Answer:
The number of electrons in the outermost shell of an element is represented in the periodic table as the group number that element is situated in the number of electrons in all shells of an element is represented in the periodic table as the element's atomic number.
Explanation:
I hope that helps
Enormous O unpredictability is in reference to the most exceedingly terrible conceivable development rate of the calculation. So O(N log N) implies that it will never keep running in some time more terrible than O(N log N). So in spite of the fact that Al's calculation scales superior to Bob's quadratic algo, it doesn't really mean it is better for ALL info sizes.
Maybe there is critical overhead in building up it, for example, making a lot of clusters or factors. Remember that even an O(N log N) calculation could have 1000 non settled circles that official at O(N) and still be viewed as O(N log N) the length of it is the most exceedingly awful part.
Answer:
Two possible compounds are shown below- one with an exocyclic double bond and another one with an endocyclic double bond
Explanation:
Reaction of alkene with
gives a complex of mercurous ion.
Then water molecule attacks this complex through
type reaction at more substituted position.
cleaves the resultant C-Hg bond and forms a C-H bond.
Two possible structures of an alkene is possible to yield 1-methylcyclohexanol which are shown below.
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. 
Major Species at B:
1. 
2. 
Major Species at C:
1. 
Major Species at D:
1. 
2. 
Major Species at E:
1. 
Major Species at F:
1. 
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK
= 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H
]/[H2
]
pH = pK
= 10.33