All categories

2 years ago

What determines the type of bond (single, double, or triple) an element will form?

Chemistry

1 answer:

Liono4ka [1.6K]2 years ago

5 0

The type of bond formed depends on the valence electrons (no: of electrons in the outermost shell)

When two atoms share one electron pair between each other, then they are said to be bonded by single covalent bond

When two atoms share two electron pairs between each other, they are said to be bonded by double covalent bond

When two atoms share three electron pairs between each other, they are said to be bonded by triple covalent bond.

Hope u understand

Please mark as the brainliest

You might be interested in

I need help ASAP

vova2212 [387]

Answer:

1.4 g/cm3

Explanation:

Density = Mass/Volume

Mass = 21g

Volume = 15cm3

Density = 21/15 = 1.4

8 0

2 years ago

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

guajiro [1.7K]

Answer:

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)

Kp = p SO₃² / ( p SO₂² x p O₂ )

The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm SO₂ O₂ SO₃

I 3.3 0.79 0

C -2x -x 2x

E 3.3 - 2x 0.79 - x 2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:

p SO₂ = 3.3 -0.47 atm = 2.83 atm

p O₂ = 0.79 - (0.47/2) atm = .56 atm

Now we can calculate Kp:

Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.

7 0

3 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

2 years ago

A force is applied but nothing happens

Lana71 [14]

If a force is applied but nothing happens, then it means that the forces are balanced. Being at such state, equal forces are acting on an object in opposite directions. Hope this answers the question. Have a nice day. Feel free to ask more questions.

7 0

2 years ago

Calculate the standard emf for the following reaction:

krek1111 [17]

In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
E = +1.47

Br(l) + 2e- = 2Br-
E = +1.065

We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V

4 0

3 years ago