Each of the following sets of quantum numbers is supposed to specify an orbital. Choose the one set of quantum numbers that does NOT contain an error.

a. n = 4, l = 3, ml =-4
b. n = 2, l = 2, ml =0
c. n = 3, l = 2, ml =-2
d. n = 2, l = 2, ml =+1

Answers

Answer 1

Answer:

n = 3, l = 2, ml =-2

Explanation:

Quantum numbers are a set of values which can be used to describe the energy and position of an electron in space.

There are four sets of quantum numbers;

1) principal quantum number

2) orbital quantum number

3) spin quantum number

4) magnetic quantum number.

The values of orbital quantum number include; -l to +l;

The set of quantum numbers without error is ; n = 3, l = 2, ml =-2


Related Questions

Which equation expresses the solubility product of Zn3(PO4)2?

a. Ksp = [Zn2+][PO43]

b. Ksp = [Zn2+]3 [PO43]2

c. Ksp = 6[Zn2+][PO43]2

d. Ksp = 108[Zn2+][PO43]2​

Answers

AnSweR : -

b. Ksp = [Zn2+]3 [PO43]2

Hope It Helps You ✌️

Suppose a 48. L reaction vessel is filled with 1.6 mol of Br2 and 1.6 mol of OCl2. What can you say about the composition of the mixture in the vessel at equilibrium

Answers

Answer:

There will be very little of BrOCl BrCl

Explanation:

Based on the equilibrium:

Br2(g) + OCl2(g) ⇄ BrOCl(g) + BrCl(g)

The equilibrium constant, Kc, is:

Kc = 1.58x10⁻⁵ = [BrOCl] [BrCl] / [Br2] [OCl2]

As Kc is <<< 1, in equilibrium, the concentration of products will remain lower regard to the concentration of the reactants. That means, right answer is;

There will be very little of BrOCl BrCl

Name the following aldehyde PLEASE PLEASE HELP

Answers

Answer:

Explanation:

Answer is D 2,5-dimethylheptanal

You should accern the lowest possible number close to the parent name

Linoleic acid is a polyunsaturated fatty acid found, in ester form, in many fats and oils. Its doubly allylic hydrogens are particularly susceptible to abstraction by radicals, a process that can lead to the oxidative degradation of the fat or oil.

a. True
b. Flase

Answers

Answer:

True.

Explanation:

The information presented in the question above regarding linoleic acid is true. Linoleic acid is, in fact, found in many oils and fats in the ester form. In addition, linoleic acid is considered a polyunsaturated fatty acid, due to the presence of two unsaturations in its composition. Its chemical formula is CH3-(CH2)4-CH=CH-CH2-CH=CH-(CH2)7COOH and it is an essential fatty acid for the human body, as it is essential in the composition of arachidonic acid that is responsible for building muscle, managing body fat thermogenesis, and regulating core protein synthesis.

define saturated and unsaturated fats​

Answers

Answer:

unsaturated fats, which are liquid at room temperature,are different from saturated fat because they contain one or more double bonds and fewer hydrogen atoms on their carbon chain.

Suppose you perform a titration of an unknown weak acid solution. You start with 4.00 mL of the weak acid and find that it takes 14.2 mL of 0.0500 M NaOH to reach the equivalence point. What is the concentration of the unknown weak acid solution

Answers

Answer:

0.1775 M

Explanation:

The reaction that takes place is:

HA + NaOH → NaA + H₂O

Where HA is the unknown weak acid.

At the equivalence point all HA moles are converted by NaOH. First we calculate how many NaOH moles reacted, using the given concentration and volume:

0.0500 M NaOH * 14.2 mL = 0.71 mmol NaOH

That means that in 4.00 mL of the weak acid solution, there were 0.71 weak acid mmoles. With that in mind we can now calculate the concentration:

0.71 mmol HA / 4.00 mL = 0.1775 M

Pls pls pls help me

Answers

True it is true :) trust

Calculate the solubility (in mol/L) of Fe(OH)3 (Ksp = 4.0 x 10^-38) in each of the following situations:
(A) Pure water (assume that the pH is 7.0 and remains constant).
(B) A solution buffered at pH = 5.0.
(C) A solution buffered at pH = 11.0.

