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Aleksandr [31]
3 years ago
15

When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1-butene is produced when potass

ium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base?

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
6 0

Answer:

In between 45% to 70%.

Explanation:

Find the given attachment.

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k0ka [10]

Answer:

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Explanation:

3 0
3 years ago
How many electrons are in the outermost principal energy level of an atom of carbon in the ground state
Agata [3.3K]

Answer:

four electrons

Explanation:

Let us attempt to write the electronic configuration of carbon in the ground state. This electronic configuration will now be;

C- 1s2 2s2 2p2

The outermost principal energy level of carbon is the n=2 level which houses the 2s2 and 2p2, making a total of four electrons in the outermost principal energy level of an atom of carbon in the ground state.

5 0
3 years ago
A laser is emitting photons with a wavelength of 639.8 nm. What is the energy for 1 mole of these photons? For Planck's constant
PSYCHO15rus [73]

Answer:

The energy for 1 mole of these photons is E = 31 × 10^{-23} \frac{KJ}{mol}  

Explanation:

Given data

Wavelength \lambda = 639.8 × 10^{-9} m

Plank constant h = 6.626 × 10^{-34} J sec

Speed of light c = 3 × 10^{8} meter per second

We know that Energy  of a photon is given by

E = \frac{h c}{\lambda}

Value of ( h c ) = 6.626 × 10^{-34}  × 3 × 10^{8} = 19.878 × 10^{-26}

\lambda = 639.8 × 10^{-9} m

Now Energy  of a photon

E =( \frac{19.878}{639.8}) 10^{-17}

E = 0.031 × 10^{-17} Joule per mole

E = 31 × 10^{-23} \frac{KJ}{mol}  

Therefore the energy for 1 mole of these photons is E = 31 × 10^{-23} \frac{KJ}{mol}  

4 0
4 years ago
An octapeptide composed of four repeating glycylalanyl units has one free amino group on a glycyl residue and one free carboxyl
SCORPION-xisa [38]

Answer:

Following are the solution to this question:

Explanation:

The octapeptides are: gly-ala-gly-ala-gly-ala-gly-ala

Its structure includes free glycine amino, and alanine residues free carboxylic. Its peptides are customarily extracted from the left side free NH2 residues as well as the right side free -COOH residues only in the same direction, the sequence is provided.

please find the structure in the attached file.

5 0
3 years ago
Express the following numbers in their expanded forms
Helga [31]
I believe its
a)224000
b)0.000000015
c)0.655
d)9000
5 0
3 years ago
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