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poizon [28]
3 years ago
15

Given: logK=nE∘0.0592 What is the value K for this redox reaction? Zn2+(aq) + 2 Cl−(aq) → Zn(s) + Cl2(

Chemistry
1 answer:
anygoal [31]3 years ago
8 0

<em>K</em> = 2.4 × 10^(-72)

<em>Step 1</em>. Determine the <em>value of n </em>

Zn^(2+) + 2e^(-) → Zn

2Cl^(-) → Cl_2 + 2e^(-)

Zn^(2+) + 2Cl^(-) → Zn + Cl_2

∴ <em>n</em> = 2

<em>Step 2</em>. Calculate <em>K</em>

log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62

<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)

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7 0
2 years ago
What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

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3 0
3 years ago
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Answer:

241 nm, it is UV light range

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The minimum energy needed required to break the oxygen-oxygen bond = 495 kJ/mol

Energy needed for 1 molecule of oxygen = 495 × 10³ J / avogadro's constant = 495 × 10³ J  / ( 6.02 × 10²³) = 8.223 × 10⁻¹⁹ J

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4 0
2 years ago
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