## Search found 23 matches

Sat Mar 18, 2017 12:52 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 2014 Final 1C
Replies: 3
Views: 488

### Re: 2014 Final 1C

For this question, the solution in the course reader says q = m*C*delta T + 1/2*m*(enthalpy of vaporization). Why is the enthalpy of vaporization multiplied by 1/2?
Fri Mar 17, 2017 1:17 am
Forum: Student Social/Study Group
Topic: Post All Chemistry Jokes Here
Replies: 7750
Views: 1066636

### Re: Post All Chemistry Jokes Here

What was the charge when NaCl was arrested?

A salt!
Fri Mar 17, 2017 1:04 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 2016 final question 3B
Replies: 4
Views: 378

### Re: 2016 final question 3B

In addition, the more positive the value is for the standard reduction potential, the more likely the substance is going to be reduced.
Fri Mar 17, 2017 12:58 am
Forum: *Haloalkenes
Topic: Green Book Chapter 2 Question 16
Replies: 2
Views: 973

### Re: Green Book Chapter 2 Question 16

When I read the Halides section of the book again, it did not say anything about the halides getting priority over the triple bond. So, that's why I think the answer is the way it is in the textbook. Hope this helps!
Thu Mar 16, 2017 8:05 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 2016 Final
Replies: 3
Views: 498

### Re: 2016 Final

I think it's because it is easier to use whole numbers instead of fractions, since half of 2x would be x. However, I think both ways should yield the same result.
Thu Mar 16, 2017 7:58 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 2016 Final
Replies: 1
Views: 262

### Re: 2016 Final

I think to write the overall rate law, the equation would be 2ClO(g)+ O2(g) -->2ClO2(g), in which the O2 from step one and the 2O2 from step two would be cancelled and become 1 O2(g). Therefore, the rate law would be rate = k[ClO2]2[O2].
Thu Mar 09, 2017 12:14 am
Forum: *Alkanes
Topic: Tert vs. Neo?
Replies: 4
Views: 552

### Re: Tert vs. Neo?

However, neo can only be used for pentane and up while tert can apply to butane.
Wed Feb 15, 2017 1:42 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Midterm 2014 Question 8
Replies: 4
Views: 424

### Re: Midterm 2014 Question 8

Ka = sqrt (K) because Ka written out is Ka = ([H][F])/[HF] and K = ([H]^2[F]^2)/[HF]^2 since the equation is 2HF(aq) -->2H(plus) + 2F(minus).
Wed Feb 15, 2017 1:37 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Book problem 14.15
Replies: 1
Views: 247

### Re: Book problem 14.15

Hi! So for part a) of this problem, the equation is AgBr --> Ag(plus) + Br (minus). So for the half reactions, you look at how the oxidation state of each reactant changes. In the case of Ag, the oxidation state changes from 0 to +1, which means that it is being oxidized since it is losing one elect...
Fri Feb 10, 2017 4:01 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Equilibrium Redox Equations and Calculating their standard cell potentials
Replies: 1
Views: 195

### Re: Equilibrium Redox Equations and Calculating their standard cell potentials

Hi! Okay so for part a of this problem you have the equation AgBr(s) -->Ag(+) + Br-(aq). So in this case you have the two half reactions of Ag(s) --> Ag(+) +e- and Br + e- -->Br-(aq). You can see that Ag is being oxidized and Br is being reduced based on the reaction either losing or gaining electro...
Thu Feb 02, 2017 2:03 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bond Enthalpies
Replies: 2
Views: 320

### Re: Bond Enthalpies

In addition, when bonds are formed, the bond enthalpies are negative, and when the bonds are broken, the bond enthalpies are positive. So, you can then just add up the new bonds formed and broken.
Sat Jan 14, 2017 8:00 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.21 Homework Problem
Replies: 1
Views: 280

### Re: 8.21 Homework Problem

Hi! So, since you know that 0 energy is lost to the surroundings, you then set up the primary equation q(copper)+ q(water) = 0, where q = m*C*delta T. However, since you are solving for temperature, the equation for copper would be q = (20.0g)*(0.38 J/(g*degrees C))*(T - 100 degrees C) and the equat...
Fri Dec 02, 2016 11:54 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Endothermic or exothermic
Replies: 1
Views: 306

