Question:
What type of product forms in the intramolecular reaction between the aldehyde portion of the glucose molecule below and its C-5 hydroxyl group?
a. disaccharide
b. carboxylic acid
c. hemiacetal
d. ester
e. stereoisomer
Answer:
hemiacetal forms in the intramolecular reaction between the aldehyde portion of the glucose molecule and its C-5 hydroxyl group
Explanation:
It is an alcohol also an ether that has been attached to the carbon molecule. Here the hydrogen has occupied the fourth bonding position. This hemiacetal has been derived from the aldehyde. Hence, hemiketal being an alcohol as well as ether has been attached to the same carbon and also to the two other carbon.
Answer:
Yes you always only see one side of the moon
The moon flag always points towards earth
It take 1 month for the moon to orbit the earth
It takes 29.5 days for the flag to rotate once in a full circle
1. true
2. false
3. false
4. true
5.false
6. It's true but can be false because a change in state can be a physical change as well.
Answer:
The percent abundances of the vanadium isotopes are 0.25% 50V and 99.75% 51V.
Explanation:
Let 50V be isotope A
Let 51V be isotope B
Mass number of isotope A (50V) = 49.9472
Mass number of isotope B (51V) = 50.9440
Abundance of isotope A (50V) = A%
Abundance of isotope B (51V) = B% = 100 — A
The atomic weight of vanadium
= 50.9415
Atomic weight = [(Mass of A x Abundance of A)/100] + [(Mass of B x Abundance of B) /100]
50.9415 = [(49.9472xA%)/100] + [(50.9440x(100 — A))/100]
50.9415 = [49.9472A%/100] + [(5094.40 — 50.9440A%)/100]
Multiply through by 100
5094.15 = 49.9472A% + 5094.40 — 50.9440A%
Collect like terms
5094.15 — 5094.40 = 49.9472A% — 50.9440A%
—0.25 = — 0.9968A%
Divide both side by — 0.9968
A% = —0.25/ — 0.9968
A% = 0.25%
B% = 100 — 0.25 = 99.75%
The percent abundances of the vanadium isotopes are 0.25% 50V and 99.75% 51V.
