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4 years ago

Draw the structure of 2,4,4,5-tetramethyl-2-hexene.

Chemistry

1 answer:

bogdanovich [222]4 years ago

3 0

Answer:

The answer to your question is given below

Explanation:

2,4,4,5-tetramethyl-2-hexene.

To draw the structure of the above compound, we must bear the following in mind:

1. The compound is an alkene i.e it contains carbon to carbon double bond (C=C)

2. The parent name of the compound is hexene i.e it contain 6 carbon atoms.

3. The double is located at carbon 2.

4. The substituent group attached to the compound is methly.

5. There are four methly group attached to the compound of which one is located at carbon 2, two at carbon 4 and one at 5.

With the above information, we can easily draw the structure of the compound.

Please see attached photo for the structure of the compound.

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Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benz

Hoochie [10]

Answer:

% ionization for benzoic acid = 0.08%

% ionization for sodium benzoate = 2.5%

The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions

Explanation:

Ka for pure water = 1.0 * 10-⁷

Ka for sodium benzoate = 6.5*10-⁵

1. For benzoic acid (C6H5COOH)

C6H5COOH ==== C6H5COO‐ + H+

0.15M 0 0

0.15-x x x

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = [X] [X] / 0.15 - X

1.0*10-⁷ = [X]² / 0.15 - x

But x is negligible compared to 0.15,

(1.0*10-⁷)*0.15 = x²

Take square root of both sides,

X = 1.22 * 10-⁴

% ionization = ( [H+] / [C6H5COOH] ) * 100

% ionization = (1.22*10-⁷ / 0.15) * 100

% ionization = 0.08%

2. For C6H5COONa

Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions

Ka for C6H5COONa = 6.5*10-⁵

6.5*10-⁵ = [X]² / (0.10 - X)

Cross-multiply both sides;

(6.5*10-⁵ * 0.10) = X²

Take square root of both side,

X= 2.5*10-³

% ionization = (2.5*10-³ / 0.10) *100

% ionization = 2.5%

5 0

3 years ago

What is the pH of a solution within a solution with pH = 4.50? [H +] = 3.25×10-6 M?

Lisa [10]

Answer: 5.48

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution.

Mathematically, pH = - log(H+)

where H+ represent the concentration of hydrogen ion

So, to get the pH of the solution with [H +] = 3.25×10-6 M:

Apply, pH = -log(H+)

pH = - log (3.25×10-6 M)

pH = - ( -5.48)

(Note that the minus signs will cancel out each other)

Therefore pH = 5.48

Now we know that the pH of the solution with hydrogen ion concentration of 3.25×10-6 M is 5.48 (i.e slightly acidic)

Thus, we can finally say 5.48 is the pH of the solution within a solution with pH = 4.50

6 0

4 years ago

Describe the composition of the nucleus of an atom. A) It is composed of protons. B) It is composed of protons and neutrons. C)

Kryger [21]

He atom<span> consists of a tiny </span>nucleus<span> surrounded by moving electrons. </span>

3 0

3 years ago

Read 2 more answers

What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300

Alexxandr [17]

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: $T_{1}$ = 237 K, $P_{1}$ = 101.33 kPa, $V_{1}$ = ?

$T_{2}$ = 300 K, $P_{2}$ = 151.99 kPa, $V_{2}$ = 0.50 L

Formula used is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L$

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

8 0

3 years ago

You are dating Moon rocks based on their proportions of uranium-238 (half-life of about 4.5 billion years) and its ultimate deca

Aliun [14]

1. Use the formula:

$-kt = ln[\frac{FinalAmount}{InitialAmount}]$

$k = ln2 / half-life$

So, $t = - \frac{4.5}{ln2} \times ln[\frac{0.55}{1}]$

t = 3.8 billion yr

2. Same formula:

$t = - \frac{4.5}{ln2} \times ln[\frac{0.63}{1}]$

t = 2.93 billion yr.

8 0

3 years ago