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.
The temperature is gonna be
The half-life in years of Neptunium-237 which was the first isotope is 2.1 
years.
Neptunium is most stable and Neptunium-237 is undergoes alpha decay, it means Neptunium-237 is decays by the emission of alpha particles . Seven alpha particles is emitted during decay of Neptunium-237. Neptunium-237 is radioactive actinide elements and first transuranium element.
The transuranium synthesis process involves creating a transuranium element through the transmutation process . The transmutation process is the process of creating heavy elements from light elements. Hence the process is the transmutation of light elements. There are two types: artificial and natural transmutation.
to learn more about transuranium element.
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Answer:
When hydrogen gas combines with nitrogen to form Ammonia the following chemical reaction will take place. Our equilibrium reaction will be N2(g) + 3H2(g) ⇔ 2NH3(g) + Heat. In this case, Hydrogen and nitrogen react together to form ammonia.
Explanation:
Actually the correct answer must be:
The limiting reactant in the reaction is the one which has
the lowest ratio of moles available
over coefficient in the balanced equation
This is because the actual mass or number of moles of the
reactant does not directly dictate if it is a limiting reactant, this must be
relative to the other reactants.
So the answer is:
e. none of the above
