Answer:
She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.
Explanation:
A supersaturated solution contains more salt than it can normally hold at a given temperature.
A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.
Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.
B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.
C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.
Answer:
The lock-and-key model:
c. Enzyme active site has a rigid structure complementary
The induced-fit model:
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
Common to both The lock-and-key model and The induced-fit model:
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
d. Substrate binds to the enzyme through non-covalent interactions
Explanation:
Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.
The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.
The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.
The proton which is easily abstracted in
1-Benzyl-3-propylbenzene is the proton which is present on carbon atom in between two phenyl rings, or the central carbon which is shared by two benzene rings.
This easy abstraction of proton is due to its high acidity. Remember those species are always more acidic whose
conjugate base is stable. Like the acidity of carboxylic acid is due to stability of the
acetate ion.
In our case the stability of conjugate base arises due to
stability of negative ion due to resonance. As shown below, the negative charge can delocalize on both rings.
I have shown the resonance of negative ion on both Phenyl rings with
Blue and
Pink colors.<span />
Answer : The value of 
of the weak acid is, 4.72
Explanation :
First we have to calculate the moles of KOH.
Now we have to calculate the value of of the weak acid.
The equilibrium chemical reaction is:
Initial moles 0.25 0.03 0
At eqm. (0.25-0.03) 0.03 0.03
= 0.22
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[HK]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BHK%5D%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the value of of the weak acid is, 4.72
