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Elina [12.6K]
2 years ago
7

How many grams of Cl2 are in 1.20 x 1024 Cl atoms?

Chemistry
1 answer:
Vaselesa [24]2 years ago
5 0
<h3>Answer:</h3>

70.906 g

<h3>Explanation:</h3>

We are given;

  • Atoms of Chlorine = 1.2 × 10^24 atoms

We are required to calculate the mass of Chlorine

  • We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
  • That is , 1 mole of an element = 6.022 × 10^23 atoms
  • Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms

But since Chlorine gas is a molecule;

  • 1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms

But, molar mass of Chlorine gas = 70.906 g/mol

Then;

70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms

                                          = 1.20 × 10^24 atoms

Thus;

For 1.2 × 10^24 atoms ;

= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)

<h3>=  70.906 g </h3>

Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g

=  

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2 years ago
A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
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Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

4 0
3 years ago
What is the redox half equation for 3Ag2S + 2Al --&gt; 6Ag + Al2, and identity which material is oxidized and which is reduced?
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Al is oxidized while Ag is reduced.

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Reduction half equation;

6Ag^+ + 6e -------> 6Ag

Overall redox reaction equation;

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Hence; Al is oxidized while Ag is reduced.

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