______ is/are usually the best conductor/s.

Which choice(s) correctly rank(s) the bonds in terms of increasing polarity?

Marina CMI [18]

Answer:

(II) only correctly rank the bonds in terms of increasing polarity.

Explanation:

Bond polarity is proportional to difference in electronegativity between bonded atoms.

Atoms Electronegativity Bond Electronegativity difference

Cl 3.0 Cl-F 1.0

Br 2.8 Br-Cl 0.2

F 4.0 Cl-Cl 0

H 2.1 H-C 0.4

C 2.5 H-N 0.9

N 3.0 H-O 1.4

O 3.5 Br-F 1.2

I 2.7 I-F 1.3

Si 1.9 Cl-F 1.0

P 2.2 Si-Cl 1.1

Si-P 0.3

Si-C 0.6

Si-F 2.1

So, clearly, order of increasing polarity : O-H > N-H > C-H

So, (II) only correctly rank the bonds in terms of increasing polarity

4 0

2 years ago

The electromagnetic spectrum includes the complete range of electromagnetic waves placed in order of: A.) increasing amplitude B

son4ous [18]

Hello the best answer is c
I love kay.

3 0

3 years ago

A sample of helium gas has a volume of 620. Ml at a temp of 500 K. If we decrease the temperature to 100K while keeping the pres

Ne4ueva [31]

Answer:

A sample of helium gas has a volume of 620mL at a temperature of 500 K. If we ... to 100 K while keeping the pressure constant, what will the new volume be?

Explanation:

3 0

2 years ago

Help pleassssse match allllll of them...

sleet_krkn [62]

Look at the letters, 3 would be MMeters

4 0

2 years ago

Read 2 more answers

In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?

klio [65]

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the cyclohexanone.

See figure 1

I hope it helps!

4 0

2 years ago