I think there are g, h, and i orbitals, there are just no elements with electrons at a ground state in any of those orbitals. The first element that has an electron that would occupy the g orbital would be element 121, which has not been discovered yet.

Hope this helps!

## Search found 153 matches

- Wed Nov 04, 2020 1:31 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: l numbers bigger than 3
- Replies:
**5** - Views:
**40**

- Wed Nov 04, 2020 1:25 pm
- Forum: Resonance Structures
- Topic: Delocalized e-
- Replies:
**5** - Views:
**51**

### Re: Delocalized e-

Delocalized electrons are basically electrons that are not associated with a specific atom, but instead are equally likely to be located at any of the bonds involved with the resonance structure. For example, in benzene (C6H6), the electrons are delocalized, meaning that there is an equal probabilit...

- Thu Oct 29, 2020 1:15 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Sapling Energy Level and Wavelength
- Replies:
**3** - Views:
**46**

### Re: Sapling Energy Level and Wavelength

I am not sure exactly what question you are referring to, but if a colored line is produced, that means it is part of the Balmer series, so n=2. If the question does not specify anything about the type of radiation emitted, you can check if the wavelength is part of the visible light region (420 to ...

- Thu Oct 29, 2020 1:33 am
- Forum: DeBroglie Equation
- Topic: Week 2,3,3 Sampling #25
- Replies:
**2** - Views:
**45**

### Re: Week 2,3,3 Sampling #25

The equations E=hv and c=vλ refer to electromagnetic radiation and therefore can only be used for photons. On the other hand, λ=h/p and ke=1/2mv^2 are for any particles that have rest mass. The second two equations can be used for electrons, neutrons, protons, etc. Since photons have no rest mass, t...

- Thu Oct 29, 2020 12:31 am
- Forum: Empirical & Molecular Formulas
- Topic: Empirical Formula from combustion
- Replies:
**1** - Views:
**44**

### Re: Empirical Formula from combustion

The reason the oxygen is unknown is because combustion reactions typically have a compound that reacts with an excess amount of oxygen to form carbon dioxide and water. Since all the carbon in the carbon dioxide and all the hydrogen in the water are from the original compound, you know that the mass...

- Thu Oct 29, 2020 12:22 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Sapling HW #11
- Replies:
**2** - Views:
**29**

### Re: Sapling HW #11

To find n2, convert the wavelength into meters, and then use the equation c=\lambda \nu to find the frequency. Once you find the frequency, you can use the Rydberg equation. Input the frequency, Rydberg's constant, and the final energy level into \nu = R (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&...

- Thu Oct 29, 2020 12:19 am
- Forum: Photoelectric Effect
- Topic: Sapling HW #11
- Replies:
**5** - Views:
**55**

### Re: Sapling HW #11

To find n2, convert the wavelength into meters, and then use the equation c=\lambda \nu to find the frequency. Once you find the frequency, you can use the Rydberg equation. Input the frequency, Rydberg's constant, and the final energy level into \nu = R (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&...

- Thu Oct 29, 2020 12:05 am
- Forum: Trends in The Periodic Table
- Topic: Sapling Hw 2, 3, 4 Question 20
- Replies:
**4** - Views:
**43**

### Re: Sapling Hw 2, 3, 4 Question 20

In the lecture, Dr. Lavelle explains how one of the anomalies in the ionization energy, why oxygen has a lower ionization energy than nitrogen, is because oxygen has 4 electrons in the 2p orbital. This means that the 2px orbital has two electrons, unlike nitrogen, which only has one electron for eac...

- Wed Oct 28, 2020 9:55 pm
- Forum: Properties of Electrons
- Topic: Sapling #10
- Replies:
**2** - Views:
**49**

### Re: Sapling #10

I had this same question, so I asked a TA about it. It is true that each intermediate change in energy would cause an individual spectral line, so there should be more intermediate lines. However, for the sake of the question, Sapling tried to keep it simple, or at least that is the conclusion we ca...

- Wed Oct 28, 2020 9:09 pm
- Forum: Significant Figures
- Topic: 0's
- Replies:
**21** - Views:
**200**

### Re: 0's

Trailing zeros are not significant when there is no decimal point. For example, 200 would only have 1 sig fig because the trailing zeros are insignificant. However, since 2.000 x 10^1 has a decimal point, the zeros are significant. This is because 200 could be a rounded number or there is some other...

- Wed Oct 28, 2020 7:44 pm
- Forum: Einstein Equation
- Topic: Solving Equations using E=hv
- Replies:
**6** - Views:
**122**

### Re: Solving Equations using E=hv

I would assume that you could use E=h\nu as you would with the frequency of a single photon because the frequency per photon gives you the energy per photon, so the frequency per mole should give you the energy per mole. The Js^-1 unit for Planck's constant should cancel with the J for the energy, a...