### Re: Endothermic or exothermic

Hi! When the Kc increases along with the temperature, the reaction is endothermic because Kc and temperature have an inverse relationship for exothermic reactions, whereas Kc and temperature have a directly proportional relationship for endothermic reactions. Hope this helps!
Fri Dec 02, 2016 6:39 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Sig Figs in pH
Replies: 2
Views: 339

### Re: Sig Figs in pH

I think when in doubt, it is always best to keep 3 sig figs, even for ph :)
Sat Nov 26, 2016 9:52 pm
Forum: Polyprotic Acids & Bases
Topic: Writing Reaction Equation for Polyprotic Acids/Bases
Replies: 2
Views: 535

### Re: Writing Reaction Equation for Polyprotic Acids/Bases

Hi! So, in order to figure out the overall reaction for polyprotic substances, it would probably be easier to use two steps in order to balance the overall reaction equation and to figure out the K overall. However, two steps are not necessary; if you know how many H+ ions the substance accepts or d...
Tue Nov 22, 2016 8:59 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework 11.67
Replies: 1
Views: 214

### Re: Homework 11.67

Hello! So, you are given the initial pressures for each of the elements in the reaction. Using these initial values, you need to set up an ICE chart to find the x values within the equilibrium pressures of each element. In order to find x, you set the K value (given) equal to the expression of K wit...
Tue Nov 15, 2016 11:22 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: #9 Quiz 3 Prep Fall 2015
Replies: 3
Views: 468

### Re: #9 Quiz 3 Prep Fall 2015

Hello!

So the picture below should demonstrate how to solve the problem-- I had to use an ICE box in order to solve the problem. Hope this helps!
Wed Nov 09, 2016 2:04 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Equilibrium Constant vs Reaction Quotient [ENDORSED]
Replies: 4
Views: 883

### Re: Equilibrium Constant vs Reaction Quotient[ENDORSED]

Hi! Although Q and K are calculated the same way, the same numbers are not used. For Q, the numbers that are used for the calculation are not at equilibrium; rather, they are measured values at a specific point in time. However, when measuring K, the values used are at equilibrium, which is why Q an...
Wed Oct 26, 2016 11:25 am
Forum: Lewis Structures
Topic: Lone Pairs in expanded Octet?
Replies: 2
Views: 629

### Re: Lone Pairs in expanded Octet?

Another thing that can help you is looking at the formal charge of each of the elements in the compound. Since you want each atom to have a formal charge of zero, you can rearrange the bonds accordingly.
Thu Oct 20, 2016 11:59 am
Forum: Ionic & Covalent Bonds
Topic: Home work question 3.39
Replies: 3
Views: 535

### Re: Home work question 3.39

Potassium Phosphide is written as K3P. 1. First, you see that there are 3 K atoms and I P atom. 2. For K, there is only 1 valence electron, so multiplying one by three, you get 3 valence electrons for K. 3. For P, there are 5 valence electrons. Adding the P valence electrons to the K, you get 8 vale...
Thu Oct 13, 2016 11:42 am
Forum: Photoelectric Effect
Topic: Rydberg Constant
Replies: 2
Views: 312

### Re: Rydberg Constant

It is not possible to convert Hz to m^-1. However, Hz is equivalent to s^-1.

Hope this helped!
Wed Oct 12, 2016 2:48 pm
Forum: Properties of Electrons
Topic: Removing 2nd Electron
Replies: 2
Views: 731

### Re: Removing 2nd Electron

But lets say you are removing the first two electrons from Magnesium since it has two electrons in the outer shell. After you remove the first electron, there is only one left; is it easier to remove that second electron because Magnesium wants to be in a balanced state? Or is it still more difficul...
Sun Sep 25, 2016 12:35 pm
Forum: Empirical & Molecular Formulas
Topic: Assessment Question #8
Replies: 2
Views: 361

### Re: Assessment Question #8

Dear Joslyn, Another issue that I found with your calculation was with the hydrogen. After you divided the 7.88g of hydrogen given by 1.008 (the molar mass of the hydrogen element), you got the number 7.30 mol, whereas I re-did the calculation and got 7.82 mol. Then, after dividing by the lowest mol...