## Search found 29 matches

Tue Mar 14, 2017 6:32 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 2014 Midterm Question 3
Replies: 3
Views: 466

### Re: 2014 Midterm Question 3

Yes you would have to choose the chemicals containing Zn and Fe in order to get full credit. To determine this you would have to look at the standard reduction potentials of Zn+2, Cu+2, Fe+2, and Fe+3 to determine the one with the lowest reduction potential and the one with the highest reduction pot...
Tue Mar 07, 2017 7:15 pm
Forum: *Alkanes
Topic: Section 1.1 pg 7
Replies: 1
Views: 333

### Re: Section 1.1 pg 7

When counting the carbons to determine the name of the parent chain, you have to look for the longest chain. In order to determine which carbons can be part of the chain you could try tracing a line with your finger between all the connected carbons. If you have to pick up your finger or go backward...
Wed Mar 01, 2017 5:10 pm
Forum: General Rate Laws
Topic: Questions from last quiz
Replies: 1
Views: 347

### Re: Questions from last quiz

Yes you cannot determine the rate law by just looking at the stoichiometry of the overall reaction. You can however determine the order of the reaction if you are given the reaction mechanism because the species involved in the slow step determine the rate of the reaction. So since you know that the...
Sun Feb 26, 2017 11:09 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Activation Energy
Replies: 3
Views: 535

### Re: Activation Energy

A reaction that produces products that are lower in energy than the reactants is exothermic and a reaction that produces products that are higher in energy than the reactants is endothermic. Both exothermic and endothermic reactions require activation energy and you would not be able to tell if the ...
Sat Feb 18, 2017 5:05 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Homework 15. 63
Replies: 2
Views: 579

### Re: Homework 15. 63

If k= the rate constant for the forward reaction and k'= the rate constant for the reverse reaction then yes K for the forward reaction = k/k'. In 15.63 though the Arrhenius equation for two temperatures is used. This formula is derived from the formula given in the course reader as ln k = ln A - (E...
Sat Feb 18, 2017 4:26 pm
Forum: General Rate Laws
Topic: 15.9 [ENDORSED]
Replies: 4
Views: 615

### Re: 15.9[ENDORSED]

Rates of reactions are given in Molarity per second (M/s) therefore, depending on the order of the reaction, the rate constant will have a different units. For example when the reaction is zero order the rate law is in the form rate=k so in order for the units of the rate to be M/s the rate constant...
Thu Feb 09, 2017 9:47 pm
Forum: Balancing Redox Reactions
Topic: 14.11b
Replies: 1
Views: 286

### Re: 14.11b

Just like platinum, C(s, graphite) is often used as an inert electrode because it is conductive, inexpensive, and often does not get involved with reactions due to those reactions with graphite being not thermodynamically favored. For example in 14.11b, the C(s, graphite) is used as an inert electro...
Thu Feb 09, 2017 9:31 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.19
Replies: 1
Views: 275

### Re: 14.19

The problem actually tells you in the second sentence that "copper was connected as the anode."
Sat Feb 04, 2017 2:03 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Units of entropy [ENDORSED]
Replies: 3
Views: 481

### Re: Units of entropy[ENDORSED]

The same rules of units apply as when we were calculating enthalpy. If the moles cancel out, then your final answer should not be J/(K * mol) unless the problem is asking for your answer per mole in which case you would divide your answer by the number of moles given in the problem or the chemical e...
Fri Jan 27, 2017 4:50 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Constant Temperature and Pressure
Replies: 2
Views: 369

### Re: Constant Temperature and Pressure

I thought about it some more and let me just clarify what I was trying to say. So because exothermic reactions generate heat, they would likely increase the temperature of the system or reaction so essentially this would mean that the temperature is not constant. To approximately maintain constant t...
Fri Jan 27, 2017 11:23 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Constant Temperature and Pressure
Replies: 2
Views: 369

### Re: Constant Temperature and Pressure

When a problem says that the reaction is occurring at constant temperature and pressure this means that the external temperature and pressure are not changing. The temperature or pressure of the system can still change but these changes would not affect the temperature or pressure of the often much ...
Fri Jan 27, 2017 10:58 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: HW 8.117
Replies: 1
Views: 342

