Equal is the answer
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Answer:
See explanation
Explanation:
The first step in this reaction is a unimolecular reaction. It involves the formation of the carbocation. This is so because tertiary alkyl halides only undergo substitution by SN1 mechanism due to sterric crowding.
The second step in the reaction is bi molecular. In this step, the carbocation now combines with the OH^- to yield the alcohol.
Net equation of the reaction is;
(CH3)3CBr + OH^- -------> (CH3)3COH + Br^-
The intermediate here is the carbocation, (CH3)3C^+
Answer:
kf = 1.16 x 10¹⁸
Explanation:
Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹
Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹
Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹
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Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r
ΔG°r = ΔG°1 + ΔG°2 + ΔG°3
ΔG°r = -42.9 - 35.8 - 24.3
ΔG°r = -103.0 kJmol⁻¹
ΔG°r = -RTlnKf
-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf
kf = e ^(-103,000/-8.31x298)
kf = e ^41.59
kf = 1.16 x 10¹⁸
Answer:
3.74 x 10²² particles
Explanation:
Given parameters:
Mass of compound = 1.43g
Molar mass of compound = 23g
Unknown:
Number of particles of sodium = ?
Solution:
To find the number of particles of Na in the compound, we need to obtain the mass of sodium from the total mass given;
Mass of sodium = 
= 
= 1.43g
Now find the number of moles of this amount of Na in the sample;
Number of moles = 
= 
= 0.062mole
Now;
1 mole of substance = 6.02 x 10²³ particles
0.062 mole of substance = 0.062 x 6.02 x 10²³ particles
= 3.74 x 10²² particles
