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Free_Kalibri [48]
4 years ago
11

What percentage of gases are in the earth?

Chemistry
1 answer:
ioda4 years ago
5 0

Answer:

By volume, dry air contains 78.09% nitrogen, 20.95% oxygen, 0.93% argon, 0.04% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% at sea level, and 0.4% over the entire atmosphere.

Explanation:

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Determine which of these properties would distinguish these two substances: (a) boiling point; (b) combustion analysis results;
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(a) boiling point

(d) density at a given temperature and pressure.

Explanation:

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3 years ago
At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration
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Answer:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.03901 mol/liter
  • [Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

  • Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
  • Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

  • Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
  • Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

            H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

I            0.0481      0.0326            0

C              -x                 -x              +x

E          0.0481-x    0.0326-x         x

<u />

<u>4. Equilibrium expression</u>

       K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}

     0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}

<u />

<u>5. Solve:</u>

            x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0

Use the quadatic formula:

x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}

The positive result is x = 0.00909

Thus the concentrations are:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
  • [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

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