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3 years ago

Base-pairing in DNA involves _____. acid-base reaction ionic bonds hydrogen bonds covalent bonds

2 answers:

6 0

The answer is hydrogen bonds.

In the DNA double helix structure nitrogen atoms are chemically bonded to hydrogen atoms. Given that nitgrogen atoms are more electronegative than hydrogen atoms, hydrogen atoms in a base-pairing will bear a partial positive charge.

That partial positive charge is atracted to the oxygen atoms (which are also more electronegative than hydrogen atoms) in the same base pairing in DNA.

That atraction constitutes the hydrogen bond.

vladimir2022 [97]3 years ago

3 0

The answer is hydrogen bonds

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Question 7 (Essay Worth 5 points)

Lubov Fominskaja [6]

The periodic table of elements is divided into columns and rows. The vertical columns represent the groups while the horizontal rows represent the periods. A group of elements usually have similar physical and chemical properties. The first column of the periodic table is made up of 7 elements including SODIUM.

1. The name of the element chosen is sodium and its chemical symbol is Na.

2. Sodium is a metal and thus it has metallic properties.

3. Sodium belongs to group one elements whose family name is ALKALI METALS.

4. The neutral atom of sodium is made up of 11 protons, 11 electrons and 12 neutrons.

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3 years ago

Read 2 more answers

1. Oxygen<br> full electron configuration

Arada [10]

Answer:

$1s^{2} 2s^{2} 2p^{4}$

Explanation:

Oxygen has eight eletrons and six valence electrons, giving it the electron configuration of $1s^{2} 2s^{2} 2p^{4}$ .

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3 years ago

how will you seperate dye from black ink? draw a neatly labelled same?? fast I will mark brainly 100%

Nonamiya [84]

Answer:

put it through a distillery

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3 years ago

Can boron atoms, B form dative bonds with nitrogen atoms, N in the compound ammonia?

expeople1 [14]

"The boron-nitrogen interaction in the studied molecules shows some similarities with the N→B bond in the H3N-BH3 molecule, formally understood as covalent-dative. ... The results show that all the studied BN bonds are triple, since three two-center orbitals have been obtained."

"Formation of a dative bond or coordinate bond between ammonia and boron trifluoride. When the nitrogen donates a pair of electrons to share with the boron, the boron gains an octet. ... In addition, a pair of non-bonding electrons becomes bonding; they are delocalized over two atoms and become lower in energy."

3 0

3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

7 0

2 years ago