- Wed Oct 28, 2020 7:41 pm
- Forum: DeBroglie Equation
- Topic: Sapling Question 22
- Replies:
**3** - Views:
**40**

### Re: Sapling Question 22

In this situation, they are giving you a wavelength and asking you to find the kinetic energy. These are related by the DeBroglie equation, so you first need to use that to find the velocity of the electron. Once you figure out the velocity, you can input that value into the kinetic energy equation ...

- Wed Oct 28, 2020 6:18 pm
- Forum: DeBroglie Equation
- Topic: sapling hw problem
- Replies:
**5** - Views:
**126**

### Re: sapling hw problem

Your new equation looks correct. I used the equation you listed and got 6.7x10^-26 J, so maybe check your calculations because all your work before that looks correct. Make sure you converted the wavelength into meters, that could just be a small error.

Hope this helps!

Hope this helps!

- Wed Oct 28, 2020 6:04 pm
- Forum: Limiting Reactant Calculations
- Topic: Cl vs Cl2
- Replies:
**9** - Views:
**96**

### Re: Cl vs Cl2

Yes, chlorine gas is Cl2. There are 7 diatomic molecules (H,O,N,Cl,Br,I,F), so whenever these are gases or "exist naturally", you should use the subscript 2.

Hope this helps!

Hope this helps!

- Wed Oct 28, 2020 5:20 pm
- Forum: SI Units, Unit Conversions
- Topic: SI Unit for Mass
- Replies:
**10** - Views:
**125**

### Re: SI Unit for Mass

Most of the energy equations require kg because Joules = kg⋅m^2⋅s^−2, so you need to use kg to make sure the units cancel. In converting from mass to moles, you just use the molar mass in grams, but otherwise, kg it typically used as it is the SI unit for mass.

Hope this helps!

Hope this helps!

- Wed Oct 28, 2020 4:08 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Sapling Question
- Replies:
**4** - Views:
**75**

### Re: Sapling Question

The main thing to remember here is that 4 quantum numbers specify a single electron. Therefore, n=6,ℓ=2,mℓ=−1, which has three quantum numbers and simply does not include the magnetic spin quantum number is referring to two electrons that are in the sixth shell in the d orbital. The mℓ number tells ...

- Wed Oct 28, 2020 10:47 am
- Forum: Balancing Chemical Reactions
- Topic: Problem for UA Workshop 1
- Replies:
**2** - Views:
**49**

### Re: Problem for UA Workshop 1

The way to balance this equation is: 3NaHCO_{3}+C_{6}H_{8}O_{7}\rightarrow Na_{3}C_{6}H_{5}O_{7} + 3H_{2}O + 3 CO_{2} I usually approach these problems by balancing any elements that are not hydrogen or oxygen first. In this case I began with Na. After balancing the Na, I balanced the C, and then th...

- Wed Oct 28, 2020 10:37 am
- Forum: DeBroglie Equation
- Topic: When to Use De Broglie
- Replies:
**16** - Views:
**129**

### Re: When to Use De Broglie

If given the wavelength of an ejected electron, you would use the DeBroglie equation because the electron has mass. You could figure out the velocity and then from there figure out the kinetic energy and relate that back to the incident light, depending on what the question asks for. You are correct...

- Tue Oct 27, 2020 11:29 pm
- Forum: Photoelectric Effect
- Topic: Work function or incoming light
- Replies:
**6** - Views:
**52**

### Re: Work function or incoming light

The energy required to remove an electron is the work function because you could have incoming light with greater energy than is required to remove the electron, which would then give you an ejected electron with kinetic energy (think about the equation E(photon)-work function= KE). The energy that ...

- Wed Oct 21, 2020 5:27 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Electrons Falling to Intermediate Energy Levels
- Replies:
**1** - Views:
**17**

### Electrons Falling to Intermediate Energy Levels

Is it possible for electrons to fall to intermediate energy levels? For example, could an electron that is excited to the energy level n=5 fall to n=3 before returning to the ground state? Additionally, since each of the changes in energy levels would have a different change in energy (such as n=5 t...

- Tue Oct 20, 2020 11:06 pm
- Forum: Properties of Light
- Topic: Atomic Spectra Module
- Replies:
**3** - Views:
**58**

### Re: Atomic Spectra Module

Since you are told that 1 meter is 1,650,763.73 wavelengths of radiation emitted by krypton-86, it is basically a 1 meter to 1,650,763.73 wavelengths ratio. To find 1 wavelength, just divide 1 meter by 1,650,763.73. If you set it up as 1m=1,650,763.73 wavelengths, then your goal is to find 1 wavelen...

