
= ![\dfrac{[H_3O^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH_3O%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
Here, is the dissociation constant of an acid.
Answer: London dispersion forces
a) 56g
<h3>Calculation:</h3>
At STP,
22.4 L of N₂ = 1 mol
We have given 44.8 L of N₂, therefore,
44.8 L of N₂ = 
= 
mol
We know that,
1 mol of N₂ = 28 g
Hence,
2 mol of N₂ = 28 × 2
= 56g
Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.
Learn more about calculation at STP here:
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Answer:
The correct option is C.
Solutions with low pH values have sour taste.
Explanation:
Low pH value show acidic property which can be define as , -log of conc. of hydrogen ion.
Acid has following properties
- Sour taste
- Turn blue litmus into red
- corrosive
-OH group: Base have OH groups. So it is not correct option so it not correct for low pH i.e. acid
turns litmus paper blue: Base turn red litmus into blue, so it not correct for low pH i.e. acid
feels slippery: Bases are mostly slippery like soap
tastes sour: Sour taste is for acid and bitter taste is for base. So it is correct option for acid
Answer:
The molarity of the formed CaBr2 solution is 0.48 M
Explanation:
Step 1: Data given
Number of moles CaBr2 = 0.72 moles
Volume of water = 1.50 L
Step 2: Calculate the molarity of the solution
Molarity of CaBr2 solution = moles CaBr2 / volume water
Molarity of CaBr2 solution = 0.72 moles / 1.50 L
Molarity of CaBr2 solution = 0.48 mol / = 0.48 M
The molarity of the formed CaBr2 solution is 0.48 M
