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FromTheMoon [43]
4 years ago
9

What is the balanced equation for the combustion of sulfur

Chemistry
1 answer:
Alex17521 [72]4 years ago
3 0

Answer:

There are three possible chemical equations for the combustion of sulfur:

  • 2S (s)  + O₂ (g)  → 2SO (g)

  • S (s) + O₂ (g) → SO₂ (g)

  • 2S (s) + 3O₂ (g) → 2SO₃ (g)

Explanation:

<em>Combustion</em> is a reaction with oxygen. The products of the reaction are oxides, and energy is released in the form of heat and light.

<em>Sulfur</em> iis a nonmetal, so the oxide formed is a nonmetal oxide.

The most common oxidation numbers of sulfur are -2, + 2, + 4, and + 6.

The combination of sulfur with oxygen may be only with the positive oxidation numbers (+2, + 4, and +6).

Then you have three different equations for sulfur combustion:

<u>1) Oxidation number +2:</u>

  • S(s) + O₂(g) → SO(g)

Which when balanced is: 2S(g) + O₂(g) → 2SO(g)

<u>2) Oxitation number +4:</u>

  • S(s) + O₂(g) → SO₂(g)

That equation is already balanced.

<u>3) Oxidation number +6:</u>

  • S(s) + O₂(g) → SO₃(g)

Which when balanced is: 2S(s) + 3O₂(g) → 2SO₃(g)

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What is the ph of a 0.25 m solution of c6h5nh2 given that its kb is 1.8 x 10-6?
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<h3>How do we calculate pH of weak base?</h3>

pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

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Initial:                     0.25                           0            0

Change:                    -x                             x             x

Equilibrium:        0.25-x                           x             x

Base dissociation constant will be calculated as:
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x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

pH = 3.13

Hence pH of the solution is 3.13.

To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361

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How to separate crystals from solution​
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If a saturated hot solution is allowed to cool, the solute is no longer soluble in the solvent and forms crystals of pure compound. Impurities are excluded from the growing crystals and the pure solid crystals can be separated from the dissolved impurities by filtration.

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