Answer:
The answer is E. All of the statements describe the anomeric carbon.
Explanation:
When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.
As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).
It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).
The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)
It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.
Answer:
The values for spin quantum number +1/2 and - 1/2
Explanation:
Principal quantum number denoted by (n) is used to describe the shell or orbits that electrons are found. Principal quantum number can assume a value of n= 1,2, 3, 4,5............ which indicates K, L, M, N, O shell respectively.
To know the maximum number of electrons in each shell, the formula (2n²) can be used. The letter 'n' denotes the values of principal quantum number 1,2,3,4
For example
- n=1 (K shell) has maximum number of 2 electrons
- n=2 (L shell) has the maximum number of 8 electrons
- n=3 (M shell) has the maximum number of 18 electrons
- n=4 (N shell) has the maximum number of 32 electrons
All the electron in each shell will have a spin quantum number of +1/2 and - 1/2. One electron in each degenerate orbital will spin up (+1/2) while the other electron will spin down (-1/2).
Answer:
106.25 mL
Explanation:
For this, we can use
C1×V1=C2×V2
C1 = 0.45
V1 = 85
C2= 0.20
V2= ?
0.45 × 85 = 0.20 × V2
V2= (0.45 × 85)/0.20
V2=191.25mL
To find the amount of water added, subtract V1 from V2
191.25 - 85 =106.25mL