Answers

Answer:

(A) 1.962x10^-10 M solubility in pure water

(B) 4.0 x 10^-33 M solubility

(C) 4.0 x 10^-27 M solubility

Explanation:

(A) Fe(OH)3 would give (Fe3+) and (3OH-)

Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

Let y = [Fe^3+]

Let 3y = [OH-]

4x10^-38 = (y)(3y)^3

4x10^-38 = 27y^4

y^4 = 4x10^-38 ÷ 27

y^4 = 1.481 x 10^-39

y = 1.962x10^-10 M solubility in pure water

(B) pH = 5.0

5.0 = - log [OH-]

-5.0 = log [OH-]

[OH-] = 10^-5.0 =  1.0 x 10^-5 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-5] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-5

= 4.0 x 10^-33 M solubility

(C) pH = 11.0

11.0 = - log [OH-]

-11.0 = log [OH-]

[OH-] = 10^-11.0 =  1.0 x 10^-11 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-11] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-11

= 4.0 x 10^-27 M solubility

Predict the products from reaction of 2-hexyne with the following reagents: (a) 2 equiv Br2 (b) 1 equiv HBr (c) Excess HBr (d) Li in NH3 (e) H2O, H2SO4, HgSO4

Answers

Answer:

See explanation and image attached

Explanation:

The reactions of the alkynes involved are shown in the image attached to this answer.

First of all, the reaction of two equivalents of bromine with the alkyne converts it to a saturated compound as shown. One equivalent of HBr converts the alkyne to alkene while excess HBr completely reduces the compound to a saturated compound.

Li/NH3 reduces the alkyne to an alkeneby anti addition to the triple bond.

Reaction of the alkyne with H2O, H2SO4, HgSO4 converts it to an aldehyde as shown.

Classify each aqueous solution as acidic, basic, or neutral at 25 °C.
Acidic
Basic
Neutral
pH - 3.41
pH = 10.25
pH = 7.00
[H+] -3.5 x 10-5
[H+] - 6.7 x 10-9
[OH-]-5.8 x 10-4
[H0] -1.0 x 10-7
[OH-] - 4.5 x 10-13

Answers

Answer:

pH - 3.41 = acidic

pH = 10.25 = basic

pH = 7.00 = neutral

[H+] -3.5 x 10-5 = acidic

[H+] - 6.7 x 10-9 = basic

[OH-]-5.8 x 10-4 = basic

[H0] -1.0 x 10-7 = neutral

[OH-] - 4.5 x 10-13 = acidic

Explanation:

Let us note that from the pH scale, a pH of;

0 - 6.9 is acidic

7 is neutral

8 - 14 is basic

But pH= - log [H^+]

pOH = -log [OH^-]

Then;

pH + pOH = 14

Hence;

pH = 14 - pOH

For [H+] -3.5 x 10-5

pH = 4.46 hence it is acidic

For [H+] - 6.7 x 10-9

pH = 8.17 hence it is basic

[OH-]-5.8 x 10-4

pH= 10.76 hence it is basic

[H0] -1.0 x 10-7

pH = 7 hence it is neutral

[OH-] - 4.5 x 10-13

pH = 1.65 hence it is acidic

What is the law of multiple proportions?
A. The proportion of elements to compounds is constant.
B. All elements are found in equal proportions in nature.
C. Different compounds may contain the same elements but may have different ratios of those elements.
D. All compounds contain the same elements in the same proportions.

Answers

Answer:

I think

(d) All compounds contain the same elements in the same properties

How much energy is required to melt 2 kg of gold? Use the table below and this equation.
a. 125.6 kJ
b. 1729 kJ
c. 10.4 kJ
d. 3440kJ

Answers

The equation for the energy required to melt 2 kg of gold is 3440 kJ.

What is energy?

Energy is the ability to do work or cause change. It is an essential part of everyday life and is present in many forms, such as thermal energy, electrical energy, chemical energy, and mechanical energy. Energy can be converted from one form to another in order to do work.