- Tue Oct 20, 2020 11:01 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Electromagnetic radiation and associated energy levels
- Replies:
**4** - Views:
**44**

### Re: Electromagnetic radiation and associated energy levels

The Paschen, Brackett, and Pfund series are all in the infrared region. I believe that their lowest energy levels are n=3, n=4, and n=5, respectively. I am assuming we will not be required to know these though.

Hope this helps!

Hope this helps!

- Tue Oct 20, 2020 10:56 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect Module Question
- Replies:
**4** - Views:
**89**

### Re: Photoelectric Effect Module Question

Since the kinetic energy is the difference between the energy of the photon and the energy required to emit a photon, you should figure out the energy that corresponds to a wavelength of 194 nm to begin 34B. Subtract the energy you found in 33A (the work function for Molybdenum) from the new energy ...

- Mon Oct 19, 2020 8:08 pm
- Forum: DeBroglie Equation
- Topic: Difference between DeBroglie and regular wavelength
- Replies:
**4** - Views:
**47**

### Re: Difference between DeBroglie and regular wavelength

The values should be the same, it just depends on what information you are given/need to find. If you are dealing with momentum (mass and/or velocity) then you should use DeBroglie's equation. If you are dealing with energy and/or frequency, you should use one of the other two.

Hope this helps!

Hope this helps!

- Mon Oct 19, 2020 7:34 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Saping HW Q.8
- Replies:
**8** - Views:
**77**

### Re: Saping HW Q.8

I would convert the wavelength into meters and then find the frequency. Once you have the frequency, plug it into Rydberg's equation. Since it appears as a blue line, we know that it is part of the visible spectrum, so the Balmer series. That means that n1=2. From there you should be able to solve f...

- Mon Oct 19, 2020 7:26 pm
- Forum: SI Units, Unit Conversions
- Topic: unit convertion
- Replies:
**4** - Views:
**41**

### Re: unit convertion

Yes, that is correct! Think about it like (5.11x10^-4 cm) x (10^-2m/cm). That way the units cancel and you get the right conversion.

Hope this helps!

Hope this helps!

- Mon Oct 19, 2020 7:24 pm
- Forum: Empirical & Molecular Formulas
- Topic: textbook F21
- Replies:
**1** - Views:
**38**

### Re: textbook F21

I got C49. I found the mass percentages and then found the moles of C out of 100. g to get 5.20 mol C. Then, I divided by the smallest amount of moles (0.637 moles N) to get 8.16 as the ratio of C. When I multiplied that by 6, I got C49. Maybe your ratio or one of the intermediate calculations was s...

- Mon Oct 19, 2020 7:12 pm
- Forum: Properties of Light
- Topic: Planck's Constant
- Replies:
**7** - Views:
**53**

### Re: Planck's Constant

It is 6.626x10^-34 Js.

- Mon Oct 19, 2020 7:05 pm
- Forum: DeBroglie Equation
- Topic: 1B27
- Replies:
**3** - Views:
**39**

### Re: 1B27

I got the same answer for that question. I would assume that a difference that small simply comes from rounding (maybe for Planck's constant or pi).

Hope this helps!

Hope this helps!

- Mon Oct 19, 2020 4:14 pm
- Forum: SI Units, Unit Conversions
- Topic: SI units
- Replies:
**9** - Views:
**46**

### Re: SI units

You have to use meters because the other values are are in meters and ultimately, the units need to cancel. For example, the speed of light is 3.00E8 m/s. To make sure the units cancel in the equation c=\lambda \nu , you would need your wavelength in meters as well, since the frequency is in Hz, or ...

- Mon Oct 19, 2020 3:19 pm
- Forum: DeBroglie Equation
- Topic: Super Slow Large Mass using de Broglie
- Replies:
**2** - Views:
**31**

### Re: Super Slow Large Mass using de Broglie

The mass tends to have a larger impact on the wavelength. If you want to test it out, you can find the wavelength of a baseball traveling at 1 m/s (or any very slow velocity). The wavelength is still way too small to be detected. Since Planck's constant is such a small number, the mass needs to be v...

- Mon Oct 19, 2020 3:07 pm
- Forum: DeBroglie Equation
- Topic: Biggest Item to have Wavelike Properties
- Replies:
**2** - Views:
**44**

### Re: Biggest Item to have Wavelike Properties

It is hard to say exactly how large an object can be since DeBroglie's equation depends on both the mass and velocity. I would say anything that you can hold is too large. If you ever need to check, you can always just input the information into the equation and any object with a wavelength smaller ...

