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3 years ago

How many methylene groups are present in 2,4-dimethylhexane?

Chemistry

2 answers:

stepan [7]3 years ago

8 0

There are zero number of methylene groups in the substance 2,4-dimethylhexane. Methylene groups are the part of a molecule that contains a carbon atom that has a double bond or =CH. Looking at the structure of 2,4-dimethylhexane, you cannot see any carbon atom that has a double bond.

dusya [7]3 years ago

4 0

Answer:

Zero methylene groups

Explanation:

Hi, on one side you have the 2.4-dimethylhexane molecule as shown in the figure.

On the other side, you have the methylene group with the following structure:

=CH2

As can be seen there isn't any methylene group in the 2.4-dimethylhexane

Important: don't confuse the methylene group with the methyl group. Methyl group is -CH3 and there are two of them in this molecule.

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Why is Anton Von Leeuwenhoek important in the cell theory?

vesna_86 [32]

Answer:

He was the first scientist to observe and describe bacteria and protozoa by looking at a drop of water from a pound under a microscope. He also was the one to build the first compound microscope.

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3 years ago

. If 12 oranges cost $36.00, what is the cost for 20 oranges?

Olegator [25]

Answer:

60 dollars!

Explanation:

So, 12 oranges cost $36 dollars.

Now, we can solve this by finding the value for only one orange, then multiplying that by 20 to get our answer!

So, the value of one orange is $36 / 12 = $3. One orange is worth $3 dollars.

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3 years ago

A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the

Sedbober [7]

Answer: The final temperature will be $52.74^oC$

Explanation:

Calculating the heat released or absorbed for the process:

$q=m\times C\times (T_2-T_1)$

In a system, the total amount of heat released is equal to the total amount of heat absorbed.

$q_1=-q_2$

$m_1\times C\times (T_f-T_1)=-m_2\times C\times (T_f-T_2)$ ......(1)

where,

C = heat capacity of water = $4.184J/g^oC$

$m_1$ = mass of water of sample 1 = 100.0 g

$m_2$ = mass of water of sample 2 = 71.0 g

$T_f$ = final temperature of the system = ?

$T_1$ = initial temperature of water of sample 1 = $27^oC$

$T_2$ = initial temperature of the water of sample 2 = $89.0^oC$

Putting values in equation 1, we get:

$100.0\times 4.184\times (T_f-27)=-71.0\times 4.184\times (T_f-89)\\\\171T_f=9019\\\\T_f=\frac{9019}{171}=52.74^oC$

Hence, the final temperature will be $52.74^oC$

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3 years ago

What is the average mass of a single silicon atom in grams?

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then

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4.66381*10^-23 g per atom

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4 years ago

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