### Re: HW 8.117

The problem is asking for the change in internal energy for 1 mol of H2 but the chemical equation shows the production of 3 mol H2. Therefore, you have to divide each of the stoichiometric coefficients by 3 mol so that the equation reads 1/3CH4(g) + 1/3H2O(g) → 1/3CO2(g) + H2(g). Now you can see tha...
Wed Jan 18, 2017 5:28 pm
Forum: Calculating Work of Expansion
Topic: Homework 8.9 [ENDORSED]
Replies: 3
Views: 589

### Re: Homework 8.9[ENDORSED]

In a reversible expansion the surrounding's pressure would not be constant and the system would have to be at equilibrium with the surroundings meaning that they would have the same pressure since a reversible expansion is defined as a process where infinitely small changes in the external or intern...
Wed Jan 11, 2017 10:08 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Chapter 8 Homework Hess's Law #57
Replies: 3
Views: 6297

### Re: Chapter 8 Homework Hess's Law #57

Although the reactions are not written out the ΔHc° is given and this is the standard enthalpy of combustion or the change in enthalpy when the compound undergoes combustion. Since combustion is a reaction with oxygen you can write out each of the reaction equations with their corresponding ΔHc° val...
Wed Jan 11, 2017 4:30 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Ch 8 Q 25
Replies: 1
Views: 378

### Re: Ch 8 Q 25

In order to determine the heat capacity of the calorimeter you have to first calibrate it by supplying a known amount of heat and measuring the temperature change. The heat capacity of the calorimeter can then be calculated using the formula Ccal=(qcal/change in temperature) because by definition he...
Mon Nov 28, 2016 10:28 am
Forum: Polyprotic Acids & Bases
Topic: Protonation Question
Replies: 1
Views: 554

### Re: Protonation Question

Polyprotic acids are acids that can donate more than more than 1 H+ ion and their conjugate bases can accept more than 1 H+ ion. You typically do not need to assume how many H+ ions will be lost/accepted as you can tell how many H+ ion the acid will donate by determining how many acidic hydrogen ato...
Mon Nov 21, 2016 10:30 pm
Forum: Equilibrium Constants & Calculating Concentrations
Replies: 1
Views: 369

### Re: Adding 2 equilibrium reactions

If you add the two equations together and write the equilibrium constant you get 2BrCl + H2 = Br2 + 2HCl with K=([Br2][HCl]^2)/([BrCl]^2[H2]). You can then compare this to the equilibrium constant expressions for the original 2 reactions K=[BrCl]^2/[Br2][Cl2] and K=[H2][Cl2]/[HCl]^2. When you look a...
Mon Nov 21, 2016 10:01 pm
Forum: Administrative Questions and Class Announcements
Topic: Chemistry Community Post Points
Replies: 2
Views: 700

### Re: Chemistry Community Post Points

Sorry I'm pretty sure that you can only earn 1 point per week so if you miss one week posting twice the next week can't make up for it.
Mon Nov 14, 2016 6:11 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Temperature Effect on Concentration
Replies: 1
Views: 1574

### Re: Temperature Effect on Concentration

You can think of heat as a reactant or product of a reaction so an increase in temperature will add heat to either the products, when it is an exothermic reaction, or the reactants when it is a endothermic reaction. According to Le Chatelier's Principle, the chemical reaction will shift to minimize ...
Mon Nov 14, 2016 6:01 pm
Forum: Naming
Topic: Quiz 3 Prep2 #5
Replies: 1
Views: 327

### Re: Quiz 3 Prep2 #5

So first make sure you balance the chemical equation. You should get N2 + 3H2 <==> 2NH3.
Then use [Products]/[Reactants] to solve for Kp:
Kp=[NH3]^2/[N2][H2]^3
Kp=[0.887]^2/[0.0561][0.168]^3
You should get Kp=2.96*10^3
Mon Nov 07, 2016 8:51 pm
Forum: Hybridization
Topic: 4.95 Composition of Bonds/Hybridization of Lone Pairs
Replies: 1
Views: 573