The equation for calculating the energy required to melt a certain mass of material is Q = m x Lf, where Q is the energy required (in joules), m is the mass of the material (in kilograms), and Lf is the latent heat of fusion (in joules per kilogram).
Using the table below, we can see that the latent heat of fusion for gold is 1760 kJ/kg. Therefore, the equation for the energy required to melt 2 kg of gold is: Q = 2 kg x 1760 kJ/kg = 3440 kJ.

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Consider the balanced chemical equation below.

3 A ⟶ C + 4 D

How many moles of D would be produced if 9 moles of A were used?

Answers

hope the picture above helps :)

Calculate the mass percent of carbon in the following molecule: Mn3[Mn(CO)4]3

Answers

Answer:

21.63 %

Explanation:

The molar mass of Mn₃[Mn(CO)₄]₃ is 665.64 g/mol.

Let's assume we have 1 mol of Mn₃[Mn(CO)₄]₃, if that were the case then we would have 665.64 grams.

There are 12 C moles per Mn₃[Mn(CO)₄]₃, with that in mind we calculate the weight of 12 C moles:

12 mol C * 12 g/mol = 144 g

Finally we calculate the mass percent of carbon:

144 g / 665.64 g * 100% = 21.63 %

Explain carefully what happen to the propanol-water system if approximately
50% of propanol by mass is fractionally distilled. What will be the distillate and
the residue?

Answers

Answer:

In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures.

Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?

Answers

Answer:

Explanation:

/ means divided by

* means multiply

1. formula is

partial pressure = no of moles(gas 1)/ no of moles(total)

0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->

.3/(.6+.3+.1) =

.3/1 =

.3 =

partial pressure of CO

2.

.3 * .8 atm = .24

khanacademy

quizlet

The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.

Dalton's Law of Partial pressure

Dalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.

Dalton's Law of partial pressure using mole fraction of gas

Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure

Now, we have to find the first mole fraction of CO

Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]

                                                                  = [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]

                                                                  = [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]

                                                                  = [tex]\frac{0.30}{1}[/tex]

                                                                  = 0.3

Now, put the value in above equation, we get that

Partial pressure of carbon monoxide (CO)

= Mole fraction of carbon monoxide (CO) × Total pressure

= 0.3 × 0.8

= 0.24 atm

Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.

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Which of the following elements is the largest in size
a. O
b. He
c. K
d. H

Answers

Answer:

C. K

i took this class before

Answer: The largest element is K

Explanation: As K has the largest radius among O,He and H

A ball is thrown straight up into the air with a speed of 21 m/s. If the ball has a mass of 0.1 kg, how high does the ball go? Acceleration due to gravity is g = 9.8 m/s^2

Answers

Answer:

answer = 22.5m

Explanation:

using

[tex]s = \frac{ {v}^{2} - {4}^{2} }{2a} [/tex]

s= (0²-21²)/2(-9.8)

s= -441/19.6

s= 22.5m

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Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules.
-OCH3 -Br -Cl -CH2OH
А B C D
(Provide your ranking through a string like abcd, starting with the one with the highest priority).

Answers

Answer:

Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules.

-OCH3 -Br -Cl -CH2OH

Explanation:

To give priorities for the substituents that are attached to chiral carbon and  to assign either R or S-configuration the following rules were proposed:

1. The atom with the highest atomic number is given first priority.

2. If the Groups attached to chiral carbon are having the same first atom, then check for the atomic number of the second atom.

Among the given groups,

-Br has the highest atomic number, so it is given first priority.

Then, -Cl.

Then, -OCH3

and the last one is -CH2OH.

Hence, the order is :

BCAD.

Which of the following two periods contain the same number of element?
A. 2 & 3
B.3 & 4
c. 4 & 6
D. 2 & 4​

Answers

Answer:

4 and 6

Explanation:

Period 4 has 18 elements and so does period 6.

Many reactions involve a change in hybridization of one or more atoms in the starting material. In this reaction, determine the hybridization of the indicated atoms in the organic starting material, and determine if they have changed hybridization during the reaction.

Answers

Answer:

Please find the complete question and its solution file in the attachment.

Explanation:

Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 moles of H2O?

Answers

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]

We have to find number of moles of O2 needed  to produce 12 moles of H2O.

