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1 year ago

For the n = 3 electron shell, which of the following quantum numbers are valid? Check all that apply.

Chemistry

2 answers:

stealth61 [152]1 year ago

7 0

Answer:

The valid quantum numbers are l=0, l=-2 and l= 2.

Explanation:

Given that,

n = 3 electron shell

Suppose, the valid quantum numbers are

l = 3

m = 3

l = 0

m = –2

l = –1

m = 2

We know that,

The value of n = 3

Principle quantum number :

Then the principal quantum number is 3. Which is shows the M shell.

So, n = 3

Azimuthal quantum number :

The azimuthal quantum number is l.

$l=0,1,2$

Magnetic quantum number :

The magnetic quantum number is

$m=-2,-1,0,1,2$

Hence, The valid quantum numbers are l=0, l=-2 and l= 2.

Ymorist [56]1 year ago

3 0

Answer: I = 0 m = -2 m = 2

Explanation: Edge.

the correct answers to the second part of this are:

n = 1, I = 0, m = 0

n = 3, I = 0, m = 0

n = 5, I = 4, m = -3

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2 years ago

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The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C

Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $460^oC$ = $5.8\times 10^{-6}s^{-1}$

$K_2$ = rate constant at $T_2$ = $4\times K_1$

$Ea$ = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $460^oC=273+460=733K$

$T_2$ = final temperature = ?

Now put all the given values in this formula, we get:

$\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]$

$T_2=791.1K$

Therefore, the final temperature would be, 791.1 K

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2 years ago

A student dissolved a sample in hexane, spotted it on to a TLC plate and eluted using ethyl acetate. After visualizing the TLC p

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1) Consider the following silica gel TLC plate of compounds A, B, and C developed in hexanes:

Determine the Rf values of compounds A, B, and C run on a silica gel TLC plate using hexanes as the solvent.Which compound, A, B, or C, is the most polar?What would you expect to happen to the Rf values if you used acetone instead of hexanes as the eluting solvent?How would the Rf values change if eluted with hexanes using an alumina TLC plate?

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2) You are trying to determine a TLC solvent system which will separate the compounds X, Y, and Z. You ran the compounds on a TLC plate using hexanes/ethyl acetate 95:5 as the eluting solvent and obtained the chromatogram below. How could you change the solvent system to give better separation of these three compounds?

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3) After a rather lengthy organic chemistry synthesis procedure, a student ran the product of the reaction on a TLC plate and obtained the result below. What might he/she have done wrong, if anything?

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4) A student spots an unknown sample on a TLC plate. After developing in hexanes/ethyl acetate 50:50, he/she saw a single spot with an Rf of 0.55. Does this indicate that the unknown material is a pure compound? What can be done to verify the purity of the sample?

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5) Consider a sample that is a mixture composed of biphenyl, benzoic acid, and benzyl alcohol. The sample is spotted on a TLC plate and developed in a hexanes/ethyl acetate solvent mixture. Predict the relative Rf values for the three compounds in the sample.

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6) Plate A, below, represents the TLC chromatogram of a compound run in hexanes. The same compound was then spotted on a large TLC plate and again run in hexanes. Which TLC plate, B, C, or D, correctly represents how far the compound would run on the longer plate?

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3 0

1 year ago

If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.

Elodia [21]

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water = Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

8 0

1 year ago

A 1.0 x 102- gram sample is found to be pure alanine, an amino acid found in proteins. How many moles of alanine are in the samp

Romashka-Z-Leto [24]

Answer:

1.123x10⁻⁴ moles of alanine

Explanation:

In order to convert grams of alanine into moles, we need to know its molecular weight:

The formula for alanine is C₃H₇NO₂, meaning its molecular weight would be:

12*3 + 7*1 + 14 + 16*2 = 89 g/mol

Then we divide the sample mass by the molecular weight, to do the conversion:

1.0x10⁻² g ÷ 89 g/mol = 1.123x10⁻⁴ moles

4 0

1 year ago