All categories

1 year ago

The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:

Chemistry

1 answer:

ololo11 [35]1 year ago

7 0

The allowable combination for the atomic orbital is n=3, l=1 or 2, $m_{l}$ =+1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; $m_{l}$ values are integers from -l to =l(l limits ml)

For n = 3, l can have three values: 0, 1, and 2. Hence, the l value must be lower than 3. Because of $m_{l}$ = +1, the l value must be higher than 0. Two l values are consistent with n and $m_{l)$ values:

l= 1 or 2

Therefore, the allowable condition is n=7, l=1 or 2, $m_{l}$ =+3.

To know more about quantum numbers, visit: brainly.com/question/16979660

#SPJ4

You might be interested in

_______ is the formation of alcohol from sugar.

Katarina [22]

The best and most correct answer among the choices provided by the question is the fourth choice "alcoholic fermentation"

Ethanol fermentation<span>, also called </span>alcoholic fermentation<span>, is a biological process which converts sugars such as glucose, fructose, and sucrose into cellular energy, producing </span>ethanol<span> and carbon dioxide as a side-effect.</span>

I hope my answer has come to your help. God bless and have a nice day ahead!

7 0

2 years ago

Read 2 more answers

Explain why the amino group of p-aminobenzoic acid does not participate in the reaction

lubasha [3.4K]

This is because amino group of p-aminobenzoic acid is an aniline and is less electrophilic than an alkyl amine.

<h3>What is an Aniline?</h3>

This is an aromatic amine which consists of a phenyl group attached to an amino group.

The amino group of p-aminobenzoic acid being an aniline makes it less electrophilic which is why an alkyl amine participates in the reaction instead.

Read more about Aniline here brainly.com/question/9982058

5 0

1 year ago

Any change in average weather conditions on Earth that lasts for at least several decades is called

sladkih [1.3K]

Answer:climate change

Explanation:

8 0

2 years ago

What the difference between a sodium atom (Na) and a sodium ion (Nal+)

vichka [17]

Answer:

So a sodium atom has the electronic configuration 2,8,1 meaning it has 1 electron on its outer shell. It has a neutral charge since the number of electrons is equal to the number of protons.

A sodium ion is one that has lost the electron on its valence shell. The electronic configuration is 2,8 and it has a positive charge because it has more protons than electrons.

7 0

2 years ago

Read 2 more answers

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0

2 years ago