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3 years ago

Which of the following contains both ionic and covalent bonding?

2 answers:

7 0

I believe the correct answer from the choices listed above is the first option. The compound that contains both ionic and covalent bonding is KOH or potassium hydroxide. It contains one covalent<span> (O-H) and one that is </span>ionic<span> (K-O). Hope this helps.</span>

pantera1 [17]3 years ago

6 0

Answer: KOH

Explanation:

Being KOH the only not binary compound it is the candidate to be the one with two different type of bondings.

In fact, KOH is a ionic compound because the cation K⁺ forms a ionic bond with the anion OH⁻.

On the other hand, the oxigen and hydrogen atoms in OH⁻ are covalently bonded.

The ionic bond is the result of the union of two atoms with a high elecronegativity difference, while the covalent bonding is the result of a combination of two atoms whose electronegativities is not so different.

Being K and alkali metal (the elements with lowest electronegativity) and OH⁻ an anion, they easily form the ionic bond.

From tables, the electronegativity of H is 2.20 and the electronegativity of O is 3.16, which leads to a difference of 3.16 - 2.20 = 0.96. This difference is not enough to form ions but covalent bonds. That is why the bond O-H is covalent.

When you analyze the bonds of the other choices, N-O, C-H, and H-Br, y ou conclude that the electronegativities lead to covalent bonds too.

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3 years ago

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2 years ago

Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc

Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH + NaOH ⇒ HCOONa + H₂O

mol HCOOH =

$\tt 20~ml\times 0.2~M=4~mlmol$

mol NaOH =

$\tt 25~ml\times 0.2~M=5~mlmol$

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH + NaOH ⇒ HCOONa + H₂O

4 5

4 4 4 4

0 1 1 1

Concentration of [OH⁻] from NaOH :

$\tt \dfrac{1~mlmol}{20+25~ml}=0.02$

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

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3 years ago