## Search found 173 matches

Sat Mar 13, 2021 12:07 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: Determining Change in Entropy
Replies: 1
Views: 16

### Determining Change in Entropy

Does the change in the number of moles affect the entropy of the system more than a change in states? For example, question 4H.7c asks to predict whether there is an increase or a decrease in entropy of the system for SO2(g) + Br2(g) + 2 H2O(l) --> H2SO4(aq) + 2 HBr(aq). The answer key states that t...
Fri Mar 12, 2021 8:48 pm
Forum: First Order Reactions
Topic: Form of a First Order Reaction
Replies: 3
Views: 29

### Form of a First Order Reaction

For homework problem 7B.3c, it gives the reaction 2A-->B+C but says that it is a first-order reaction. I thought this was the form of a second-order reaction since there are two molecules of A colliding with each other, but maybe I am not thinking about it correctly?

Any help would be appreciated!
Fri Mar 12, 2021 1:51 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Sapling Week 9/10 #13
Replies: 6
Views: 38

### Re: Sapling Week 9/10 #13

For this question, you need to use the pre-equilibrium approach. Since the rate is determined by the slow step, rate=k[HClO][I-]. However, HClO is an intermediate, so you need to find a way to rewrite it. We can use the fast equilibrium in step 1 to replace HClO. For step 1, K=[HClO][OH-]/[ClO-] whi...
Fri Mar 12, 2021 1:19 pm
Forum: Student Social/Study Group
Topic: 6L.3
Replies: 2
Views: 19

### Re: 6L.3

There is a chart of standard reduction potentials that can be found in the textbook that lists the E naught values for various reduction reactions. These values will be provided on an exam and they cannot be calculated in any way other than figuring them out experimentally.

Hope this helps!
Fri Mar 12, 2021 1:16 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Galvanic/ Concentration Cells
Replies: 4
Views: 23

### Re: Galvanic/ Concentration Cells

The main difference between a Galvanic cell and a concentration cell is that a concentration cell has the same electrodes, while a Galvanic cell has different electrodes, An example of a concentration cell is Ag(s)|Ag+(aq)(0.1 M)||Ag(aq)+(0.01 M)|Ag(s)|.

Hope this helps!
Fri Mar 12, 2021 12:45 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: How does a salt bridge work?
Replies: 3
Views: 27

### Re: How does a salt bridge work?

In a salt bridge, the anions flow from the cathode to the anode to reduce the positive charge and the cations flow from the anode to the cathode to reduce the negative charge. This helps keep each half-cell electrically neutral.

Hope this helps!
Fri Mar 12, 2021 12:40 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Textbook Problem 5.33 Parts A and D
Replies: 2
Views: 13

### Re: Textbook Problem 5.33 Parts A and D

For part A, you know the reaction is endothermic because breaking a bond requires energy and in this situation, the X-X bond is being broken to form 2X. For part D, adding a catalyst will speed up the rate of the reaction, but it will not cause the reaction to change if it was not already going to u...
Fri Mar 12, 2021 12:35 pm
Forum: Balancing Redox Reactions
Topic: Textbook 6K.5a
Replies: 6
Views: 30

### Re: Textbook 6K.5a

Since it is a redox reaction, you know you can add water and H+ to balance out any existing oxygens and hydrogens, so you can split the reaction up. The half-reactions would be O3-->O2 and the Br- -->BrO3- because the molecules with Br must be in a single half-reaction, and you can add waters to bal...
Sun Mar 07, 2021 11:19 pm
Forum: First Order Reactions
Topic: Order of Reactions
Replies: 9
Views: 42

### Re: Order of Reactions

The order of the reaction is not based on the number of reactants in the equation but on the order of each individual reactant. You can add the exponents on all of the reactants to figure out the order of the reaction. The order of the reaction basically tells you how the rate constant is affected b...
Sun Mar 07, 2021 11:11 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: K and Activation Energy
Replies: 5
Views: 19

