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bixtya [17]
3 years ago
8

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

Chemistry
2 answers:
Semmy [17]3 years ago
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 <u>between metallic atoms</u> like Cu, Al, Au, etc.
  • Ionic bond: this type of bond is formed <u>between ions</u>, that means that it is necessary the presence of <u>a cation</u> (ion with positive charge) and <u>and an anion </u>(ion with negative charge) and when the atoms has a <u>very high difference of electronegativity</u> (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 <u>atoms share one or more pairs of electrons</u>, this happens between nonmetals, e.g.: the molecule of chlorine gas (Cl₂).

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

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

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

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

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Density=mass/volume
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Density=890 g/100 cm³=8.9 g/cm³

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

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\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

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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

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\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⁻

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\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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