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3 years ago

Which one of these complementary base pairs can form three hydrogen bonds?

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

7 0

Answer:

Option D is the correct answer

Explanation:

There are three hydrogen bonds formed between Guanine and Cytosine base.

The first hydrogen bond is formed between the oxygen atom on Carbon-6 (C-6) of guanine, and one of the hydrogen atoms attached to the Nitrogen atom (i.e the one of the hydrogen atoms in the amino group) on the Carbon-4 (C-4)of the cytosine base.

The second hydrogen bond is formed between the hydrogen atom on Nitrogen-1 (N-1) of the guanine base and Nitrogen-3 (N-3) of the cytosine base.

The third hydrogen bond exist between one of the hydrogen atoms in the amino group attached to the second Carbon (C-2) of the guanine base, and the oxygen atom attached to the second Carbon (C-2) of the cytosine base.

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What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

Fed [463]

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$

$4.2 + log\frac{0.05}{0.045} = 4.245$

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

$pH = pKa + log\frac{base}{acid}$

$4.245 = 3.75 + log\frac{base}{acid}$

$log\frac{base}{acid} = 0.5$

$\frac{base}{acid} = 3.162$

Now we must solve the equation above. This will be done using the following values

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L$

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

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2 years ago

In part 3 of the lab, you discovered the relationship between temperature (T) and volume (V) of a gas.

rewona [7]

$\frac{V1}{T1}$ = k

Explanation:

The relation between volume, V of gas and Temperature, T of a gas is related by Charles Law.

This law states that the volume of a given amount of gas held at a constant pressure is directly proportional to the Kelvin temperature

Thus,

$\frac{V}{T}$ = k

where k is a constant

Therefore,

$\frac{V1}{T1}$ = $\frac{V2}{T2\\}$ = $\frac{V3}{T3}$ ...

This shows, as the volume of a gas goes up, the temperature also goes up and vice-versa.

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3 years ago

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tatuchka [14]

Answer:qui

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gogolik [260]

Answer:

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Explanation:

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otez555 [7]

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