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xeze [42]
3 years ago
5

What are the structural features of purines and pyrimidines?

Chemistry
1 answer:
Alla [95]3 years ago
7 0

The purines in DNA are adenine and guanine, the same as in RNA. The pyrimidines in DNA are cytosine and thymine; in RNA, they are cytosine and uracil. Purines are larger than pyrimidines because they have a two-ring structure while pyrimidines only have a single ring

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Luba_88 [7]
The answer is D. easier, higher
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What type of elements make up a metallic bond
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Explanation:

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7 0
2 years ago
A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4
morpeh [17]

<u>Answer:</u> The molecular formula for the given organic compound is C_2H_2O_4

<u>Explanation:</u>

  • The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=21.33g

Mass of H_2O=4.366g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, \frac{12}{44}\times 21.33=5.82g of carbon will be contained.

  • For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, \frac{2}{18}\times 4.366=0.485g of hydrogen will be contained.

  • Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = \frac{0.485}{0.485}=1

For Hydrogen  = \frac{0.485}{0.485}=1

For Oxygen  = \frac{0.969}{0.485}=1.99\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is C_1H_{1}O_2=CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

  • The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Thus, the molecular formula for the given organic compound is C_2H_2O_4

4 0
3 years ago
The skeletal structure in line‑angle (line‑bond) mode of 2‑isopropyltoluene is shown. Identify the number of hydrogen atoms boun
Shalnov [3]

Answer:

Fourteen hydrogen atoms are bounded in total to the carbon atoms in the structure

Explanation:

The boxes that show "one" indicate that there is only one hydrogen atom bonded to that particular carbon atom while those that show "zero" shows there are no hydrogen atoms bonded to that particular carbon atom. Those that show "three" indicate that there are three hydrogen atoms bonded to that particular carbon.

There are 10 carbon atoms in the structure.

NOTE that each of these carbon atoms must be surrounded with four bonds; which was how the number of hydrogen atoms (numbers in the boxes) weree determined.

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What will the volume be of 50 moles of helium at 25°C and 2.5 atm?​
yulyashka [42]
The answer for this question would be 6 L
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