### Re: 4.95 Composition of Bonds/Hybridization of Lone Pairs

To describe the composition of the bonds you first determine if its a sigma or a pi bond. Single bonds are sigma bonds, double bonds have both a sigma and a pi bond, and triple bonds have 2 pi and 1 sigma. Next you determine the hybridization of each atom and apply that to each of the bonds. The ene...
Thu Nov 03, 2016 12:03 am
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: Bonding vs. Anti-bonding
Replies: 2
Views: 1859

### Re: Bonding vs. Anti-bonding

Molecular orbital theory is just another theory about bonding like Lewis structures and the valence-bond theory. When you think of orbitals as wavefunctions then you can consider that when two orbitals or wavefunctions interact they can interfere constructively or destructively. So when the atomic o...
Wed Oct 26, 2016 4:59 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Shape of S2O3 2-
Replies: 1
Views: 3800

### Re: Shape of S2O3 2-

In S2O3 2- the central sulfur atom is bonded to 1 sulfur and 3 oxygens with no lone pairs so there are 4 regions of electron density which must be arranged tetrahedrally. And yes the atom with the lowest ionization energy is generally the central atom although there are exceptions like Hydrogen, whi...
Wed Oct 26, 2016 4:45 pm
Forum: Ionic & Covalent Bonds
Topic: Bond Angles?
Replies: 2
Views: 631

### Re: Bond Angles?

Bond angles can be calculated using the geometry of the shape for molecules with no lone pairs. So for linear molecules if you think about the unit circle the terminal atoms are at 0 and 180 so they are 180 apart. For trigonal planar molecules the 3 terminal atoms are spread out evenly in a 360 degr...
Sat Oct 22, 2016 6:15 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Removing an electron from Au (problem2.47)
Replies: 2
Views: 344

### Re: Removing an electron from Au (problem2.47)

When forming cations, the outermost electrons are removed. In Au the outermost electron is found in the 6s state since typically the 6s state is considered to be higher in energy and therefore farther away from the nucleus than the 5d state. This means that it is easier to remove electrons from the ...
Thu Oct 13, 2016 8:14 pm
Forum: Trends in The Periodic Table
Topic: Chapter 2, Homework Problem 2.93
Replies: 2
Views: 442

### Re: Chapter 2, Homework Problem 2.93

I think you’re right and there was an error in the solutions manual because there is a document on the class website that is called Solution Manual Errors and Ch. 2 Homework Problem 2.93 is listed with the actual answer of A=Cl; B=Na; C=Cl-; D=Na+. This makes sense though because sodium should have ...
Thu Oct 06, 2016 11:24 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Balmer/Lyman Series [ENDORSED]
Replies: 3
Views: 567

### Re: Balmer/Lyman Series[ENDORSED]

When Bohr's model of the H atom was established, free electrons (not part of the atom) were considered to have 0 energy. Electrons closer to the nucleus have less energy than those that are farther away. Since free electrons are the farthest away (n=∞) and have 0 energy then in comparison electrons ...
Thu Oct 06, 2016 11:10 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Homework 1.25 part b [ENDORSED]
Replies: 1
Views: 292

### Re: Homework 1.25 part b[ENDORSED]

The energy that you calculated in part a is the energy per photon (3.37 X 10^-19 J). 1 photon corresponds to 1 excited sodium atom. So in part b you need to convert the 5.00 mg of sodium atoms into number of Na atoms so that you can use the 3.37 X 10^-19 J per atom to solve for the total energy of t...
Thu Sep 29, 2016 9:48 pm
Forum: Photoelectric Effect
Topic: Photoelectric Effect Lecture Question
Replies: 4
Views: 614

### Re: Photoelectric Effect Lecture Question

The energy needed to eject an electron depends on the metal and its work function or threshold energy. For example the threshold energy for sodium metal is around 2.3 eV (1 eV = 1.60 X 10^-19). So approximately E threshold = 3.68 X 10^-19 J. Using the formula λ = (hc)/E you find that the wavelength ...