Answer:
a) Ka= 7.1 × 10⁻⁴; This is a weak acid because the acid is not completely dissociated in solution.
Explanation:
Step 1: Write the dissociation reaction for nitrous acid
HNO₂(aq) ⇄ H⁺(aq) and NO₂⁻(aq)
Step 2: Calculate the acid dissociation constant
Ka = [H⁺] × [NO₂⁻] / [HNO₂]
Ka = 0.022 × 0.022 / 0.68
Ka = 7.1 × 10⁻⁴
Step 3: Determine the strength of the acid
Since Ka is very small, nitrous acid is a weak acid, not completely dissociated in solution.
Answer:
francium
Explanation:
the atomic radius increases from top to bottom in a group, and decreases from left to right across a period.
According to Bronsted-Lowry reaction- an acid is any substance that donates a proton (H+ ion) to another substance hence these two substance are acid aspirin (acetylsalicyclic acid) and hydrochloric acid (HCl). And there are two pairs - an acid with a corresponding conjugate base and a base with a corresponding conjugate acid. These pairs are called conjugate acid-base pairs.
The balanced equation is 2
AlI
3
(
a
q
)
+
3
Cl
2
(
g
)
→
2
AlCl
3
(
a
q
)
+
3
I
2
(
g
)
.
<u>Explanation:</u>
- Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- .
Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.
- Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.
Balancing the equation, we get:
2AlI
3( aq ) + 3Cl2
( g ) → 2AlCl3
( aq )
+ 3
I
2 ( g )
-
Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.
-
Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.
-
We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.