Primary alkyl halides tend to undergo the SN2 reaction mechanism in nucleophilic substitution since there is less steric hindrance for nucleophilic attack and the carbocations that they form are not as stable as those formed from tertiary alkyl halides.
1-bromopentane > 1-bromo 2-methylbutane > <span>1-bromo-3-methylbutane</span>> 2-bromo 2-methylbutane
Answer:
Br - C ≡ N
Explanation:
To draw the Lewis line-bond structure we need to bear in mind the octet rule, which states that in order to gain stability each <em>atom tends to share electrons until it has 8 electrons in its valence shell</em>.
- C has 4 e⁻ in its valence shell so it will form 4 covalent bonds.
- Br has 7 e⁻ in its valence shell so it will form 1 covalent bond.
- N has 5 e⁻ in its valence shell so it will form 3 covalent bonds.
The most stable structure that respects these premises is:
Br - C ≡ N
It does not have any H atom.
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
2. Because a cumulonimbus cloud is a towering vertical cloud
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
