BIOCHEMISTRY QUESTION 3.)Which of the following pairs of functional groups are p

Q: BIOCHEMISTRY QUESTION 3.)Which of the following pairs of functional groups are p

O.D.12.48 Addition of 200 mmol of NaOH (400 mL ´X 0.5 M) to 50 mmol of acetic acid (100 mL X 0.5 mM) completely titrates the acid so that it can no longer act as a buffer and leaves 150 mmol of NaOH dissolved in 500 mL, an [OH–] of 0.3 M. Given [OH–], [H+] can be calculated from the water constant: [H+][OH–] = 10–14 [H+] = 10–14 M2 / 0.3 M pH is, by definition, log (1/[H+]) pH = log (0.3 M /10–14 M2) = 12.48. 2.c. P

ID: 144048 • Letter: B

Question

BIOCHEMISTRY QUESTION

3.)Which of the following pairs of functional groups are present in the molecule below? NOTE: C is grey, N is blue, O is red, H is white

A-Carboxylic acid; amine

B-Aldehyde; carboxylic acid

C-aldehyde;amine

D-anhydride;hydroxyl

E-carboxylic acid; anhydride

2.)Which of the following has the elements in the correct order of increasing electronegativity?

A- O<N<P<C

B- C<P<N<O

C- P<C<N<O

D- C<N<P<O

E- C<N<O<P

1.) Fructose 1,6-bisphosphate (F1,6BP) is converted to D-Glyceraldehyde-3-phosphate (GADP) and Dihydroxyacetone phosphate (DHAP) during glycolysis, with a standard free-energy change (delta G'°) of 23.8 kJ/mol. Which of the following conditions encountered in a normal cell will result in a spontaneous reaction in the forward direction?

A- Standard conditions. Enough energy is released for the forward reaction to be spontaneous.

B- No conditions. The reaction cannot proceed spontaneously under any conditions due to the high positive delta G'°.

C- A high concentration of DHAP and GADP relative to the concentration of F1,6BP.

D- A high concentration of F1,6BP relative to the concentration of DHAP and GADP.

E- High concentrations of the enzyme that catalyzes the reaction.

0.) You add 100mL of 0.5 M acetic acid to 400 mL of 0.5 M NaOH. The final pH is:

note:- The pKa of acetic acid is 4.76.

A- 4.76

B- 5.76

C- 9.82

D- 12.48

E- NOT ENOUGH INFORMATION

Explanation / Answer

O.D.12.48

Addition of 200 mmol of NaOH (400 mL ´X 0.5 M) to 50 mmol of acetic acid (100 mL X 0.5 mM) completely titrates the acid so that it can no longer act as a buffer and leaves 150 mmol of NaOH dissolved in 500 mL, an [OH–] of 0.3 M.

Given [OH–], [H+] can be calculated from the water constant:

[H+][OH–] = 10–14

[H+] = 10–14 M2 / 0.3 M

pH is, by definition, log (1/[H+])

pH = log (0.3 M /10–14 M2) = 12.48.

2.c.

P<C<N<O.

Navigate

BIOCHEMISTRY Please help with what you can. Will give high rating. Studying sugg

BIOCHEMISTRY QUESTION 5. In 1958, Meselson and Stahl published an elegant experi

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.

BIOCHEMISTRY QUESTION 3.)Which of the following pairs of functional groups are p

Question

Explanation / Answer

Related Questions

Navigate