Hi!

I would say when the problem involves a gas and gives you the values for enough units like pressure, temperature, volume, or moles then you should implement the Ideal Gas Law!

## Search found 18 matches

- Sun Jan 24, 2021 4:05 pm
- Forum: Ideal Gases
- Topic: How can we identify when to use the ideal gas law?
- Replies:
**11** - Views:
**67**

- Sun Jan 24, 2021 2:02 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Sapling week 2 #8
- Replies:
**10** - Views:
**96**

### Re: Sapling week 2 #8

Hi Lexy! This personally how I went about solving this problem: 1. Since Kb is given to you in this problem and the constant for Kw is known as 1.0 x 10^-14, I solved for Ka using the equation: Ka x Kb = Kw. 2. Once you calculate the value for Ka, you should notice that it is significantly small; th...

- Sat Jan 23, 2021 11:43 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling week 2 #7
- Replies:
**2** - Views:
**26**

### Re: Sapling week 2 #7

Hi Lexy! I'm not sure what all your given numbers are for your version of this problem, but at this point of the problem your Kb should = [HClO] [OH-] / [ClO-]. This should tell you whether you plugged in your initial concentration correctly, as well as creating the chemical equation. From this poin...

- Sat Jan 23, 2021 11:09 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling week 2 #4
- Replies:
**3** - Views:
**24**

### Re: Sapling week 2 #4

Hi Lexy!

If I understood correctly, it appears that your value (1.5 x 10^-3) is the value for [OH-]. This means that when you plug that value into -log you would be solving for your pOH. Hence the equation: pOH = -log[OH-].

I hope that helped!

If I understood correctly, it appears that your value (1.5 x 10^-3) is the value for [OH-]. This means that when you plug that value into -log you would be solving for your pOH. Hence the equation: pOH = -log[OH-].

I hope that helped!

- Sat Jan 23, 2021 11:02 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Sapling week 2 #5
- Replies:
**2** - Views:
**44**

### Re: Sapling week 2 #5

Hi! Here's how I went about solving this problem! 1. Write out the chemical equation and the equilibrium concentration = Kb. 2. Find the pOH using the given pH. 3. Then use the equation [OH-] = 10^-pOH. The concentrations of both [OH-] and [BH+] from the products side of the chemical equation that y...

- Sun Jan 17, 2021 6:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Question 4
- Replies:
**5** - Views:
**46**

### Re: Sapling Question 4

Hi! This is how I went about solving this problem: Also, my Kp was 363 and my initial pressure was 0.0290 bar. I started off this problem by writing out the equilibrium concentration for Kp equal to 363. I then created an ICE table // don't forget to include your given initial pressure for PCl5 in t...

- Sun Jan 17, 2021 6:28 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: 5% rule
- Replies:
**5** - Views:
**52**

### Re: 5% rule

Hi Haley!

You can do so by checking to see if your Ka or Kb value is less than 10^-3, or 10^-4 to be safe;

if it is then you can go ahead and use the 5% rule.

You can do so by checking to see if your Ka or Kb value is less than 10^-3, or 10^-4 to be safe;

if it is then you can go ahead and use the 5% rule.

- Sun Jan 17, 2021 12:32 pm
- Forum: General Science Questions
- Topic: class 1/18/21
- Replies:
**13** - Views:
**104**

### Re: class 1/18/21

Hi Lexy!

There will be no class on Monday, aka tomorrow.

The statement I have pasted is taken directly from an email Dr. Lavelle sent out on Wednesday, January 13th:

"Monday is a holiday: no class, no peer learning sessions, no office hours."

I hope that helps!

There will be no class on Monday, aka tomorrow.

The statement I have pasted is taken directly from an email Dr. Lavelle sent out on Wednesday, January 13th:

"Monday is a holiday: no class, no peer learning sessions, no office hours."

I hope that helps!

- Sun Jan 17, 2021 12:45 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: X less than 5 percent
- Replies:
**11** - Views:
**119**

### Re: X less than 5 percent

Hi!

If I am understanding your question correctly, that is a reason why we use the 5% rule!