- Mon Oct 19, 2020 11:38 am
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Question 18 from Heisenberg Uncertainty Module
- Replies:
**3** - Views:
**30**

### Re: Question 18 from Heisenberg Uncertainty Module

I started this question by figuring out the indeterminacy in position, so I converted the radius of the hydrogen atom into meters and then found 1% of that value for delta x. I inputted this value into Heisenberg's Indeterminacy Equation to find delta p, and then I divided that value by the mass of ...

- Mon Oct 19, 2020 11:12 am
- Forum: DeBroglie Equation
- Topic: No Rest Mass of a Photon
- Replies:
**4** - Views:
**56**

### No Rest Mass of a Photon

In the lecture, Dr. Lavelle says that a photon is not a particle with rest mass, but it has momentum. I am confused as to how this is possible because p=mv, so wouldn't that mean anything with momentum must have mass? The lecture includes that a photon has energy but cannot be broken down into E=mv^...

- Sat Oct 17, 2020 11:03 am
- Forum: Photoelectric Effect
- Topic: Photoelectric effect post assessment 31
- Replies:
**1** - Views:
**28**

### Re: Photoelectric effect post assessment 31

You do have to use both equations. You can use the energy to find the frequency, and then use the frequency to find the wavelength, or you can combine both equations to form , and then directly use the energy to solve for the wavelength.

Hope this helps!

Hope this helps!

- Sat Oct 17, 2020 11:00 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Topic 1A #15
- Replies:
**5** - Views:
**43**

### Re: Topic 1A #15

For this question, I converted the given wavelength into meters and then used the c=\lambda \nu to find the frequency. Using the equation \nu =R(\frac{1}{n_{1}^{2}}+\frac{1}{n_{2}^{2}}) , I inputted the frequency and Rydberg's constant. Since the line is in the ultraviolet spectrum, n_{1} mu...

- Fri Oct 16, 2020 5:12 pm
- Forum: Properties of Light
- Topic: Textbook Problem 1B.9
- Replies:
**2** - Views:
**21**

### Re: Textbook Problem 1B.9

The step you did was correct. However, that gives you the total energy of the lamp, so to find the number of photons, you first need to find the energy per photon. To do this, simply convert the given wavelength into meters and then use the equation E=h(\frac{c}{\lambda }) to find the energy...

- Fri Oct 16, 2020 4:49 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectra Module #29
- Replies:
**2** - Views:
**22**

### Re: Atomic Spectra Module #29

Since you have the wavelength in nanometers, covert it to meters and then use E=h(\frac{c}{\lambda }) to find the energy per photon. Since you are given that the bulb emits 11 J of energy in 1.0 s, divide 11 J by the energy per photon to find the number of photons the lamp generates in 1.0 s...

- Fri Oct 16, 2020 4:00 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Hydrogen Spectra
- Replies:
**2** - Views:
**30**

### Re: Hydrogen Spectra

I looked it up and the Paschen, Brackett, and Pfund series are all in the infrared region. I am assuming we will not be required to know these though.

Hope this helps!

Hope this helps!

- Fri Oct 16, 2020 3:22 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Sapling Homework #2
- Replies:
**2** - Views:
**29**

### Re: Sapling Homework #2

You can convert the wavelength to frequency and then use {\displaystyle {\nu}=R_{\text{}}\left({\frac {1}{n_{1}^{2}}}-{\frac {1}{n_{2}^{2}}}\right),} to find the energy levels. The blue line indicates that you are in the visible light region, so you are looking at the Balmer series where n1=...

- Fri Oct 16, 2020 1:59 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: E=h*V explanation in atomic spectra
- Replies:
**3** - Views:
**29**

### Re: E=h*V explanation in atomic spectra

You can figure out the energy of the photon by calculating the difference in energy between the energy level the electron begins at the final energy level to which it falls. The difference in energy would be represented at delta E. Since the photon needed to have exactly the amount of energy equal t...

- Fri Oct 16, 2020 12:49 pm
- Forum: Properties of Light
- Topic: bohr frequency condition
- Replies:
**4** - Views:
**65**

### Re: bohr frequency condition

The Bohr Frequency Condition basically states that light will be absorbed when the frequency of the incoming light is equal to the energy difference for that electron transition. This means that the difference is energy levels divided by Planck's constant will give you the frequency, giving rise to ...

