<u><em>The process of how we would obtain </em></u><u><em>ethanal</em></u><u><em> </em></u><u><em>free</em></u><u><em> from ethanol is described in the explanations below. </em></u>
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- In Chemistry, Ethanol undergoes oxidation in the presence of sodium dichromate plus sulphuric acid to yield ethanal and water.
The procedure for achieving this in the laboratory is as follows;
- Step 1; Measure a quantity of a solution of sodium dichromate acidified in a dilute sulphuric acid and pour into a test tube.
- Step 2; Add excess <em>ethanol</em>. This is because if we don't do so there will be plenty of oxidizing agent to carry out a second operation which changes the aldehyde to ethanoic acid. However, we need only the aldehyde.
- Step 3; When the aldehyde ethanal begins to form which will be evident by the change in the colour of solution from <em>orange to green</em>, then the mixture should be distilled from the test tube and tbethe aldehyde collevted so that it doesn't undergo additional oxidation into ethanoic acid.
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Answer:
By visiting other households with cats.
Explanation:
This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.
Answer:
40.95 L
Explanation:
We'll begin by calculating the number of mole in 12.3 g of O₂. This can be obtained as follow:
Mass of O₂ = 12.3 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mole of O₂ =?
Mole = mass / Molar mass
Mole of O₂ = 12.3 / 32
Mole of O₂ = 0.384 mole
Next, we shall determine the volume occupied by 0.384 mole of O₂. This can be obtained as follow:
Number of mole (n) of O₂ = 0.384 mole
Pressure (P) = 1 atm
Temperature (T) = 273 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of O₂ =?
PV = nRT
1 × V = 0.384 × 0.0821 × 273
V = 0.384 × 0.0821 × 273
V = 8.6 L
Thus, the volume of O₂ is 8.6 L
Finally, we shall determine the volume of air that contains 8.6 L of O₂. This can be obtained as follow:
Volume of O₂ = 8.6 L
Percentage of O₂ in air = 21%
Volume of air =?
Percentage of O₂ = Vol of O₂ / Vol of air × 100
21% = 8.6 / Vol of air
21 / 100 = 8.6 / Vol of air
Cross multiply
21 × Vol of air = 100 × 8.6
21 × Vol of air = 860
Divide both side by 21
Volume of air = 860 / 21
Volume of air = 40.95 L
Therefore, the volume of air is 40.95 L.
