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3 years ago

Do polar molecules conduct energy as well as ionic molecules?

Chemistry

2 answers:

castortr0y [4]3 years ago

6 0

Answer:

Polar molecules typically do not conduct electricity as well as ionic molecules. This is because charges in polar molecules due to unequal sharing of electrons are not as strong as the charges on ions

Mkey [24]3 years ago

3 0

Answer:

Polar covalent compounds can not conduct electricity. But when they are dissolved in water, then they are capable of conducting electricity as the electrons will be free to conduct electricity.

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What is the density of a substance that has a mass of 20 g and a volume of 10 mL?

Paha777 [63]

Answer:

That unknown substance is water

Explanation:

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3 years ago

If 11.00 mL of a standard 0.2831 M NaOH solution is required to neutralize 26.86 mL of H2SO4, what is the molarity of the acid s

MrRa [10]

Answer:

0.1159 M

Explanation:

Using the formula below:

CaVa = CbVb

Where;

Ca = concentration/molarity of acid (M)

Cb = concentration/molarity of base (M)

Va = Volume of acid (mL)

Vb = volume of base (mL)

According to this question, the following information were given:

Ca = ?

Cb = 0.2831 M

Va = 26.86 mL

Vb = 11.00 mL

Using CaVa = CbVb

Ca × 26.86 = 0.2831 × 11

26.86Ca = 3.1141

Ca = 3.1141 ÷ 26.86

Ca = 0.1159

The molarity of the acid (H2SO4) solution is 0.1159 M

3 0

3 years ago

A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H 4, 3ClCH

bearhunter [10]

Answer:

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = 1.36 x 10^-3

Explanation:

The reaction

CICH2COOH ⇄ H+ (aq) + CICH2COO- (aq)

The initial concentration of CICH2COOH is 0.10 M , chloroacetic acid is 11.0% ionized.

let us calculate Ka

First , find change in concentration

since , 11% ionized

change in concentration = 0.10 X 11% = 0.011 M

Initial Concentration of CICH2COOH = 0.10 M

change in concentration of CICH2COOH = - 0.011 M

Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m

Initial Concentration of CICH2COO- = 0 M

change in concentration of CICH2COO- = + 0.011 M

Equilibrium Concentration of CICH2COO- = 0.011 M

Initial Concentration of H+ = 0 M

change in concentration of H+ = + 0.011 M

Equilibrium Concentration of H+ = 0.011 M

Therefore,

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = [H+][CICH2COO-] /[CICH2COOH]

Ka = (0.011 * 0.011) / (0.089)

Ka = 1.36 x 10^-3

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