Search found 20 matches
- Sun Jul 30, 2017 8:38 pm
- Forum: Lewis Acids & Bases
- Topic: Homework problem 12.25
- Replies: 2
- Views: 476
Re: Homework problem 12.25
Yes, you know that Kw is 1.0x10^-14 and you have OH- so you can find H3O+ by using Kw=[H3O+][OH-]
- Sun Jul 30, 2017 6:35 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 12.49
- Replies: 5
- Views: 1307
Re: 12.49
Hypobromite ion is the conjugate base of hypobromous acid. From the table 12.1, the pKa of hypobromous acid is 8.69. We know that pKa+pKb=14. To get pKb, take 8.69 from 14 and you get 5.31. From table 12.2, we know that the pKb for C17H19O3N is 5.79.The pKb of hypobromite is smaller, which means tha...
- Sun Jul 30, 2017 6:09 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Relationship between pKa and pKb
- Replies: 2
- Views: 422
Re: Relationship between pKa and pKb
Yes, you can also think that as the pKa gets higher, the acid gets weaker and same for bases.
- Sun Jul 30, 2017 5:17 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: pKa vs Ka
- Replies: 3
- Views: 703
Re: pKa vs Ka
Why do we use pKa instead of Ka?
- Sun Jul 30, 2017 4:32 pm
- Forum: Bronsted Acids & Bases
- Topic: HCl vs. HF
- Replies: 2
- Views: 492
HCl vs. HF
We know that HCl, HBr, and HI are strong acids but why is HF a weak acid?
- Sun Jul 30, 2017 1:52 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3007663
Re: Post All Chemistry Jokes Here
I make bad chemistry puns.... but only periodically
- Sat Jul 29, 2017 5:16 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: calculating K
- Replies: 2
- Views: 435
calculating K
Is there a difference in calculation when calculating atm or bar when finding K?
- Sun Jul 23, 2017 7:27 pm
- Forum: Lewis Structures
- Topic: Energy in lewis structures
- Replies: 3
- Views: 560
Energy in lewis structures
How can you tell which structure has higher or lower energy? For example in 3.54, we have to identify which structure has the lowest energy.
- Sun Jul 16, 2017 4:25 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Quick Question about SO4
- Replies: 8
- Views: 1292
Re: Quick Question about SO4
Structure B is better because it shows that S has a lower formal charge when the double bonds are created which makes it more stable.
- Sun Jul 16, 2017 4:00 pm
- Forum: Ionic & Covalent Bonds
- Topic: Polar vs. nonpolar
- Replies: 2
- Views: 489
Re: Polar vs. nonpolar
I know that when there is a dipole, there is a negative charge with the atom with more electronegativity and more of a positive charge on the opposite side. That's when it becomes a polar bond.
- Sun Jul 16, 2017 3:08 pm
- Forum: Ionic & Covalent Bonds
- Topic: Bonds
- Replies: 2
- Views: 553
Re: Bonds
I think you can determine if a bond is ionic if the electronegativity difference is greater than 2. If the electronegativity difference is less than 1.5, it's a covalent bond.
- Sun Jul 16, 2017 3:03 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: formal charge vs. octet
- Replies: 3
- Views: 662
formal charge vs. octet
When do we know when finding the formal charge is more important than using the octet rule?
- Sun Jul 09, 2017 6:23 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Exercise 2.29
- Replies: 10
- Views: 3554
Re: Exercise 2.29
In 2.29 part c), I don't understand why the maximum number of electrons is 8 if n=2. Can someone please explain?
- Sun Jul 09, 2017 5:25 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Question 2.39
- Replies: 1
- Views: 447
Re: Question 2.39
In a) and in b), the configuration doesn't follow Hund's rule so it must be at an excited state. In c), the configuration doesn't follow Aufbau's principle which means that it would also be at an excited state. d) is the only configuration that follows Aufbau's principle and Hund's rule which means ...
- Sat Jul 08, 2017 11:20 pm
- Forum: SI Units, Unit Conversions
- Topic: Molar Mass
- Replies: 7
- Views: 1578
Re: Molar Mass
I think molecular weight and molar mass have different units because molecular weight comes from the total amount of the atomic mass in a molecule and molar mass is the amount of mass per mole of a molecule.
- Sat Jul 08, 2017 11:00 pm
- Forum: Limiting Reactant Calculations
- Topic: Excess reactant
- Replies: 1
- Views: 480
Excess reactant
Can someone please explain how after you calculate the amount of excess reactant, you would find the amount of limiting reactant needed to react with the excess reactant that is not used?
- Sun Jul 02, 2017 7:01 pm
- Forum: Properties of Light
- Topic: Amplitude of the wave [ENDORSED]
- Replies: 7
- Views: 1387
Re: Amplitude of the wave [ENDORSED]
I know from physics, that amplitude and energy are proportional. So the bigger the amplitude, the more energy. I'm not completely sure if it relates to the energy of a photon but that could get you thinking.
- Sun Jul 02, 2017 6:39 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: chapter 1 question 57
- Replies: 2
- Views: 487
chapter 1 question 57
The question asks: Lines of the Balmer series of the hydrogen spectrum are observed at 656.3, 486.1, 434.0, and 410.2 nm. What is the wavelength of the next line in the series?
I know that you can use Rydberg's formula to solve this but how could you solve this without using the Rydberg's formula?
I know that you can use Rydberg's formula to solve this but how could you solve this without using the Rydberg's formula?
- Sun Jul 02, 2017 4:47 pm
- Forum: Properties of Light
- Topic: Chapter 1 Number 19 [ENDORSED]
- Replies: 2
- Views: 526
Re: Chapter 1 Number 19 [ENDORSED]
So how I thought of it was that the question is asking how much energy is in that 1.00 mol of sodium. After you multiply avogadro's constant with the energy from part A, you get 203x10^5 J/mol. But if you think about what the question is asking for, it says how much energy is per that mole. So reall...
- Fri Jun 30, 2017 8:42 pm
- Forum: Properties of Light
- Topic: Exercise 1.23 Pg 27 Ch 1
- Replies: 2
- Views: 514
Re: Exercise 1.23 Pg 27 Ch 1
So after you get 2.2513x10^-14 J, that is your E in the formula. You can use the two formulas given: E=h(nu) and c=(lambda)(nu). Since nu is in both formulas, you can substitute them. when you substitute for nu, the new equation you get is E=hc/lambda. In order to find lambda, you can make the equat...