Answer:
H =2; I = 2; J = 2
Explanation:
Carbon is element 6 in the Periodic Table.
Start at element 1 (H) and count from left to right until you reach element 6 (C).
You get the electron configuration
C: 1s² 2s²2p².
Thus,
H =2; I = 2; J = 2
Answer:
I would expect to extract the acetic acid.
Explanation:
In the first step, since we are adding a concentrated acid,<u> it will react with the bases present in the mixture (diethylamine and ammonia) </u><u>forming salts</u><u>, </u><u>which are soluble in water</u>. Therefore, after draining the aqueous layer, we will have phenol and acetic acid left in the organic layer.
In the second step, we are adding a diluted base, so it will react with a strong acid. This compound is acetic acid, and its salt will be present in the aqueous layer. Phenol will be left on the organic layer.
Answer:
The density of acetic acid at 30°C = 1.0354_g/mL
Explanation:
specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)
Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)
= 0.9956 g/mL × 1.040
= 1.0354_g/mL
Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.
Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure
Acetic acid is 60.05 grams/mole. In 1 liter of vinegar or 1000 ml there would be 0.046% of acetic acid = 46 ml x 1.005g/ml = 46.23 grams/60.05 grams= 0.77 moles per litre of vinegar.This then would be the concentration of acetic acid in for example 1 liter of vinegar.
Answer:
3 different elements are present in the compound.