If we were to calculate a percentage greater than 5% we must apply the quadratic equation instead because the approximation method was invalid.

If I am understanding your question correctly, that is a reason why we use the 5% rule!

If we were to calculate a percentage greater than 5% we must apply the quadratic equation instead because the approximation method was invalid.

- Sun Jan 17, 2021 12:36 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Solvents in Equilibrium Constants
- Replies:
**3** - Views:
**16**

### Re: Solvents in Equilibrium Constants

Hi Samantha!

To answer your first question, we do not have to put solvents in equilibrium constant calculations because they would end up canceling out on both sides of the reaction anyway. This is because their effect or change is insignificant.

To answer your first question, we do not have to put solvents in equilibrium constant calculations because they would end up canceling out on both sides of the reaction anyway. This is because their effect or change is insignificant.

- Sat Jan 16, 2021 7:56 pm
- Forum: Ideal Gases
- Topic: Hw Question #4
- Replies:
**5** - Views:
**56**

### Re: Hw Question #4

Hi Brandon! I started off this problem by writing out the equilibrium concentration for Kp equal to 363. Then I created an ICE table; and of course don't forget to include the given initial pressure for PCl5. From these steps, I was able to get the equation 363x^2 + x -0.0290 = 0, which will be used...

- Sat Jan 16, 2021 1:32 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Acid and Base
- Replies:
**3** - Views:
**27**

### Re: Acid and Base

Hi! Karl is completely right in their explanation; I would also like to add that it would be a great idea to search up a list of the strong/weak acids and bases online. Personally this helped me identify them more easily and efficiently, and it might help you visualize and assess their differences a...

- Sun Jan 10, 2021 6:06 pm
- Forum: Administrative Questions and Class Announcements
- Topic: missing section
- Replies:
**7** - Views:
**74**

### Re: missing section

Hi Kylie! I pulled this straight from the syllabus: "TA Discussion Sections: Attend only your TA discussion section you are enrolled in." This statement can be found on page 1 of the syllabus. Although, my TA told us that discussion attendance is not mandatory, and that it will not affect ...

- Wed Jan 06, 2021 8:29 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 1B Module Question 15
- Replies:
**1** - Views:
**44**

### Re: 1B Module Question 15

Hi Jaden! So in this problem we are going to use Kc = [P] / [R] to find the equilibrium constant for this reaction. The chemical equation given to us is already balanced; sometimes for these problems we are given chemical equations that are not balanced -- when that is the case, you must 1st balance...

- Wed Jan 06, 2021 1:54 pm
- Forum: General Science Questions
- Topic: step up sessions
- Replies:
**16** - Views:
**209**

### Re: step up sessions

Hi! https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Chem14UA_PL_ALL.pdf I believe you can find all the appropriate zoom links there! This can also be found in the recent email chain Dr. Lavelle has sent out under the blue link: "Peer Learning Sessions and Step-Up Program".

- Wed Jan 06, 2021 11:14 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Endothermic vs Exothermic Stabilities
- Replies:
**8** - Views:
**126**

### Re: Endothermic vs Exothermic Stabilities

Hi Sid! In an exothermic reaction, the change in enthalpy is always negative and energy is being released. When looking at an energy diagram for an exothermic reaction, you will see that the reactants are at a higher energy level while the products are at a visibly lower energy level. The reasoning ...

- Wed Jan 06, 2021 12:09 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Part 2 module review #19
- Replies:
**3** - Views:
**45**

### Re: Part 2 module review #19

Yes, you're correct, as the others stated! You would omit the solid; I just wanted to add: do not forget to balance the chemical equation given: your ultimate product / reactant would appear as such: [H2]^3 / [AsH3]^2, aka the equilibrium constant. And also! For this problem do not forget to convert...

- Tue Jan 05, 2021 11:25 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Video Module Part 2
- Replies:
**6** - Views:
**65**

### Re: Video Module Part 2

Hi Vivian! To my understanding, because there are in fact more Reactants than Products, in other words [R] > [P] for this particular example, this then relates/leads to R proceeding towards the Product, P, hence R --> P. Reactants become Products. When [P] > [R], than there are more products so it t...