Answer:
A.) 4.0
Explanation:
The general equilibrium expression looks like this:
![K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%20%7D)
In this expression,
-----> K = equilibrium constant
-----> uppercase letters = molarity
-----> lowercase letters = balanced equation coefficients
In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.
<----- Equilibrium expression
![K = \frac{[2 M] [2 M]}{[1 M] [1 M] }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5B2%20M%5D%20%5B2%20M%5D%7D%7B%5B1%20M%5D%20%5B1%20M%5D%20%7D)
<----- Insert molarities
<----- Multiply
<----- Divide
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Answer:
a. 3-methylbutan-2-ol
b. 2-methylcyclohexan-1-ol
Explanation:
For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>
In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.
For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.
See figure 1
I hope it helps!
Answer:
shorter wavelength = alpha wave
Explanation:
Given that,
The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz.
We need to compare the wavelengths of these two waves.
For alpha wave,
For beta wave,
From the above calculations, we find that the wavelength of the alpha wave is shorter than the wavelength of the beta wave.
Answer:
N,N-dimethylacetamide is formed.
Explanation:
- It is an example of a nucleophilic addition-elimination reaction. Here dimethylamine acts as a nucleophile.
- In the first step, dimethyl amine gives nucleophilic addition reaction at carbonyl center of acetyl chloride.
- In the second step, removal of Cl atoms occurs.
- In the third step, deprotonation takes place from amino group to produce N,N-dimethylacetamide.
- Full reaction mechanism has been shown below.
