According to the beer-lambert law when its concentration increases so does the absorbance increase.
beer-lambert law states that the amount of energy absorbed or transferred by a solution is proportional to the molar absorptivity of the solution and the concentration of the solute.
This means that concentrated solutions absorb more light than dilute solutions.
so the According to the beer-lambert law concentration increases, so does the absorbance. Therefore, absorbance is directly proportional to concentration.
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Answer:
E = 147000 J
Explanation:
Given that,
The mass of meteor, m = 50 kg
The altitude of the meteor, h = 300 m
We need to find the potential energy of the meteor. The formula for the potential energy is given by :
Put all the values,
So, the required potential energy is equal to 147000 J.
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.
Answer:
a. Values are - (1) 3,1,-1,1/2 (2) 2,1,-1,-1/2 (3) 3,0,0,1/2 (4) 4,3,3,-1/2
(5) 3,2,2,1/2
b. Order be- I > Sn > Xe
Explanation:
a.
Valid quantum numbers are -
'l' value should be less than 'n' value and 'ml' value should be '-l' to 'l'
'ml' value should not more than 'l' or 'n'.
Values are -
3,1,-1,1/2
2,1,-1,-1/2
3,0,0,1/2
4,3,3,-1/2
3,2,2,1/2
b.
Given that-
I [Kr]4d¹⁰5s²5p⁵
Sn [Kr]4d¹⁰5s²5p²
Xe [Kr]4d¹⁰5s²5p⁶
Order be-
I > Sn > Xe
Xe is least because it is completely filled outer shell (5s²5p⁶