The area is 60.2, to find area it's length times width.
the answer would be B) Moon
Answer:
3 e⁻ transfer has occurred.
Explanation
This is a redox reaction.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
- [Ne]= (1s²) (2s² 2p⁶)
A combination of both the reactions( Half-reactions) leads to a redox reaction.
Let us look at initial configurations of Al and Cl
[Al]= 1s² 2s² 2p⁶ 3s² 3p¹
[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵
Hence, Al can lose 3 electrons to achieve octet config.
and, Cl can gain 1e to achieve nearest noble gas config. [Ar]
This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.
Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃
Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)
Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.
1,3-butadiene is the simplest conjugated diene and undergoes 1,4 addition reaction in acidic environment.
Chemical reaction: CH₂=CH-CH=CH₂ + H₂O → CH₃-CH=CH-CH₂-OH.
CH₂=CH-CH=CH₂ - 1,3-butadiene.
CH₃-CH=CH-CH₂-OH - 2-buten-1-ol.
Diene<span> or </span>diolefin<span> is a </span>hydrocarbon<span> that has two </span>carbon double bonds<span>.</span>
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.
