Search found 12 matches

by Quinn_Simpson_3D
Sun Oct 28, 2018 2:40 pm
Forum: Trends in The Periodic Table
Topic: Atomic Radius
Replies: 7
Views: 139

Re: Atomic Radius

When an electron is removed, the remaining electrons feel a greater attraction to the nucleus as there is more protons than electrons. This is makes it harder to take other electrons away.
by Quinn_Simpson_3D
Sun Oct 28, 2018 2:31 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron Affinity [ENDORSED]
Replies: 7
Views: 166

Re: Electron Affinity [ENDORSED]

Electron affinity does not have as much as a defined trend as ionization energy and atomic radius. However, the gases in the top right corner of the periodic table have a high electron affinity because they want to gain electrons to create a stable, filled shell, like a noble gas configuration. Meta...
by Quinn_Simpson_3D
Sun Oct 28, 2018 2:26 pm
Forum: Trends in The Periodic Table
Topic: Ionization Energy [ENDORSED]
Replies: 7
Views: 144

Re: Ionization Energy [ENDORSED]

Ionization energy as you go across a period because there are more electrons in the valence shell. Since a full shell is the most stable, elements with less valence electrons are more readily able to lose their electrons because they will be closer to a full shell. And since elements with close to a...
by Quinn_Simpson_3D
Sun Oct 21, 2018 1:16 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Q 42 on post assessment
Replies: 5
Views: 85

Re: Q 42 on post assessment

The Rydberg equation is 1/lambda = R((1/n^2(initial))-(1/n^2(final)), where R is Rydberg's constant. You have the frequency, which means you can find the wavelength using c=(wavelength)(frequency). Once you have the wavelength, you know the wavelength and the final electron state, which is 4. Since ...
by Quinn_Simpson_3D
Sun Oct 21, 2018 1:06 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Test 2 Material
Replies: 9
Views: 91

Re: Test 2 Material

For test 2, we need to know everything in the Quantum World chapter up to what we learned the Friday before the test, which was naming the state of an electron using the n, l, m, and positive/negative orientation of the electron. What we learn on the Monday before the test will not be tested on.
by Quinn_Simpson_3D
Sun Oct 21, 2018 1:03 pm
Forum: DeBroglie Equation
Topic: DeBroglie wavelike properties question
Replies: 4
Views: 108

Re: DeBroglie wavelike properties question

All objects give off wavelength particles, as De Broglie's equation can apply to any object with momentum, or mass and velocity. However, some objects are way too big to give off noticeable wavelike properties, as the wavelengths are often extremely small. Professor Lavelle stated that the "thr...
by Quinn_Simpson_3D
Wed Oct 10, 2018 10:11 am
Forum: Einstein Equation
Topic: Homework Question 1B7b
Replies: 2
Views: 153

Re: Homework Question 1B7b

In order to solve this problem, you need to find the amount of atoms of Na. To do this convert the 5.00 mg to .005 g. Then divide by 22.99 g.mol^-1 (Na molar Mass) to find the number of moles, which is 2.17 x 10^-4. You then multiply by Avogadro's constant (6.022 x 10^23) to find the number of atoms...
by Quinn_Simpson_3D
Wed Oct 10, 2018 9:55 am
Forum: Ideal Gases
Topic: Reading the textbook [ENDORSED]
Replies: 81
Views: 52064

Re: Reading the textbook [ENDORSED]

Reading the textbook can be helpful because even if you don't fully understand it, you will have an idea of what is going to be taught in class. For me, this helps create a deeper understanding than when I just listen to the lecture and then do the homework.
by Quinn_Simpson_3D
Wed Oct 10, 2018 9:50 am
Forum: Properties of Light
Topic: Photoelectric Effect PostMod #30
Replies: 1
Views: 43

Re: Photoelectric Effect PostMod #30

In order to solve this problem, you need to know the mass of an electron, which is 9.11 x 10^-31 kg. You also know the velocity of the electron, which is 6.51 x 10^5 m/s. Since you have both these values, you can plug them into the equation (1/2)mv^2. The value you get is 1.99 x 10^-19 J . Once you ...
by Quinn_Simpson_3D
Wed Oct 03, 2018 10:31 pm
Forum: General Science Questions
Topic: Rusty on High School Chem [ENDORSED]
Replies: 113
Views: 66353

Re: Rusty on High School Chem [ENDORSED]

I took AP chem in high school, and you have to put in a lot of work to do well. That being said if you read the textbook (possibly reading some sections multiple times) and do more problems than the required homework, you can learn it well. Also, Dr. Lavelle's Audio-Video Modules are very helpful. I...
by Quinn_Simpson_3D
Wed Oct 03, 2018 10:27 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Isotopes [ENDORSED]
Replies: 12
Views: 366

Re: Isotopes [ENDORSED]

Isotopes are chemical element variants with a different number of neutrons than the original proton amount. This is why the molar mass of elements isn't exactly the same as the protons and neutrons added together. Based on their percentage on Earth, the different element variations are averaged out,...
by Quinn_Simpson_3D
Wed Oct 03, 2018 10:20 pm
Forum: Limiting Reactant Calculations
Topic: Finding theoretical yield
Replies: 6
Views: 129

Re: Finding theoretical yield

You have to know the molar ratio between both reactants in order to find which one is the limiting reagent. Once you find the moles of each product, you have to use the molar ratios when calculating the amount of product for each given mole. Once you have the mass of both products, you can see which...

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