What is the mass in grams of 2.6 moles of angelic acid?
1 answer:
Answer:
260.34g
Explanation:
First, you need to know what angelic acid is comprised of. It is written as C₅H₈O₂.
In order to solve for the mass of 2.6 moles of angelic acid, you need the mass of 1 mole of angelic acid. This can be found by adding the masses from the periodic table, like shown below:
5 carbon atoms = (5)(12.01g) = 60.05g
8 hydrogen atoms = (8)(1.01) = 8.08g
2 oxygen atoms = (2)(16) = 32g
angelic acid = 60.05 + 8.08 + 32 = 100.13g
Then, set up a basic stoichiometric equation and solve. The units should cancel out.
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Answer:
1. When observing a positive test for the jones reagent and negative for the Lucas test, it indicates that it is in the presence of a primary alcohol.
Jones reagent behaves like a strong oxidant, where it transforms the primary alcohols into carboxylic acids and the secondary alcohols into ketones. Tertiary alcohols do not react.
With the Lucas test, tertiary alcohols react immediately producing turbidity, while secondary alcohols do so in five minutes. Primary alcohols do not react significantly with Lucas reagent at room temperature.
2. No reaction (See the attached drawing)
3. (see the attached drawing)
Answer:
M
Explanation:
Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.
Partial pressure = Henry's constant x molar concentration
0.21 = Hx1.38x
H =
H = 152.17 atm/M
For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:
P = 0.21*0.875 = 0.1837 atm
Then, the molar concentration [O2], will be:
P = Hx[O2]
0.1837 = 152.17x[O2]
[O2] = 0.1837/15.17
[O2] = M