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3 years ago

Chlorine gas (cl2) forms when two chlorine atoms share an electron. what type of bonding is present in chlorine gas?

Chemistry

2 answers:

Masja [62]3 years ago

7 0

Is there some choises

zhenek [66]3 years ago

5 0

Explanation:

When a chemical bond is formed by sharing of electrons then it is known as a covalent bond.

A covalent bond is always formed between two non-metals.

For example, atomic number of chlorine is 17 and its electronic distribution is 2, 8, 7.

As a result, it needs one more electron to complete its octet. So, when it chemically combines with another chlorine atom then there will occur sharing of electrons.

Hence, $Cl_{2}$ is formed which is a covalent compound.

On the other hand, when a chemical bond is formed by transfer of electrons from one atom to another then it is known as an ionic bond. An ionic bond is always formed between a metal and a non-metal.

For example, LiCl is an ionic compound.

Thus, we can conclude that in a chlorine gas covalent bonding is present when two chlorine atoms share an electron.

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5.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 7

slega [8]

Answer:

The molecular formula = C6H6

Explanation:

Step 1: Data given

Mass of compound X = 5.00 grams

Mass of products =

CO2 = 16.39 grams

H2O = 3.46 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16. 0g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles CO2 = 16.93 grams / 44.01 g/mol

Moles CO2 = 0.385 moles

Moles C = 1* 0.385 = 0.385 moles

Moles H2O = 3.46 grams / 18.02

Moles H2O = 0.192 moles

Moles H = 2* 0.192 = 0.384 moles

Step 3: Calculate mass

Mass = moles * molar mass

Mass C = 0.385 moles *12.0 g/mol

Mass C = 4.62 grams

Mass H = 0.39 grams

Mass O = 5.00 - 4.62 -0.38 moles

Mass O = 0 grams

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.385 moles / 0.384 = 1

H: 0.384 moles / 0.384 = 1

The empirical formula is CH

This molecular formula is 13 g/mol

We have to multiply the empirical formula by n

n = 78 g/mol / 13 g/mol

n = 6

The molecular formula = 6*(CH) = C6H6

4 0

3 years ago

• Explain why each of the following pairs is not likely to form an ionic bond.

Olegator [25]

Answer:

A)Chlorine and Bromine:

They are both non metal hence they form a covalent bond due to covalent bonding.

B)Potassium and Helium:

Helium ion has a small cationic radius and distorted by the potassium ion due to polarization.

C)Sodium and Lithium:

Both are metals hence they form a metallic bond since they share electrons to the electron cloud.

4 0

3 years ago

Sketch two amino acids side-by-side, on one of them label the functional groups, then show how the two can be joined together.

Alinara [238K]

Answer : The sketch of two amino acids side by side are shown below.

Explanation :

Amino acid : Amino acid are the acid that contains two functional groups which are carboxylic group, $-COOH$ and ammine group, $-NH_2$ .

When the two or more that two amino acids join together with the help of peptide bond, they produces polypeptide chain or protein.

The bond present between the two amino acid is called a peptide bond.

The peptide bond is a chemical bond that is formed between the two molecules when the nitrogen of one amino acid react with the carbon of another amino acid by releasing a water molecule.

Hence, the sketch of two amino acids side by side are shown below.

6 0

3 years ago

The bonds in the compound MgSO4 can be described as

Butoxors [25]

C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond

7 0

3 years ago

Give all possible ml values for orbitals that have each of the following: (a) l = 2; (b) n = 1; (c) n = 4, l = 3.

Olin [163]

Answer:

(a) ml = 0, ±1, ±2

(b) ml = 0

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So in our exercise,

(a) l = 2; equivalent with with sublevel d

-l ≤ ml ≤ l, ml = 0, ±1, ±2, equivalent with dxy, dxz, dyz, dx2-y2, dz2

(b) n = 1;

n = 1, only 01 level

l = 0, equivalent with sublevel s

ml = 0

l = 3, equivalent with sublevel f

ml = 0, ±1, ±2, ±3, ±4

7 0

3 years ago