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2 years ago

A Review Constants Periodic Table

Chemistry

1 answer:

soldi70 [24.7K]2 years ago

8 0

Answer:

C. 1, 4–dimethyl–1–cyclohexene

D. 4, 6–dichloro–1–heptene

Explanation:

C. Determination of the name of the compound.

To name the compound given above we must observe the following:

1. Determine the functional group of the compound. In this case, the functional group is the double bond (–C=C–).

2. Determine the parent name by counting the number of carbon that makes up the ring. In this case, 6 carbon makes up the ring. Hence the parent name is cyclohexene.

3. Identify the substituent group attached to the compound. In this case, two methyl group i.e –CH₃ are attached.

4. Locate the position of the substituent group attached by giving the functional group the lowest possible count. In this case, the functional group is at carbon 1, the first methyl group is at carbon 1 and the 2nd methyl group is at carbon 4.

5. Combine the above to obtain the name. Thus, the name of the compound is:

1, 4–dimethyl–1–cyclohexene.

D. Determination of the name of the compound.

To name the compound given above we must observe the following:

1. Determine the functional group of the compound. In this case, the functional group is the double bond (–C=C–).

2. Determine the parent name by counting the number of carbon that makes up the chain. In this case, 7 carbons makes up the chain. Hence, the parent name of the compound is heptene.

3. Identify the substituent group attached to the compound. In this case, two chlorine (Chloro) i.e –Cl are attached.

4. Locate the position of the substituent group attached by giving the functional group the lowest possible count. In this case, the functional group is at carbon 1, the first chlorine is at carbon 4 and the 2nd is at carbon 6.

5. Combine the above to obtain the name. Thus, the name of the compound is:

4, 6–dichloro–1–heptene

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Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by 'n'.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

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Answer,

For example, silver ion can be precipitated with hydrochloric acid to yield solid silver chloride. Because many cations will not react with hydrochloric acid in this way, this simple reaction can be used to separate ions that form insoluble chlorides from those that do not.

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Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element under the same conditions. Estimate the other el

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Answer:

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Explanation:

What is effusion?

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

Graham's law of effusion of gas

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

$\boxed{ \frac{Rate_1}{Rate_2} = \sqrt{ \frac{M_2}{M_1} } }$

Calculations

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

$\frac{Rate_1}{Rate_2} = 1.7$

$1. 7 = \sqrt{ \frac{M_2}{28} }$

Square both sides:

$2.89 = \frac{M_2}{28}$

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

Identity of gas

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

Thus based on the above reasoning, its probable identity is Krypton.

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