### Re: K and Activation Energy

The rate constant, k, only changes based on temperature and due to the presence of a catalyst, so I do not think you could determine anything about k from the activation energy directly. However, the Arrhenius equation shows that the rate constant increases exponentially when the activation energy d...
Sun Mar 07, 2021 10:52 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Determining The Favored Side of an Equilibrium Product
Replies: 12
Views: 42

### Re: Determining The Favored Side of an Equilibrium Product

If Q is less than K, then the equilibrium shifts to the right or makes more products. On the other hand, if Q is greater than K, then the equilibrium shifts to the left or makes more reactants. This is because if Q is less than K, that means there is currently a higher concentration of reactants tha...
Sun Mar 07, 2021 7:58 pm
Forum: Zero Order Reactions
Topic: Catalysts
Replies: 6
Views: 53

### Re: Catalysts

Catalysts are mostly used with zero-order reactions because zero order reactions do not depend on the concentration of reactants, they only depend on the rate constant k, which is dictated by the catalyst. When the catalyst is saturated, that determines the rate of the reaction and k. Hope this helps!
Sun Mar 07, 2021 7:46 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Sapling Week 9/10 HW #6
Replies: 2
Views: 33

### Re: Sapling Week 9/10 HW #6

The graphs for the first-order and second-order reactions have very similar shapes, but in reality, they are slightly different. In a second-order reaction, the amount of reactant is decreasing at a faster rate, which is displayed by the graph. However, for the sake of this problem, both first and s...
Sun Mar 07, 2021 7:33 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Inert Conductors
Replies: 5
Views: 33

### Re: Inert Conductors

I think platinum is the most common inert conductor, but graphite can also be used and gold (for most occasions). Iodine is a poor conductor because it is a halogen and has 7 valence electrons. The best conductors have only 1 valence electron.

Hope this helps!
Sun Feb 28, 2021 12:09 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Sapling Week 7/8 #7
Replies: 3
Views: 19

### Re: Sapling Week 7/8 #7

The question tells you that the electrode on the left is the anode and the electrode on the right is the cathode. This means that Pb is being oxidized and Ag is being reduced. The anode is always written on the far left and the cathode is written on the far right of the shorthand notation. Then, you...
Sun Feb 28, 2021 11:55 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: HW Question
Replies: 2
Views: 12

### Re: HW Question

Reduction happens at the cathode, so the species that is being reduced is the cathode. Since Ni goes from an oxidation number of 2+ to an oxidation number of 0, we know it is being reduced, so it is the cathode. Zn goes from an oxidation number of 0 to +2, so it is being oxidized and is the anode. H...
Sun Feb 28, 2021 11:51 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Sapling question 6 week 7/8
Replies: 1
Views: 6

### Re: Sapling question 6 week 7/8

Since the X electrode more readily reacts to form a cation than Y does, that means that the solution of X^2+ will become very positive charged. Therefore, the anions must flow through the salt bridge to the X electrode to balance out this positive charge so that the electrons can continue to flow. H...
Sun Feb 28, 2021 11:45 am
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Sapling Week 7 and 8 HW Question 17
Replies: 3
Views: 32

### Re: Sapling Week 7 and 8 HW Question 17

For this problem, just use the Nernst equation. The E naught value is 0 and the value of Q can be found by using 4.5 atm as the partial pressure of the product and 0.42 M as the concentration of the reactants, so Q=4.5/0.42^2.

Hope this helps!
Sun Feb 28, 2021 1:28 am
Forum: Balancing Redox Reactions
Topic: Sapling Week 5 Second Section
Replies: 2
Views: 34

### Re: Sapling Week 5 Second Section

To solve this question, divide the overall reaction into its two half-reactions. For the half-reaction involving ClO−, add water to balance out the oxygens and then add hydrogens to the other side to balance out the hydrogens from the water. Since it is in a basic solution, add enough OH- to the sid...
Sun Feb 28, 2021 1:20 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Equation for E°
Replies: 3
Views: 40

### Re: Equation for E°

The equation you use depends on what values of E° you are using. If you are given the standard reduction potentials, then you should use E° cell = E° cathode - E° anode. The negative sign before the E° anode basically switches the value to be the standard potential of the oxidation reaction. If you ...
Sun Feb 21, 2021 8:53 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Sapling #9, Week 7/8
Replies: 8
Views: 53

