Search found 110 matches
- Tue Oct 08, 2019 7:02 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Spectral Lines
- Replies: 3
- Views: 157
Re: Spectral Lines
Location on the electromagnetic spectrum doesn't exactly determine which series a line belongs to. Each line represents energy released by the movement of a hydrogen electron from one energy level to a lower energy level. Each energy level is represented by what's called the principal quantum number...
- Tue Oct 08, 2019 6:14 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Fundamentals E17
- Replies: 2
- Views: 339
Re: Fundamentals E17
The question is asking for the sample with more moles. In part a, you would convert both the 75g indium and 80g tellurium into their respective moles by dividing the substance molar mass. 75g \textrm{In} \div 114.82g/mol = 0.65 mol \textrm{In} 80g \textrm{Te} \div 127.60g/mol = 0.63 mol \textrm{Te} ...
- Tue Oct 08, 2019 5:42 pm
- Forum: Photoelectric Effect
- Topic: Topic 1B Quantum Theory Example
- Replies: 2
- Views: 106
Re: Topic 1B Quantum Theory Example
The mass of an electron is 9.109 x 10^(-31) kg. If we needed to use this constant on a test, I'm sure we would be given the value on a sheet that has all of the formulas and constants.
- Tue Oct 08, 2019 5:23 pm
- Forum: Properties of Electrons
- Topic: Electron Gaining/Losing energy
- Replies: 5
- Views: 687
Re: Electron Gaining/Losing energy
Just to add to the topic, when an electron goes down to lower energy levels, it releases energy in the form of photons.
- Thu Oct 03, 2019 6:02 pm
- Forum: Limiting Reactant Calculations
- Topic: Alternate way of solving for limiting reactant
- Replies: 2
- Views: 170
Re: Alternate way of solving for limiting reactant
It doesn't matter which product you use. As long as you convert each reactant into the same product, you will be able to compare how much product each reactant produced. Whichever reactant produced less of the product is the limiting reactant.
- Tue Oct 01, 2019 12:38 am
- Forum: Limiting Reactant Calculations
- Topic: Limiting Reactant Calculation post-module question
- Replies: 4
- Views: 169
Re: Limiting Reactant Calculation post-module question
Since C6H9Cl3 is the limiting reactant and AgCl is the product we want to calculate, we need to focus on the molar ratio between them, which is 1 to 1. Because the molar ratio is 1 to 1 (meaning they have the same coefficients in the balanced equation), 0.004 mol C6H9Cl3 produces 0.004 mol AgCl. The...
- Tue Oct 01, 2019 12:22 am
- Forum: Limiting Reactant Calculations
- Topic: How to Find a Mass of A Product
- Replies: 4
- Views: 764
Re: How to Find a Mass of A Product
We can't find the mass of a product with only molar mass. We need to be given some more information, like moles of reactants or products for example. If moles of a product is given, we multiply that number by the molar mass of that product to find the mass in grams. If moles of reactants are given, ...
- Mon Sep 30, 2019 11:57 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Accuracy vs Precision
- Replies: 10
- Views: 405
Re: Accuracy vs Precision
A simple way for me to figure out the difference between these two terms is to think of precision as consistency and accuracy as the bullseye of a target.
- Mon Sep 30, 2019 11:45 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Help on G25
- Replies: 3
- Views: 241
Re: Help on G25
I see this as a dilution problem where we can use MV = MV. To represent the volume doubling 90 times, we can use this expression: (0.10)(2^90)L. The equation should be (0.10M)(0.01L) = (? M) [(0.10)(2^90)L] . By isolating the final molarity, we can find its value. Then multiply that value by 0.01L t...
- Mon Sep 30, 2019 11:07 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Advice from a Medical Student - Part II [ENDORSED]
- Replies: 298
- Views: 299157
Re: Advice from a Medical Student - Part II [ENDORSED]
Thank you for sharing your path to becoming a doctor. I feel overwhelmed with how long and difficult the journey through medicine is, but these types of stories help give me hope and inspiration.