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3 years ago

Which compound is formed from a network of bonded ions?

Chemistry

2 answers:

amid [387]3 years ago

8 0

Answer: D) salt, NaCl

Explanation: Bonded Ions are held together by the electrostatic force of attarction between the oppositely charged ions.

Thus Sodium Chloride is the only otion out of the given compounds that forms a network of bonded ions.

Sugar doesnot form this work becuase it gets breaks down ino CO2 and H2O which are uncharged molecules.

Methane is a molecule with covalent bonding thus cannot form ion bonded network.

TEA [102]3 years ago

4 0

N<span>etwork of bonded ions is established</span> between two atoms due to transfer of electrons from one atom to the other.
therefore,
Sodium chloride is formed from a network of bonded ions.
hence the right answer is <span>D) salt, NaCl</span>

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3 years ago

What stress will shift the following equilibrium system to the right?

Morgarella [4.7K]

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2 years ago

1.25mole Nocl was placed in a2.50l reaction chamber at 427c. After equilibrium was reached .1.10moles of Nocl remained.Calculate

Dmitry [639]

Answer:

Kc → 5.58×10⁻⁴

Explanation:

Equilibrium reaction is:

2NOCl (g) ⇄ 2NO (g) + Cl₂(g)

Initially we have 1.25 moles of NOCl

After the equilibrium, we have 1.10 moles. So, during the process:

(1.25 mol - 1.1 mol) = 0.15 moles have reacted.

As ratio are 2:2, and 2:1, 0.15 moles of NO and (0.15 /2) = 0.075 moles of chlorine, were produced in the equilibrium.

Finally in equilibrium we have: 1.10 moles of NOCl, 0.15 moles of NO and 0.075 moles of Cl₂. But these amount are not molar, so we need molar concentration in order to determine Kc:

1.10 mol /2.50L = 0.44 M

0.15 mol / /2.50L = 0.06 M

0.075 mol /2.50L = 0.03 M

Let's make expression for Kc → [Cl₂] . [NO]² / [NOCl]²

Kc = (0.03 . 0.06²) / 0.44² → 5.58×10⁻⁴

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3 years ago

What is the pH of a solution that results when 0.010 mol HNO3 is added to 500 mL of a solution that is 0.10 M in aqueous ammonia

kakasveta [241]

Here is the full question.

What is the pH of a solution that results when 0.010 mol HNO3 is added to 500 mL of a solution that is 0.10 M in aqueous ammonia and 0.20 M in ammonium nitrate? Assume no volume change. The Kb of ammonia is 1.8*10^-5.

Answer:

8.82

Explanation:

Volume = 500 mL = 0.500 L

Number of moles of NH₃ = 0.10 mole × 0.500 L = 0.050 moles

Number of moles of NH₄⁺ = 0.20 mole × 0.500 L = 0.10 moles

NH₃ + H⁺ ----------> NH₄⁺

In 0.010 mole of HNO₃ ;

Number of moles of NH₃ = 0.050 moles - 0.010 moles

= 0.040 moles

Number of moles of NH₄⁺ = 0.10 moles + 0.010 = 0.11 moles

Concentration of NH₃ = $\frac{number of moles}{volume}$