What type of hydrocarbon is 2-octene, and what about the structure allows 2-octene to be classified as this type of hydrocarbon?

Hello! I need a bit of help and Im kinda in a bit of a hurry so please help! Thank you :D

Scorpion4ik [409]

The first one is substance 3
The second one is Oxygen, Helium, and carbon dioxide
The third one is the oil floats on top of the water

8 0

2 years ago

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Mitosis creates

g100num [7]

B Im sirloin steak so I think that’s the answer

6 0

3 years ago

These statements describe three different reactions.

Naily [24]

An atom gains an electron from another atom. Hence, option B is the correct answer.

An atom is a particle of matter that uniquely defines a chemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons.

When an atom shares electrons with another atom then it results in the formation of a covalent bond.

Whereas when an atom transfer electrons from one atom to another then it results in the formation of an ionic bond.

When the nucleus of an atom splits then it represents a nuclear fission reaction and energy is released during this process.

Hence, option B is the correct answer.

Learn more about the atom here:

brainly.com/question/1566330

#SPJ1

5 0

1 year ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

2 years ago

Nistu is cooking rice in a pot. When she leaves the pot on the stove for a while, the temperature of the rice increases. What ha

Mila [183]

Answer:

Their average kinetic energy increases

Explanation:

The average kinetic energy of the rice molecules increases as the pot is left on the cooking stove.

Heat is transferred to the pot by conduction from the heat source. The heat is then transferred to the rice in the cooking pot by convection.

As the water in the pot heats up.
The rice gains thermal energy.
This causes the molecules of the rice particles to start vibrating.
As the molecules vibrate about their fixed position, their thermal energy continues to increase.
Therefore, the amount of heat absorbed by the rice increases with time and this actually cooks the food.

5 0

3 years ago

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