- Tue Oct 06, 2020 1:43 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectra Post Mod Q
- Replies:
**1** - Views:
**28**

### Re: Atomic Spectra Post Mod Q

Since you are told that 1 meter is 1,650,763.73 wavelengths of radiation emitted by krypton-86, it is basically a 1 meter to 1,650,763.73 wavelengths ratio. To find 1 wavelength, just divide 1 meter by 1,650,763.73. If you set it up as 1m=1,650,763.73 wavelengths, then your goal is to find 1 wavelen...

- Mon Oct 05, 2020 10:55 pm
- Forum: Photoelectric Effect
- Topic: Electron Mass
- Replies:
**2** - Views:
**16**

### Re: Electron Mass

The mass of an electron is always the same so we can just input 9.11x10^-31 kg. I think you need this mass to solve the rest of the problem so I do not know how else you would find it.

Hope this helps!

Hope this helps!

- Mon Oct 05, 2020 5:50 pm
- Forum: Balancing Chemical Reactions
- Topic: HW H.21
- Replies:
**5** - Views:
**52**

### Re: HW H.21

Oxygen is a diatomic molecule and therefore, its gaseous form (which occurs in nature) must always be O2. Any time oxygen is naturally occurring, it must be the molecule, so O2. I would assume only in lab settings (which would be specified in the problem) would you use O instead of O2. Hope this hel...

- Mon Oct 05, 2020 2:52 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Fundamentals G23 [ENDORSED]
- Replies:
**3** - Views:
**51**

### Re: Fundamentals G23 [ENDORSED]

I think your method is correct, there may just be an error in the math. I got 0.0086 moles Cl- from NaCl and 0.0040 mol Cl- from KCl. I added the moles up and then divided by the volume to get an answer of 0.13 M Cl-. Your method is almost the same and when I tried it, I got the same answer, so mayb...

- Mon Oct 05, 2020 2:34 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect Module
- Replies:
**6** - Views:
**75**

### Re: Photoelectric Effect Module

Given that the energy of the photon is high enough to result in an electron being ejected, it should be a 1 to 1 ratio of high energy photons to electrons ejected (ideally).

Hope this helps!

Hope this helps!

- Mon Oct 05, 2020 2:30 pm
- Forum: Limiting Reactant Calculations
- Topic: Fundamentals M5
- Replies:
**7** - Views:
**83**

### Re: Fundamentals M5

Since you know there has to be a ratio of 6 moles ClO2 to moles 2 BrF3, with 12 moles ClO2 you would need 4 moles BrF3 (to maintain the same ratio). Since you have 5 moles BrF3, when you only need 4, BrF3 is the excess reactant.

Hope this helps!

Hope this helps!

- Mon Oct 05, 2020 2:16 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Molar mass for O2 vs O
- Replies:
**11** - Views:
**119**

### Re: Molar mass for O2 vs O

The molar mass of O2 would be 32 g/mol because O2 technically has 2 moles of oxygen. Since oxygen itself has a molar mass of 16 g/mol, you would need to double it to get the molar mass of O2.

Hope this helps!

Hope this helps!

- Mon Oct 05, 2020 2:14 pm
- Forum: Significant Figures
- Topic: M.3
- Replies:
**4** - Views:
**43**

### Re: M.3

You are given that there is 17.5 g of CO2 produced from the thermal decomposition of CaCO3. Since you use 17.5g of CO2 in the percent yield calculation, which only has three significant figures, your final answer would also only have three significant figures, so it would be 93.1%.

Hope this helps!

Hope this helps!

- Mon Oct 05, 2020 12:03 pm
- Forum: Balancing Chemical Reactions
- Topic: L.35
- Replies:
**9** - Views:
**113**

### Re: L.35

This question actually has a typo. The compound FeBr2 should be Fe3Br8. There is also a link to solution manual errors on Dr. Lavelle's website that lists this.

Hope this helps!

Hope this helps!

- Mon Oct 05, 2020 11:59 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Process of Molarity Calculation
- Replies:
**6** - Views:
**78**

### Re: Process of Molarity Calculation

Since the moles of KCL are the same throughout the entire process and all the water is retained (50 mL are added to the initial 75mL, no portion of the 75 mL are transferred elsewhere), you can divide directly by 0.125 L to find the molarity.

Hope this helps!

Hope this helps!

- Fri Oct 02, 2020 11:32 am
- Forum: Limiting Reactant Calculations
- Topic: Fundamental L.39 [ENDORSED]
- Replies:
**3** - Views:
**175**

### Re: Fundamental L.39 [ENDORSED]

I began this problem by subtracting the weight of the crucible from the weight of the crucible and product together to find the weight of the product alone. Since there was 1.50g of metallic tin at the beginning, divide this number by the weight of the final product (that you found during the previo...