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.
Reactant A must be second order and reactant B first order. Overall = add them so 3rd order.
Answer:
19.9 atoms
Explanation:
Grams --- Moles --- Atoms
You're converting from atoms (molecules) to moles.
You do not have to calculate the mass of "di phosphorus pentoxide."
Since you're already given 1.2x10^25 atoms, you start with that. You need to cancel out the atoms, so you need Avogadro's number as shown in the image.
(This has nothing to do with the problem) But in case if you're wondering, the "di" in phosphorus means there's 2 phosphorus and the "pent" means that there are 5 oxygens. So P2O5. Go to your periodic table, multiply their respective atomic masses. You would multiply phosphorus twice and oxygen 5 times. And add them up to get the overall mass.
I hope this helped!
The light bends enough to separate the colors, the object causes the light to change directions.
Answer:
Explanation:
Hello.
In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:
Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:
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