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3 years ago

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CH3-CH2 CH2 CH2-CH3 clasification?

1 answer:

adoni [48]3 years ago

6 0

Answer:

CH2-CH4

Explanation:

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Identify the catalyst in this reaction, explain how

telo118 [61]

Question:

Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:

equations

Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.

Answer:

NO

It is present but not consumed

NO Lowers the activation energy of the reaction

Explanation:

A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction

Therefore, whereby NO is not consumed, it is the catalyst

It functions by lowering the activation energy

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3 years ago

Water from a riverbed carries sediment downstream. The water pressure cuts deep into the riverbed, creating a deep, narrow chann

atroni [7]

Answer:

A canyon has formed.

Explanation:

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2 years ago

During a workout, you set the treadmill to a pace of 55.0m/min. How many minutes will you walk if you cover a distance of 7500ft

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55 m=180.446 ft.
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3 years ago

What shape is formed from water?

alisha [4.7K]

Answer:

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Explanation:

.....................…...…

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3 years ago

Read 2 more answers

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

3 years ago