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3 years ago

If two nonmetals with the same electronegativity bond, what type of bond will form?

2 answers:

7 0

If two nonmetals with the same electronegativity bond, what type of bond will form then Non polar covalent bond (option D) will form.

AlladinOne [14]3 years ago

6 0

Answer:

Two nonmetals with the same electronegativity will form a non polar covalent bond.

Explanation:

The type of bond between atoms is classified in 3 big groups:

Metallic bond: this type of bond only take place between metallic atoms like Cu, Al, Au, etc.

Ionic bond: this type of bond is formed between ions, that means that it is necessary the presence of a cation (ion with positive charge) and and an anion (ion with negative charge) and when the atoms has a very high difference of electronegativity (more that 2), that makes the ionic bond always polar, because there will be always a positive pole (cation) and a negative pole (anion). This is common between a metal and a nonmetal, for example: sodium chloride (NaCl).
Covalent bond: this type of bond occurs when atoms share one or more pairs of electrons, this happens between nonmetals, e.g.: the molecule of chlorine gas (Cl₂).

Apart from that, depending on the electronegativity difference, the covalent bonds are clasified in polar and non polar:

- Polar covalent bond: the difference of electronegativity is important but less than an ionic bond (between 0 and 2).

- Non polar covalent bond: this occurs when the atoms forming bonds have the same electronegativity.

So, analyzing the statement, if we have two nonmetals it is a covalent bond, and if the two nonmetals atoms have the same electronegativity the bond will be non polar.

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A piece of copper has a volume of 100.0cm^3 (3power) if the mass of the copper is 890g, what is the density of copper??

Bas_tet [7]

Density=mass/volume
Data:
mass=890 g
volume=100 cm³

Density=890 g/100 cm³=8.9 g/cm³

answer: the density of the copper is 8.9 g/cm³

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mr_godi [17]

Answer:

b. Disulfide bond

Explanation:

Covalent linkage or bonding is a type of bond in which electrons are shared between atoms in a compound or molecule. Each atom contributes to the shared electron.

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Determine the new pressure of a sample of hydrogen gas if the volume and moles are constant, the initial pressure is

masha68 [24]

Answer:

6.3 atm

Explanation:

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2 years ago

What makes a colloidal dispersion stable?

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16. Which of the following aqueous solutions contains the greatest number of 10ns.

gavmur [86]

Answer:

$\boxed{\text{b) 300.0 mL of 0.10 mol/L CaCl}_{2}}$

Explanation:

a) 400.0 mL of 0.10 M NaCl

(i) Moles of NaCl

$\text{ Moles of NaCl } = \text{0.4000 L NaCl} \times \dfrac{\text{0.10 mol NaCl}}{\text{1 L NaCl}} = \text{0.040 mol NaCl}$

(ii) Moles of ions

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We get 2 mol of ions from 1 mol of NaCl

$\text{Moles of ions } = \text{0.040 mol NaCl} \times \dfrac{\text{2 mol ions}}{\text{1 mol NaCl}} = \text{0.080 mol ions}$

b) 300.0 mL of 0.10 M CaCl₂

(i) Moles of CaCl₂

$\text{ Moles of CaCl}_{2} =\text{0.3000 L CaCl}_{2} \times \dfrac{\text{0.10 mol CaCl}_{2}}{\text{1 L CaCl}_{2}} = \text{0.030 mol CaCl}_{2}$

(ii) Moles of ions

CaCl₂(s) ⟶ Ca²⁺(aq) + 2Cl⁻(aq)

We get 3 mol of ions from 1 mol of CaCl₂

$\text{Moles of ions } = \text{0.030 mol CaCl}_{2} \times \dfrac{\text{3 mol ions}}{\text{1 mol CaCl}_{2}} = \text{0.090 mol ions}$

c) 200.0 mL of 0.10 M FeCl₃

(i) Moles of FeCl₃

$\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}$

(ii) Moles of ions

FeCl₂(s) ⟶Fe³⁺(aq) + 3Cl⁻(aq)

We get 4 mol of ions from 1 mol of FeCl₃

$\text{ Moles of FeCl}_{3} =\text{0.2000 L FeCl}_{3} \times \dfrac{\text{0.10 mol FeCl}_{3}}{\text{1 L FeCl}_{3}} = \text{0.020 mol FeCl}_{3}$

d) 200.0 mL of 0.10 M KBr

(i) Moles of KBr

$\text{ Moles of KBr} = \text{0.2000 L KBr} \times \dfrac{\text{0.10 mol KBr }}{\text{1 L KBr}} = \text{0.020 mol KBr}$

(ii) Moles of ions

KBr(s) ⟶ K⁺(aq) + Br⁻

We get 2 mol of ions from 1 mol of KBr

$\text{Moles of ions } = \text{0.020 mol KBr} \times \dfrac{\text{2 mol ions}}{\text{1 mol KBr}} = \text{0.040 mol ions}\\\\\text{We get the most ions ions from }\boxed{\textbf{300.0 mL of 0.10 mol/L CaCl}_{2}}$

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