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2 years ago

Select all of the statements that correctly describe DNA structure:

Chemistry

1 answer:

GalinKa [24]2 years ago

3 0

Answer:

Hydrogen bonds will hold two DNA strands together.
The sugars and phosphate groups are located on the exterior of the helix.
Base pairing will always be between a purine base and a pyrimidine base.

I hope this helps you

:)

You might be interested in

What is an alloy? Answer question 2 if you want.

Anna11 [10]

Answer:

An alloy is a combination of metals or metals combined with one or more other elements. For example, combining the metallic elements gold and copper produces red gold, gold and silver becomes white gold, and silver combined with copper produces sterling silver. Elemental iron, combined with non-

Explanation:

that's what an alloy is all the rocks including gold silver combined with copper and sterling silver

7 0

3 years ago

Read 2 more answers

BaCl2(aq) + Na2CO3(aq) BaCO3(s) + NaCl(aq

Vitek1552 [10]

This is the equation balanced:

BaCl2(aq) + Na2CO3(aq) = BaCO3(s) + 2 NaCl(aq)

Then the coefficient in front of Na Cl is 2.

Answer: 2

7 0

3 years ago

Rank the following elements by effective nuclear charge, Zeff, for a valence electron. F LI Be B N

Stels [109]

Answer:

Rank in increasing order of effective nuclear charge:

Li < Be < B < N < F

Explanation:

This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.

1) Effective nuclear charge definitions

While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.

Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.

2) Z eff for a F valence electron:

F's atomic number: Z = 9
Total number of electrons: 9 (same numer of protons)
Period: 17 (search in the periodic table or do the electron configuration)
Number of valence electrons: 7 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
Zeff = Z - S = 9 - 2 = 7

3) Z eff for a Li valence eletron:

Li's atomic number: Z = 3
Total number of electrons: 3 (same number of protons)
Period: 1 (search on the periodic table or do the electron configuration)
Number of valence electrons: 1 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
Z eff = Z - S = 3 - 2 = 1.

4) Z eff for a Be valence eletron:

Be's atomic number: Z = 4
Total number of electrons: 4 (same number of protons)
Period: 2 (search on the periodic table or do the electron configuration)
Number of valence electrons: 2 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
Z eff = Z - S = 4 - 2 = 2

5) Z eff for a B valence eletron:

B's atomic number: Z = 5
Total number of electrons: 5 (same number of protons)
Period: 13 (search on the periodic table or do the electron configuration)
Number of valence electrons: 3 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
Z eff = Z - S = 5 - 2 = 3

6) Z eff for a N valence eletron:

N's atomic number: Z = 7
Total number of electrons: 7 (same number of protons)
Period: 15 (search on the periodic table or do the electron configuration)
Number of valence electrons: 5 (equal to the last digit of the period's number)
Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
Z eff = Z - S = 7 - 2 = 5

7) Summary (order):

Atom Zeff for a valence electron

Li 1

Be 2

Conclusion: the order is Li < Be < B < N < F

6 0

3 years ago

A buffer solution is prepared by dissolving equal amounts of ascorbic acid and sodium ascorbate, the conjugate base, in water. T

Gnom [1K]

THANKS

$\color{pink} \rule{500pt}{100000000pt}$

8 0

2 years ago

Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2

baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) $Cu_{(s)}$ → $Cu^{2+}_{(aq)} +2e$ E°=0.337 v

(reduction) $MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}$ → $MnO_{2 (aq)}+2H_{2}O$ E°=1.679 v

(overall) $2MnO_{4 (aq)}+3Cu_{(s)}$ +8H^{+}_{(aq)}→ $3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O$ E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

$E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}$

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

$E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346$

E=1.346

5 0

3 years ago