Ethyl Butanoate when treated with a base looses a proton which is more acidic in nature. In this case ethyl butanoate acts as a lowery bronsted acid. It donated the more acidic proton to lowery bronsted base.
Among the protons attached to different carbon atoms the hydrogen atoms next to carbonyl functional group (labelled as red in attached picture) are more acidic in nature and are readily donated on treatment with strong base. These hydrogen atoms are also called alpha hydrogen name after their position.
Acidity of Alpha Hydrogens:
The driving force behind the acidity of alpha hydrogens is the formation of enolates. The enolate formed is resonance stabilized. This stability is the main reason for the said acidity. The pKa value of said protons is approximately 20-25. Hence, the enolate formed is infact the conjugate base and can act as neucleophile.
The question asks about the average kinetic energy so it is not related with mass. We only need to compare the temperature. The higher temperature is, the higher kinetic energy is. So the answer is (2).
Molybdenum Arsenide
I think that’s right but not %100 sure
According to the equation of molarity:
Molarity= no.of moles / volume per liter of Solution
when we have the molarity=0.58 M and the beaker at 150mL so V (per liter) = 150mL/1000 = 0.150 L
by substitution:
∴ No.of moles = Molarity * Volume of solution (per liter)
= 0.58 * 0.150 = 0.087 Moles
Answer:
Explanation:
<u>1) Balanced chemical equation:</u>
<u>2) Mole ratio:</u>
- 2 mol S : 3 mol O₂ : 2 mol SO₃
<u>3) Limiting reactant:</u>
n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂
n = 7.0 g / 32.065 g/mol = 0.2183 mol S
Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859
Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5
Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.
<u>4) Calcuate theoretical yield (using the limiting reactant):</u>
- 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃
- x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃
<u>5) Yield in grams:</u>
- mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g
<u>6) </u><em><u>Percent yield:</u></em>
- Percent yield, % = (actual yield / theoretical yield) × 100
- % = (7.9 g / 10.0 g) × 100 = 79%
