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Viefleur [7K]
3 years ago
9

When DNA is replicated, it is necessary for the two strands to "unzip" temporarily. Choose which bonding type is most appropriat

e for holding the strands together in this way.
A) ionic bonding
B) hydrogen bonding
C) non-polar covalent bonding D) polar covalent bonding
Chemistry
1 answer:
wariber [46]3 years ago
5 0

Answer: B) hydrogen bonding

Explanation:

A) An ionic bond is formed when an element completely transfers its valence electron to another element.

B) Hydrogen bonds are special type of dipole dipole forces which are formed when hydrogen bonds with an electronegative element.

DNA strands are held together by hydrogen bonds between the bases on one strand and those on the other. Adenine and thymine have two hydrogen bonds between them while cytosine and guanine have three hydrogen bonds between them.

C) Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms.

D) A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

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7 0
3 years ago
What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
bulgar [2K]

Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

\Delta T_b = I \times K_b \times (w/M)/ x

\Delta T_b = I \times K_b \times w/Mx

\Delta T_b = 1 \times 0.51 \times1.5/(0.250 \times 58.44) = 0.052

\Delta T_f = M \times K_f = 1.86 \times 1.5 = 2.79

4 0
3 years ago
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nirvana33 [79]

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6 0
3 years ago
A 25.0-mL sample of 0.150 M butanoic acid is titrated with a 0.150 M NaOH solution. What is the pH before any base is added? The
hoa [83]
HA ⇄ H⁺ + A⁻ 
so:
\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}
and now:
\frac{(x)(x)}{(0.150-x)} = 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83
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