Answer:
The molarity of the HCl solution should be 4.04 M
Explanation:
<u>Step 1:</u> Data given
volume of HCl solution = 10.00 mL = 0.01 L
volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L
<u>Step 2:</u> The balanced equation
HCl + NaOH → NaCL + H2O
Step 3: Calculate molarity of HCl
n1*C1*V1 = n2*C2*V2
Since the mole ratio for HCl and NaOH is 1:1 we can just write:
C1*V1 =C2*V2
⇒ with C1 : the molarity of HCl = TO BE DETERMINED
⇒ with V1 = the volume og HCl = 10 mL = 0.01 L
⇒ with C2 = The molarity of NaOH = 1.6 M
⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L
C1 * 0.01 = 1.6 * 0.02524
C1 = (1.6*0.02524)/0.01
C1 = 4.04M
The molarity of the HCl solution should be 4.04 M
Comment above 100 percent correct
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.