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Sergio [31]
1 year ago
5

propane (c3h8) is burned with air. for each case, obtain the balanced reaction equation for complete combustion a. with the theo

retical amount of air. b. with 20% excess air. c. with 20% excess air, but only 90% of the propane being con- sumed in the reaction.
Chemistry
1 answer:
andrew-mc [135]1 year ago
8 0

Combustion of propane is the exothermic reaction

Combustion reactions are reactions in which a substance reacts with an excess of air or oxygen to form products. the combustion of hydrocarbons in the presence of stoichiometric or excess levels of oxygen forms carbon dioxide and water and with the release of heat the molecular formula of propane is C₃H₈. one mole of propane needs five moles of oxygen for complete oxidation to form carbon dioxide and water.

The balanced equation for combustion of propane is:

C₃H₈(g) + 5O₂(g)→3CO₂(g)+4H₂O(g)

The products are 3 moles of CO₂ and 4 moles of H₂O for every mole of the alkane burnt completely. propane combust completely

Know more about combustion of propane

brainly.com/question/28463327

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Answer:- Frequency is 4.80*10^1^4Hz .

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where, \nu is frequency, c is speed of light and \lambda is the wavelength.

Value of speed of light is \frac{3.0*10^8m}{s} . We need to convert the given wavelength from nm to m.

1nm=10^-^9m

So, 625nm(\frac{10^-^9m}{1nm})

= 6.25*10^-^7m

Let's plug in the values in the formula and calculate the frequency:

\nu =\frac{3.0*10^8m.s^-^1}{6.25*10^-^7m}

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The primary gas in a volcano is: water vapor carbon dioxide sulfur dioxide nitrogen
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In a solution of pure water, the dissociation of water can be expressed by the following: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) T
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Answer:

  • [H₃O⁺] = 2.90 × 10⁻¹⁰ M

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1)<u><em> Ionization equilibrium equation: given</em></u>

  • H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>

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<u>3) Stoichiometric mole ratio:</u>

As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:

  • [H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M

  • ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷  × 1.0 × 10⁻⁷  = 1.0 × 10⁻¹⁴ M

<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>

Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:

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Then you can substitute the known values and solve for the unknown:

  • 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M

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As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.

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