From given equation

We can see that

6 moles of   H2O produced by Oxygen =7 moles

1 mole of   H2O produced by Oxygen=[tex]\frac{7}{6}[/tex]moles

12 moles of H2O produced by Oxygen=[tex]\frac{7}{6}\times 12[/tex]moles

12 moles of H2O produced by Oxygen=[tex]7\times 2[/tex]moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

The amount of oxygen required for the combustion of ethane to produce 12 moles of water is 14 moles.

How are the moles produced in reaction calculated?

The moles of oxygen produced in the reaction can be given from the stoichiometric law of the balanced chemical equation.

The balanced chemical equation for the combustion of ethane is:

[tex]\rm 2\;C_2H_6\;+\;7\;O_2\;\rightarrow\;4\;CO_2\;+\;6\;H_2O[/tex]

The 6 moles of water are produced from 7 moles of oxygen. The moles of oxygen required to produce 12 moles of water are:

[tex]\rm 6\;mol\;H_2O=7\;mol\;Oxygen\\12\;mol\;H_2O=\dfrac{7}{6}\;\times\;12\;mol\;O_2\\ 12\;mol\;H_2O=14\;mol\;O_2[/tex]

The moles of oxygen required to produce 12 moles of water are 14 moles.

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a laser emits light with a frequency of 4.69 x 10 to the 14th power s - 1 calculate the wavelength of this light.

Answers

Answer:

6.40x10^-7

Explanation:

answer with work is attached.

Which of the following is the correct way to balance the following chemical question:
2SnO2 + 4H2 -> 2Sn + 4H2O
SnO2 + 2H2 -> Sn + 2H2O
a. Both equation I and II are balanced, but equation I is the correct way to write the balanced equation.
b. Can you divide equation II by another factor and still have it be correct? Why or why not?
c. In a complete sentence, write down a method you could use to determine if an equation is written in the correct way.

Answers

Answer:

i have no answer for part A

part B

the one that has a 4 can be divided by 2 because reducing

part c

you can determine if an equation is written in the correct way by balancing the equation as if it had not been done already.

A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g

Answers

Answer:

pH=8.676

Explanation:

Given:

0.75 M [tex]NH_{3}[/tex]

0.20 M [tex]NH_{4}[/tex]

The objective is to calculate the pH of the buffer using the kb for [tex]NH_3[/tex]

Formula used:

[tex]pOH=pka+log\frac{[salt]}{[base]}\\[/tex]

pH=14-pOH

Solution:

On substituting salt=0.75 and base=0.20 in the formula

[tex]pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324[/tex]

pH=14-pOH

On substituting the pOH value in the above expression,

pH=14-5.324

Therefore,

pH=8.676

What is the cell potential of an electrochemical cell that has the half-reactions shown below?
Ag⁺ + e⁻ → Ag
Fe → Fe³⁺ + 3e⁻

Answers

Answer:

E°(Ag⁺/Fe°) = 0.836 volt

Explanation:

3Ag⁺ + 3e⁻ => Ag°;            E° = +0.800 volt

Fe° => Fe⁺³ + 3e⁻ ;             E° = -0.036 volt

_________________________________

Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...    

E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt

Using the following reaction:

H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)

Calculate the molarity of the H2SO4 solution if 14.92 mL of NaOH was necessary to reach the endpoint of a titration. The molarity of the NaOH solution was 0.83 M and 25.18 mL of H2SO4 was added to the Erlenmeyer flask.

Answers

Answer:

The molar concentration of the H₂SO₄ solution is 0.28 M

Explanation:

Molar concentration = number of moles / volume in litres

Number of moles = molar concentration × volume

From the equation of reaction, molar ratio of acid to base = 1 : 2

Using the formula; Na/Nb = CaVa/CbVb

Where Na is the number of moles of acid; Nb = number of moles of base; Ca = concentration of acid; Va = volume of acid; Cb = concentration of base; Vb = volume of base; Na/Nb = mole ratio of acid to base

Substituting the given values in the equation:

1/2 = Ca × 25.18 / 0.93 × 14.92

Ca = 0.93 × 14.92/ 25.18 × 2

Ca = 0.28M

Therefore, the molar concentration of the H₂SO₄ solution is 0.28 M

3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample

Answers

Answer:

[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]

Explanation:

