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abruzzese [7]
2 years ago
9

Which type of molecule is made from nucleotides?

Chemistry
1 answer:
murzikaleks [220]2 years ago
5 0

Answer:

Nucleic acid is made from nucleotides

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The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at
nalin [4]

Answer:

The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>

Explanation:

The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.

<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.

Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.

Hence , the answer is true .

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2 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

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3 years ago
Question 1
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