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xeze [42]
3 years ago
13

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the format

ion of gaseous sulfur trioxide. Assume- ing that the reaction goes to completion, what is the ra- tio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?
Chemistry
1 answer:
Aneli [31]3 years ago
7 0

Answer:

Ratio of the final volume of the gas mixture to the initial volume of the gas mixture is 2 : 3

Explanation:

Before proceeding, we have to write out the balanced chemical equation and this is given as;

sulfur dioxide gas + oxygen gas → gaseous sulfur trioxide

SO₂ + O₂  → SO₃

Upon balancing the equation, we have;

2SO₂ + O₂  → 2SO₃

From the equation, we can tell that 2 moles of SO₂ react with 1 moles of O₂ to give 2 moles of SO₃.

Applying gay lusaac's law;

Gay Lussac's Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

This means 2 vol of SO₂ react with 1 vol of O₂ to give 2 vol of SO₃.

Initial volume of gas mixture (No product has been formed) = 2 vol of SO₂ + 1 vol of O₂ = 3 vol

Final volume of gas mixture (Reactants has been exhausted and product formed) = 2 vol of SO₃

Ratio (Final : Initial);

2 : 3

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A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb
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We are given the number of moles:

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A 0.180 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilib
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Answer:

1.09 x 10⁻⁴ M

Explanation:

The equation of the reaction in given by

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From the equation,

1 mole of Ni²⁺ reacts with 6 moles of aqueous NH₃ to give 1mole of Ni(NH3)

therefore

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                       K  = [Ni(NH₃)₆} / [Ni²⁺] 6[NH₃]

   5.5 x 10⁸          =  0.18 M / [Ni²⁺] [0.12]⁶

                   [Ni²⁺]= 0.18 M / (5.5 x 10⁸) (2.986 x 10⁻⁶)

                            =0.18 M / 0.00001642

                           = 1.09 x 10⁻⁴ M

[Ni]²⁺                  =   1.09 x 10⁻⁴ M

Hence the concentration of Ni²⁺ is  1.09 x 10⁻⁴ M

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