Answer:
Both reactions share a common intermediate and differ only in the leaving group
Explanation:
The elimination reaction of tertiary alkyl halides usually occur by E1 mechanism. In E1 mechanism, the substrate undergoes ionization leading to the loss of a leaving group and formation of a carbocation.
Loss of a proton from the carbocation completes the reaction mechanism yielding the desired alkene.
In the cases of t-butanol and t-butyl bromide, the mechanism is the same. The both reactions proceed by E1 mechanism. The leaving groups in each case are water and chloride ion respectively.
Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate
Answer:
2
Explanation:
1. The dew is formed when the water vapor at the atmosphere contacts the leaves, which are at a low temperature, so, the vapor temperature decreases, and the liquid is formed. So, it's a gas to liquid change.
2. Ice cubes are at the solid-state, thus this transformation is solid to a liquid change.
3. The cold juice is at a low temperature, so when the water vapor of the air contacts with the glass, its temperature decreases, and its change to a liquid phase. So, it's a gas to liquid change.
4. The evaporated water from the Earth's surface goes to the atmosphere, and, at high altitudes, the temperature is low, so the water vapor condenses and the drops get closer together forming the clouds. So, it's a gas to a liquid change.
Answer:
5.702 mol K₂SO₄
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Compounds
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 993.6 g K₂SO₄
[Solve] moles K₂SO₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of K: 39.10 g/mol
[PT] Molar Mass of S: 32.07 g/mol
[PT] Molar mass of O: 16.00 g/mol
Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 4 sig figs.</em>
5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄
