All categories

2 years ago

Which of the following are likely to form a covalent bond?

Chemistry

2 answers:

ohaa [14]2 years ago

6 0

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

A covalent bond is defined as the bond which is formed when sharing of electrons takes place between the atoms forming a compound. This type of bond is usually formed between two non-metals.

From the given options:

<u>Option A:</u> two gold atoms

Gold atom is a metal and thus, will not form covalent bond.

<u>Option B:</u> two neon atoms

Neon atom is a noble gas and is unreactive atom. Thus, it will not form covalent bond.

<u>Option C:</u> hydrogen and oxygen atoms

Both the atoms are non-metals and thus, will lead to the formation of a covalent bond.

<u>Option D:</u> magnesium and bromine atom

Magnesium atom is a metal and bromine atom is a non-metal. Thus, they will not lead to the formation of covalent bond but will form ionic bond.

Hence, the correct answer is Option C.

Alex17521 [72]2 years ago

5 0

C) hydrogen and oxygen atoms

You might be interested in

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

Westkost [7]

Answer:

a) $K_{2} S$ and $NH_{4} Cl$ :

There are no insoluble precipitate forms.

b) $Ca Cl_{2}$ and $(NH_{4} )_{2} Co_{3}$ :

There are the insoluble precipitates of $CaCo_{3}$ forms.

c) $Li_{2}S$ and $MnBr_{2}$ :

There are the insoluble precipitates of $MnS$ forms.

d) $Ba(No_{3} )_{2}$ and $Ag_{2} So_{4}$ :

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e) $Rb_{2}Co_{3}$ and $NaCl$ :

There are no insoluble precipitates forms.

Explanation:

Solubility rule suggests:- $K_{2} S$ ⇒ soluble, $NH_{4} Cl$ ⇒ soluble.

KCl ⇒ soluble, $(NH_{4})_{2} S$ ⇒ soluble.

There are no insoluble precipitate forms.

Solubility rule suggests:- $Ca Cl_{2}$ ⇒ soluble, $(NH_{4} )_{2} Co_{3}$ ⇒ soluble.

$CaCo_{3}$ ⇒ insoluble, $NH_{4} Cl$ ⇒ soluble.

There are the insoluble precipitates of $CaCo_{3}$ forms.

Solubility rule suggests:- $Li_{2}S$ ⇒ soluble, $MnBr_{2}$ ⇒ soluble.

$LiBr$ ⇒ soluble, $MnS$ ⇒ insoluble.

There are the insoluble precipitates of $MnS$ forms.

Solubility rule suggests:- $Ba(No_{3} )_{2}$ ⇒ soluble, $Ag_{2} So_{4}$ ⇒insoluble.

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

Solubility rule suggests:- $Rb_{2}Co_{3}$ ⇒ soluble, $NaCl$ ⇒ soluble.

$RbCl$ ⇒ soluble, $Na_{2} Co_{3}$ ⇒ soluble.

There are no insoluble precipitates forms.

6 0

2 years ago

An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic fi

arsen [322]

Explanation:

In a magnetic field, the radius of the charged particle is as follows.

r = $\frac{mv}{qB}$

where, m = mass, v = velocity

q = charge, B = magnetic field

Therefore, q will be calculated as follows.

q = $\frac{mv}{rB}$

= $\frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}$

= $\frac{34.98 \times 10^{-22}}{0.01092}$

= $3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}$

= +2e

Thus, we can conclude that the charge of the ionized atom is +2e.

5 0

2 years ago

M=<br> V=432.6 mL<br> D=8.4 g/ml

Nitella [24]

Answer:

Explanation:

M=D times V

Answer-3,633.84g

Rounded Answer (correct sig figs)- 3600g

6 0

3 years ago

Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m

mina [271]

Answer:

<h3>I don't know what is the answer of your question sorry never mind..</h3>

Explanation:

<h3>And please marks me as brainliest... </h3>

3 0

2 years ago

Acetic acid, ethanol, acetaldehyde and ethane form a series of 2-carbon molecules which differ in their extent of oxidation. Con

yanalaym [24]

Explanation:

Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.

$CH_3CH_2OH$ → $CH_3CHO$ [oxidation by loss of hydrogen]

An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.

An oxidizing agent used along with dilute sulphuric acid for acidification.

Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.

The oxidation of aldehydes to carboxylic acids can be done by the two-step process.

In the first step, one molecule of water is added in the presence of a catalyst that is acidic.

There is a generation of a hydrate. (geminal 1,1-diol).

$CH_3CHO$ → $CH_3CH_2COOH$ [reduction by the gain of electrons]

Here, the oxidizing agent used is $Cr_3O$ in the presence of acetone.

7 0

3 years ago