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.
Answer:
Here's what I get
Explanation:
1. Sugar
(a) Dissolving in water
The white solid dissolves in water to give a colourless solution. There is no evidence that a new substance is being produced.
(b) Addition of sodium hydroxide
Adding the colourless solution of sodium hydroxide to the colourless sugar solution gives a colourless solution. There is no evidence that a new substance is being produced.
2. Magnesium sulfate
(a) Dissolving in water
The colourless crystals dissolve in water to give a colourless solution. There is no evidence that a new substance is being produced.
(b) Addition of sodium hydroxide
Adding the colourless solution of sodium hydroxide to the colourless solution of magnesium sulfate gives a white precipitate (see image). This is evidence that a new substance is being produced.
Answer:
of HA is 6.80
Explanation:
Acid dissociation constant () of HA is represented as-
Where species inside third bracket represents equilibrium concentrations
Now, plug in all the given equilibrium concentration into above equation-
So,
Hence
Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.
Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.
The reaction is as shown in the image.
The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid is shown in the image.
m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.