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3 years ago

1. What is polarity?

Chemistry

1 answer:

xeze [42]3 years ago

7 0

Polarity is an ionic bond with unequal electrons

Answer: Option A

<u>Explanation: </u>

In polar covalent bonds, the atoms possess an uneven attraction of electrons, which causes an uneven sharing. This sometimes simply called the polar bond, the electron distributions around the molecule are asymmetrical. As the electronegativity variation higher, the resonance increasingly supports the contribution of ions.

If the difference in electronegativity is very large, for example between the electropositive atoms like sodium and electronegative atom such as fluorine, resonance dominates in the ionic structure, and the bond can be considered ionic. As the electronegativity variation between the two bound elements increases, the non-polar bond becomes a polar bond, which then becomes as an ionic bonds.

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p32p32 is a radioactive isotope with a half-life of 14.3 days. if you currently have 63.163.1 g of p32p32 , how much p32p32 was

AlekseyPX

Answer:

92.93 g

Explanation:

Number of half lives that have elapsed in eight days =8/14.3 = 0.559

Fraction of the radioactive nuclide that remains after 0.559 half lives is given by

N/No=(1/2)^0.559

Where N= mass of radioactive nuclides remaining after a time t

No= mass of radioactive nuclides originally present

N/No=(1/2)^0.559= 0.679

Mass of nuclides present eight days before= 63.1g/0.679

Mass of nuclides present eight days before=92.93 g

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4 years ago

Which of the following values are not equal to 1 mole?

Gekata [30.6K]

Answer:

none of them are equal to one mole

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3 years ago

Which substance is the limiting reactant when 4.0 g of sulfur reacts with 6.0 g of oxygen and 8.0 g of sodium hydroxide accordin

Marysya12 [62]

C, NaOH(aq) is the answer

6 0

3 years ago

Help?

Ahat [919]

Answer:

$KClO_3$

Explanation:

Hello!

In this case, as we know the mass of the total sample, we can first compute the mass of oxygen:

$m_O=22.9g-7.33g-6.65g=8.92g$

Next, we compute the moles of each element:

$n_K=\frac{7.33g}{39.9g/mol}= 0.184mol\\\\n_{Cl}=\frac{6.65g}{35.45g/mol}=0.188mol \\\\n_O=\frac{8.92g}{16.00g/mol} =0.5575mol$

Now, we divide the moles by 0.184 moles, the fewest ones, to obtain:

$K=\frac{0.184}{0.184}=1.0 \\\\Cl=\frac{0.187}{0.184}=1.0\\\\O=\frac{0.5575}{0.184} =3.0$

Therefore, the empirical formula is:

$KClO_3$

Regards!

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3 years ago

:answer question number 2

egoroff_w [7]

Explanation:

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3 years ago