Hello,
Here is your answer:
The proper answer to this question is option D "<span>sodium hydroxide".
Here is how:
Sodium Hydroxide its a white substance that is a </span><span>electrolyte.
Your answer is D.
If you need anymore help feel free to ask me!
Hope this helps!</span>
Gain or lose.
The exchange of electrons in chemical bonding seeks to fulfill the octet rule. There are some exceptions, such as with hydrogen and helium, whose valence shells have a capacity of two electrons.
Answer:
C.
Explanation:
The mass of the reactants should not change, in fact it would be equal because the only thing that changes is the form in which your products are in. The reactants will still have the same amount of mass from the products as no products were removed or added, the structure changed, the mass did not.
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.
