The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Answer:
24.9 L Ar
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Aqueous Solutions</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 40.0 g Ar
[Solve] L Ar
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Ar - 39.95 g/mol
[STP] 22.4 L = 1 mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
24.9235 L Ar ≈ 24.9 L Ar
Answer:
The nervous system helps all the parts of the body to communicate with each other. It also reacts to changes both outside and inside the body. The nervous system uses both electrical and chemical means to send and receive messages
Explanation:
Answer:
Rate constant k = 1.57*10⁻⁵ s⁻¹
Explanation:
Given reaction:

Expt [A] M [B] M Rate [M/s]
1 3.40 4.16 1.82*10^-4
2 4.59 4.16 3.32*10^-4
3. 3.40 5.46 1.82*10^-4
![Rate = k[A]^{x}[B]^{y}](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D)
where k = rate constant
x and y are the orders wrt to A and B
To find x:
Divide rate of expt 2 by expt 1
![\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2](https://tex.z-dn.net/?f=%5Cfrac%7B3.32%2A10%5E%7B-4%7D%20%7D%7B1.82%2A10%5E%7B-4%7D%20%7D%20%3D%5Cfrac%7B%5B4.59%5D%5E%7Bx%7D%20%5B4.16%5D%5E%7By%7D%20%7D%7B%5B3.40%5D%5E%7Bx%7D%20%5B4.16%5D%5E%7By%7D%20%7D%5C%5C%5C%5Cx%20%3D2)
To find y:
Divide rate of expt 3 by expt 1
![\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0](https://tex.z-dn.net/?f=%5Cfrac%7B1.82%2A10%5E%7B-4%7D%20%7D%7B1.82%2A10%5E%7B-4%7D%20%7D%20%3D%5Cfrac%7B%5B3.40%5D%5E%7Bx%7D%20%5B5.46%5D%5E%7By%7D%20%7D%7B%5B3.40%5D%5E%7Bx%7D%20%5B4.16%5D%5E%7By%7D%20%7D%5C%5C%5C%5Cy%20%3D0)
Therefore: x = 2, y = 0
![Rate = k[A]^{2}[B]^{0}](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BA%5D%5E%7B2%7D%5BB%5D%5E%7B0%7D)
To find k
Use rate for expt 1:
![k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7BRate1%7D%7B%5BA%5D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B1.82%2A10%5E%7B-4%7DM%2Fs%20%7D%7B%5B3.40%5D%5E%7B2%7D%20%7D%20%3D1.57%2A10%5E%7B-5%7D%20s-1)