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3 years ago

Which of the following bonds is least reactive? C—C H—H O—H H—Cl

2 answers:

8 0

The correct option is H - H
Compare to other type of bonds given above, the hydrogen to hydrogen bond is very unreactive. This is because the bond is very stable. Each of the hydrogen atom in the bond donate their single electron to form a covalent bond, which is quite stable.

amm18123 years ago

7 0

$\boxed{{\text{H}} - {\text{H bond}}}$ is the least reactive among the given bonds.

Further explanation:

<u>Chemical bond: </u>

The attraction between atoms, molecules or ions that is responsible for the formation of chemical compounds is known as a chemical bond. It is formed either due to electrostatic forces or by the sharing of electrons. There are many strong bonds such as ionic bonds, covalent bonds, and metallic bonds while some weak bonds like dipole-dipole interactions, London dispersion forces, and hydrogen bonding also exist.

Octet rule:

According to this rule, the elements have the tendency to bond with other elements in order to acquire eight electrons in their valence shells. This results in achieving a stable noble gas configuration. Generally, all elements obey octet rule but there are some exceptions to this rule. For example, Be and B do not follow the octet rule.

C-C bond

The atomic number of carbon is 6 so its electronic configuration is $1{s^2}2{s^2}2{p^2}$ . It has four valence electrons in its valence shell. It can form four covalent bonds with other elements. But in the C-C bond, there is only one bond formed between both the carbon atoms, leading to their incomplete octet. So C-C bond is not stable and thus highly reactive in nature.

H-H bond

The atomic number of hydrogen is 1 and its electronic configuration is $1{s^1}$ . It has one valence electron in it. It can form one covalent bond with any other element in order to complete its duplet. One bond is present between two hydrogen atoms in the H-H bond. Also, two 1s orbitals are overlapped in order to form this bond and this overlapping is highly effective in nature. So H-H bond is stable and thus not at all reactive in nature.

O-H bond

The atomic number of oxygen is 8 so its electronic configuration is $1{s^2}2{s^2}2{p^4}$ . It has six valence electrons in it. The atomic number of hydrogen is 1 and its electronic configuration is $1{s^1}$ . It has only one valence electron in its valence shell. Oxygen needs two electrons to complete its octet while hydrogen needs only one electron. But there is one single bond between O and H in the O-H bond. So O-H bond is not stable and thus reactive in nature.

H-Cl bond

The atomic number of chlorine is 17 and its electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}$ . It has seven valence electrons in it. The atomic number of hydrogen is 1 and its electronic configuration is $1{s^1}$ . It has only one valence electron in its valence shell. The octet of both atoms is complete in H-Cl bond but 1s and 3p orbitals are overlapping in this case that is not so effective. So H-Cl bond is somewhat reactive in nature.

Therefore H-H bond is the least reactive among the given bonds.

Learn more:

1. Which molecule cannot be adequately described by a single Lewis structure/ brainly.com/question/6786947

2. Do carbon dioxide and water have the same geometry? brainly.com/question/2176581

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical bonding and molecular structure

Keywords: octet rule, chemical bond, H-H bond, H-Cl bond, C-C bond, O-H bond, atomic number, 1, 17, 6, 4, attraction, overlapping.

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Which species is classified as a Arrenhius base? 1. CH3OH 2. LiOH 3. PO43- 4. CO32-

Otrada [13]

Answer:

2. LiOH

Explanation:

An Arrhenius base is a substance or chemical compound which increases the number of OH- ions when added in water.

LiOH or lithium hydroxide is one of the strong Arrhenius base as it ionizes completely or near-completely in solution. When LiOH is added to water , it dissociates into lithium (Li+) and hydroxide (OH−) ions.

Hence, the correct answer is 2. LiOH.

4 0

3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

3 years ago

0.836 m

Nataly_w [17]

The force exerted on the Kite is 49.6 N.

<u>Explanation:</u>

Force is nothing but the product of pressure exerted by the wind on the kite and the area of the kite.

F = P × A

Where,

F is the Force in Newton = ?

P is the Pressure = 113 Pa = 113 N/m²

A is the Area of the Kite

A = 0.836 m × 0.525 m = 0.4389 m²

Now, we can find the force as,

F = 113 N/m² × 0.4389 m²

= 49.6 N [ ∵ In the units, m² gets cancelled]

So the force exerted on the Kite is 49.6 N.

6 0

3 years ago

A 50.0 mL sample of a 0.200 M aqueous solution of Na3PO4 was added to 50.0 mL of a 0.100 M aqueous solution of BaCl2. The mixtur

sukhopar [10]

Answer:

1.02 grams

Explanation:

The reaction between Na₃PO₄ and BaCl₂ is:

2Na₃PO₄(aq) + 3BaCl₂(aq) → Ba₃(PO₄)₂(s) + 6NaCl(aq)

We need to find the number of moles of Na₃PO₄ and BaCl₂:

$n_{Na_{3}PO_{4}} = C*V = 0.200 M*0.050 L = 0.01 moles$

$n_{BaCl_{2}} = C*V = 0.100 M*0.050 L = 0.005 moles$

Now, we need to find the limiting reactant:

$n_{Na_{3}PO_{4}} = \frac{2}{3}*0.005 moles = 0.0033 moles$

We have that it is needed 0.0033 moles of Na₃PO₄ to react with BaCl₂ and we have 0.01 moles of Na₃PO₄, so the limiting reactant is BaCl₂.

The number of moles of Ba₃(PO₄)₂ is:

$n_{Ba_{3}(PO_{4})_{2}} = \frac{0.005}{3} = 0.0017 moles$

Finally, the mass of Ba₃(PO₄)₂ is:

$m = n_{Ba_{3}(PO_{4})_{2}}*M = 0.0017 moles*601.93 g/mol = 1.02 g$

Therefore, should be obtained 1.02 grams of the precipitate.

I hope it helps you!

3 0

3 years ago

Which of the following correctly describes a compound

Reptile [31]

combination of elements on the periodic table to create a substance. so I think I understand about this okay but I have a 98% on that type of subject

8 0

3 years ago