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3 years ago

Name the following compound: 2-ethyl-4-methylheptene 3,5-dimethyl-2-octene 2-ethyl-4-methylheptane 3-methyl-5-propyl-2-hexene

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

5 0

Answer:

3,5-dimethyl-2-octene

Explanation:

The parent chain will be choosen based on the highest value. In this case, if we count from top to bottom, we'll get seven carbon, however if we count from the second carbon, going left and then down, we'll get eight carbon. So the parent chain is octene

The double bond is located at the second carbon and the methyl groups are located on carbon 3 & 5. Since there are two methyl groups, we add di- in front of methyl to indicate two methyl groups present.

Note: The functional group has to be prioritise and it needed to be a part of the parent chain. In this case, the functional group is the double bond. (alkene)

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Fill in the blanks for the following statements: The rms speed of the molecules in a sample of H2 gas at 300 K will be _________

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Answer : The rms speed of the molecules in a sample of $H_2$ gas at 300 K will be four times larger than the rms speed of $O_2$ molecules at the same temperature, and the ratio $\mu _{rms}(H_2)/\mu _{rms}(O_2)$ constant with increasing temperature.

Explanation :

Formula used for root mean square speed :

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where,

$\mu _{rms}$ = rms speed of the molecule

R = gas constant

T = temperature

M = molar mass of the gas

At constant temperature, the formula becomes,

$\mu _{rms}=\sqrt{\frac{1}{M}}$

And the formula for two gases will be,

$\frac{\mu _{H_2}}{\mu _{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$

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Now put all the given values in the above formula, we get

$\frac{\mu _{H_2}}{\mu _{O_2}}=\sqrt{\frac{32g/mole}{M_{2g/mole}}}=4$

Therefore, the rms speed of the molecules in a sample of $H_2$ gas at 300 K will be four times larger than the rms speed of $O_2$ molecules at the same temperature.

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3 years ago

A scuba diver that ascends to the surface too quickly can experience decompression sickness, which occurs when nitrogen that dis

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Answer:

The factor that will change the volume of the diver's lungs upon reaching the surface is 4

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Given data:

Pressure increases 1 atm = 101.325 kPa

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