Search found 20 matches

by 404768057
Sun Jul 30, 2017 8:38 pm
Forum: Lewis Acids & Bases
Topic: Homework problem 12.25
Replies: 2
Views: 175

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-]
by 404768057
Sun Jul 30, 2017 6:35 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: 12.49
Replies: 5
Views: 308

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...
by 404768057
Sun Jul 30, 2017 6:09 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Relationship between pKa and pKb
Replies: 2
Views: 173

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.
by 404768057
Sun Jul 30, 2017 5:17 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: pKa vs Ka
Replies: 3
Views: 306

Re: pKa vs Ka

Why do we use pKa instead of Ka?
by 404768057
Sun Jul 30, 2017 4:32 pm
Forum: Bronsted Acids & Bases
Topic: HCl vs. HF
Replies: 2
Views: 153

HCl vs. HF

We know that HCl, HBr, and HI are strong acids but why is HF a weak acid?
by 404768057
Sun Jul 30, 2017 1:52 pm
Forum: Student Social/Study Group
Topic: Post All Chemistry Jokes Here
Replies: 7337
Views: 789146

Re: Post All Chemistry Jokes Here

I make bad chemistry puns.... but only periodically
by 404768057
Sat Jul 29, 2017 5:16 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: calculating K
Replies: 2
Views: 158

calculating K

Is there a difference in calculation when calculating atm or bar when finding K?
by 404768057
Sun Jul 23, 2017 7:27 pm
Forum: Lewis Structures
Topic: Energy in lewis structures
Replies: 3
Views: 214

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.
by 404768057
Sun Jul 16, 2017 4:25 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Quick Question about SO4
Replies: 8
Views: 626

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.
by 404768057
Sun Jul 16, 2017 4:00 pm
Forum: Ionic & Covalent Bonds
Topic: Polar vs. nonpolar
Replies: 2
Views: 187

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.
by 404768057
Sun Jul 16, 2017 3:08 pm
Forum: Ionic & Covalent Bonds
Topic: Bonds
Replies: 2
Views: 244

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.
by 404768057
Sun Jul 16, 2017 3:03 pm
Forum: Formal Charge and Oxidation Numbers
Topic: formal charge vs. octet
Replies: 3
Views: 224

formal charge vs. octet

When do we know when finding the formal charge is more important than using the octet rule?
by 404768057
Sun Jul 09, 2017 6:23 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Exercise 2.29
Replies: 9
Views: 653

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?
by 404768057
Sun Jul 09, 2017 5:25 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Question 2.39
Replies: 1
Views: 222

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 ...
by 404768057
Sat Jul 08, 2017 11:20 pm
Forum: SI Units, Unit Conversions
Topic: Molar Mass
Replies: 7
Views: 646

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.
by 404768057
Sat Jul 08, 2017 11:00 pm
Forum: Limiting Reactant Calculations
Topic: Excess reactant
Replies: 1
Views: 244

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?
by 404768057
Sun Jul 02, 2017 7:01 pm
Forum: Properties of Light
Topic: Amplitude of the wave [ENDORSED]
Replies: 7
Views: 456

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.
by 404768057
Sun Jul 02, 2017 6:39 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: chapter 1 question 57
Replies: 2
Views: 224

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?
by 404768057
Sun Jul 02, 2017 4:47 pm
Forum: Properties of Light
Topic: Chapter 1 Number 19 [ENDORSED]
Replies: 2
Views: 252

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...
by 404768057
Fri Jun 30, 2017 8:42 pm
Forum: Properties of Light
Topic: Exercise 1.23 Pg 27 Ch 1
Replies: 2
Views: 201

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...

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