When the product formation is decreased if a substance B is added to an enzyme reaction and more substrate being added would not increase the amount of produce formed, then we assume that substance b could be a noncompetitive inhibitor. This type of inhibitor would be one that would bind to the enzyme with or without the presence of a substrate in different sites at the same time. It would change the conformation of the enzyme and also the active sites. As a result, the substrate would not be able to bind to the enzyme more effectively than the usual. The overall efficiency would decrease.
Answer:
An excellent experiment is to heat sodium thiosulfate in a water bath. The solid crystals will dissolve into the water in the hydrated crystals forming a supersaturated solution. ... Placing a small crystal in the supersaturated solution will cause the liquid to turn solid.
Answer:
Number of moles of sodium dissolved = 6.0 *10^23
Explanation:
The image for the question is attached
Solution
a) Total 181 ions of Na are dissolved
b)
The number of moles of sodium dissolved = 181/6.023 *10^23
Number of moles of sodium dissolved = 5.987 * 10^23
Number of moles of sodium dissolved = 6.0 *10^23
Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹
Explanation:
Given -
- An organic compound gives H₂ gas with Na
- On treatment with alkaline iodine it gives yellow ppt.
- On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)
To Find -
- Name the compound and write the reaction involved
Now,
Let A be the organic compound.
Then,
- A + Na → + H₂↑
- A + I₂ → CHI₃ (yellow ppt.)
- A + CrO₃ + H⁺ → C₂H₄O
Now,
Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.
- Functional group of aldehyde = —CHO
And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).
Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.
It means,
We know that 1° alcohol on oxidation gives aldehyde.
Here it gives 2 Carbon aldehyde.
It means,
Here 2 Carbon and 1° alcohol is used.
Now,
Its cleared that Compound A is Ethanol.
Reaction Involved -
- CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
- CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
- CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO
