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2 years ago

Which compound contains both covalent and ionic bonds? a. NCl3 b. NaF c. HCN d. NH4OH

Chemistry

1 answer:

Marina86 [1]2 years ago

8 0

<h2> $NH_4OH$ contains both covalent and ionic bonds.</h2>

Explanation:

A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals.

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and the element which accepts the electrons is known as electronegative element or non metal.

a. $NCl_3$ contain covalent bonds as they are made up of non metals only.

b. $NaF$ contain ionic bonds as they are made up of sodium metal and fluorine non metal.

c. $HCN$ contain covalent bonds as they are made up of non metals only.

d. $NH_4OH$ contain ionic bonds between $NH_4$ and $OH$ and covalent between N and H in $NH_4^+$

Learn more about ionic and covalent bonds

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Which of the following is not part of the proper protocol for using acids and bases?

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Explanation:

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4.How element gold is different from oxygen?

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Answer:

Gold is a metal, more specifically a transition metal, whereas Oxygen is a nonmetal, more specifically a reactive nonmetal. Using this information, you can compare and contrast metals, nonmetals, and metalloids.

Metals are:

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Mostly silver or gray in color

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Malleable – able to be hammered into a thin sheet

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Explanation:

Reccomend this site for questions llike these: https://ptable.com/#Properties

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2 years ago

To what category of elements does an element belong if it's a poor conductor of electricity

Karo-lina-s [1.5K]

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3 years ago

A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resu

Deffense [45]

Answer:

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown non-electrolyte = 2.17 grams

Mass of chloroform = 225.0 grams

The freezing point of the resulting solution is –64.2 °C

The freezing point of pure chloroform is – 63.5°C

kf = 4.68°C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ ΔT = The freezing point depression = T (pure solvent) − T(solution) = -63.5°C + 64.2 °C = 0.7 °C

⇒i = the van't Hoff factor = non-electrolyte = 1

⇒ kf = the freezing point depression constant = 4.68 °C/m

⇒ m = molality = moles unknown non-electrolyte / mass chloroform

0.7 °C = 1 * 4.68 °C/m * m

m = 0.150 molal

Step 3: Calculate moles unknown non-electrolyte

molality = moles unknown non-electrolyte / mass chloroform

Moles unknown non-electrolyte = 0.150 molal * 0.225 kg

Moles unknown non-electrolyte = 0.03375 moles

Step 4: Calculate molecular mass unknown non-electrolyte

Molar mass = mass / moles

Molar mass = 2.17 grams / 0.03375 moles

Molar mass = 64.3 g/mol

The molar mass of the unknown non-electrolyte is 64.3 g/mol

6 0

3 years ago