From the question we are told that:

Mass of mixture [tex]m=3.455g[/tex]

Mass of Barium [tex]m_b=0.2815g[/tex]

Equation of Reaction is given as

[tex]Ba2+ + H2SO4 => BaSO4 + 2 H+[/tex]

Generally the equation for Moles of Barium  is mathematically given by

Since

 [tex]Moles of Ba^{2+} = Moles of BaSO_4[/tex]

Therefore

 [tex]Moles of Ba^{2+} = \frac{mass}{molar mass of BaSO4}[/tex]  

 [tex]Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol[/tex]

Generally the equation for Mass of Barium  is mathematically given by

 [tex]Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}[/tex]  

 [tex]Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g[/tex]

Therefore

 [tex]Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%[/tex]    

 [tex]Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%[/tex]

 [tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]

All --- is a good conductor of heat​

Answers

all metals is a good conductor of heat

Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 35. g of hydrochloric acid is mixed with 73.0 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Round your answer to significant digits.

Answers

Answer:

No mass of HCl could be left over by the chemical reaction because is the limting reactant and it is all consumed.

Explanation:

Our reactants are: HCl and NaOH

Products are: NaCl and H₂O

This is a neutralization reaction that can also be called an acid base reaction, an acid and a base react to produce water and a neutral salt, in this case where we have strong acid and base.

Ratio is 1:1. We convert mass to moles:

35 g . 1 mol / 36.45 g = 0.960 moles of HCl

73 g . 1 mol / 40 g = 1.82 moles of NaOH

As ratio is 1:1, for 0.960 moles of HCl we need 0.960 moles of NaOH and for 1.82 moles of NaOH, we need 1.82 moles of acid.

As we only have 0.960 moles of HCl and we need 1.82 moles, no acid remains after the reaction goes complete. HCl is the limiting reactant, so the acid, it is all consumed.

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similarities of trial and pre-trial In a standard cost accounting system, the entry to record purchase of raw materials on account for $13500 when the standard cost is $12620 includes:______.a. debit to Raw Materials Inventory for 12,750, debit to Materials Price Variance for $750 and credit to Accounts Payable for $13,500. B. debit to Materials Price Variance for S7S0 and credit to Accounts Payable for $750. c. debit to Raw Materials Inventory for $13,500 and credit to Accounts Payable of $13,500. d. debit to Raw Materials Inventory for $12,750 and credit to Accounts Payable of $12,750. Suppose that the inflation rate is 2% and the real terminal value of an investment is expected to be $82,500 in 4 years. Calculate the nominal terminal value of the investment at the end of year 4. What is the complementary DNA strand for this DNA strand : GTTACGAACT The a oranial price of a skate broad was reduced by 15 dollars .the new price is 49 dollars if p=the stakebroads oranail price in dollars what mathematical sentence expresses the informationa.49-p=15b.15-p=49c.15+p=49d.p-15=49 Explanation of the poem you and today by Ella Wheeler Wilcox Mustang Corporation reports the following for the month of April:Finished goods inventory, April 1$32,600 Finished goods inventory, April 30 26,600 Total cost of goods manufactured 123,800The cost of goods sold for April is:a. $61,900.b. $124,500.c. $112,900.d. $173,700.e. $150,000 Evaluate the expression when c = 3 and x= -5, -C+5x How many molecules are present in 1296 g of dinitrogen pentoxide (N2O5) what are all the factos of 36 Unvoiced and voiced word what is the value of acceleration due to gravity at the equatorial region of the earth Which idea was the English Bill of Rights most directly responsible for? why worklife balance is important? Find the distance between the points ( 4, 7) and (-1,2 Explain how you can write a quadratic function in a factored form that would have a vertex with an x-coordinate of 3 and two distant roots [tex]factories : - \\ \\ p {}^{2} + 8p - 84[/tex] conclusion de la Moral A caterer is agrranging 72 ham sandwiche and 48 pimento cheese sandwiches on serving trays. He would like to arrange the sandwiches so tht each tray has the same number of each kind of sandwich. The number of ham sandwiches on a tray does not need to equal the number of pimento cheese sandwiches on the tray. What is the maximum number of traysthe caterer can use? . Provide details about Meiosis.