Identify the intermediate leads to the major product for the reaction of 2-methyl-2-butene with hydrogen bromide in the presence
1 answer:
Answer:
II
Explanation:
We must have a good idea of the fact that there are two mechanisms that come into play when we are discussing about the addition of hydrogen halides to alkenes. The first is the ionic mechanism and the second is the radical mechanism.
The ionic mechanism is accounted for by the Markovnikov rule while the radical mechanism occurs in the presence of peroxides and is generally referred to as anti Markovnikov addition.
The intermediate in anti Markovnikov addition involves the most stable radical, in this case, it is a tertiary radical as shown in the images attached. The most stable radical is II hence it leads to the major product shown in the other image.
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<h3>Answer:</h3>
2100 g Fe₂(SO₄)₃
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Step 1: Define</u>
5.26 mol Fe₂(SO₄)₃
<u>Step 2: Identify Conversions</u>
Molar Mass of Fe - 55.85 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Fe₂(SO₄)₃ - 2(55.85) + 3(32.07) + 12(16.00) = 399.91 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2103.53 g Fe₂(SO₄)₃ ≈ 2100 g Fe₂(SO₄)₃
Answer:
The number of moles in 89 grams of KCl is approximately 1.193 moles
Explanation:
A mole of an element or compound, is the quantity of the element or compound that contains 6.022 × 10²³ molecules of the element or compound, such that the molar mass of one mole is dependent on the mass of each molecule of the element or compound that make up the 6.022 × 10²³ particles
The given parameters are;
The mass of the KCl = 89 grams
The Molar mass, MW, of potassium, K = 39.1 g/mole
The Molar mass, MW, of chlorine, Cl = 35.5 g/mole
Therefore, given that KCl, has one K and one Cl, per mole, we have;
The Molar mass, MW, of potassium chloride, KCl = (39.1 + 35.5) g/mole = 74.6 g/mole
Therefore, the number of moles of KCl in 89 g, 'n' is given as follows;
n = (89 g)/(74.6 g/mole) ≈ 1.193 moles
The number of moles in 89 grams of KCl, n ≈ 1.193 moles.