Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
While metallic bonds have the strong electrostatic force of attractions between the cation or atoms and the delocalized electrons in the geometrical arrangement of the two metals. ... Metallic bonds are malleable and ductile, while covalent bonds and ionic bonds non-malleable and non-ductile.
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:
Follows this order: B=> A => C.
Explanation:
NB: kindly check the attachment for the diagram of compounds A, B and C.
Elution is a very important concept in chromatography separation techniques. It deals with the use of eluent in the removal of an adsobate from an adsorbent. The principle behind Elution is just about how polar the solvent is.
So, in this question Compound B will go with the Elution first because of its polarity. Compound B has lesser polarity as compared to Compounds A and B.
Compound A will then elutes second because of its polarity too as resonance increases its polarity.
Last, compound C elutes because it has the highest polarity which is caused by electronegative atoms.