### Re: Sapling #9, Week 7/8

For this question, start by looking up the standard reduction potentials for the two half-reactions. Since the reaction in a galvanic cell is spontaneous, the cell potential must be positive. Therefore, using the equation E°(cell)= E°(cathode) - E°(anode), we know that the half-reaction with the lar...
Sun Feb 21, 2021 8:41 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: HW Question
Replies: 1
Views: 27

### Re: HW Question

For this question, start by looking up the standard reduction potentials for the two half-reactions. Since the reaction in a galvanic cell is spontaneous, the cell potential must be positive. Therefore, using the equation E°(cell)= E°(cathode) - E°(anode), we know that the half-reaction with the lar...
Sun Feb 21, 2021 8:08 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Different Standard Reduction Potentials
Replies: 2
Views: 12

### Different Standard Reduction Potentials

What causes one compound to have a higher standard reduction potential than another?
Sun Feb 21, 2021 7:21 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: E Cell Equation
Replies: 2
Views: 25

### Re: E Cell Equation

I believe the salt bridge or the porous disk is used to solve this problem. You are correct in understanding that the electron transfer to the right beaker will make it more negative, eventually causing the current to stop. However, the salt bridge allows ions to flow in between the two solutions an...
Sun Feb 21, 2021 6:30 pm
Forum: Balancing Redox Reactions
Topic: Easy way to remember reduction/oxidazing agents?
Replies: 10
Views: 49

### Re: Easy way to remember reduction/oxidazing agents?

The easiest way to know which compound is reduced vs oxidized is just to understand the oxidation numbers. The compound that loses electrons (and has an increase in oxidation number) is oxidized while the compound that gains electrons (and has a decrease in oxidation number) is reduced. The phrase &...
Sun Feb 21, 2021 6:20 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Standard Hydrogen Electrode [ENDORSED]
Replies: 2
Views: 45

### Re: Standard Hydrogen Electrode[ENDORSED]

The cell refers to the Galvanic cell where the electrical current is generated by the redox reactions. We do not have to do anything to compare the e- transfer to the standard hydrogen electrode, it is just important to know that the standard reduction potentials, or E naught values, are given with ...
Sun Feb 14, 2021 3:31 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Sapling 3
Replies: 13
Views: 86

### Re: Sapling 3

Any endothermic process has a positive delta H value as heat is entering the system. Therefore, going from a solid to a liquid or a liquid to a gas is associated with a positive delta H value. Similarly, an increase in entropy indicates an increase in the possible microstates or orientations of a mo...
Sun Feb 14, 2021 3:23 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Sapling wk 5/6 #7
Replies: 2
Views: 33

### Re: Sapling wk 5/6 #7

To begin this problem, it is important to understand that the heat from the heater can be converted into J/s and then multiplied by the time the sample is heated, to find the total heat supplied. ΔHvap can be calculated by dividing the heat supplied by the moles of vapor, so subtract the initial mas...
Sun Feb 14, 2021 3:14 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Sapling 5/6 #19
Replies: 4
Views: 60

### Re: Sapling 5/6 #19

For this question, start by using the partial pressures to find the reaction quotient: Q=([NH3]^2/[N2]*[H2]^3)
Once you have the value for Q, you can use the equation ΔG= ΔG∘ + RT(ln(Q)), where R is the gas constant 8.314 J/K*mol, to find ΔG.

Hope this helps!
Sun Feb 14, 2021 3:10 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Sapling #15
Replies: 3
Views: 32

### Re: Sapling #15

You can treat this problem like the Hess's Law questions. Combine the two equations in a way that results in the final reaction A+B-->2C. If you need to flip the products and reactants of one of the equations, make sure to change the sign of the entropy and the enthalpy, and if you multiply the equa...
Sun Feb 14, 2021 1:20 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: Residual Entropy
Replies: 1
Views: 9

### Re: Residual Entropy

Residual entropy is the entropy of a compound at 0 K, and it is based on the orientation of the molecules in a crystal/lattice. Molecules with various potential microstates (basically different positions/orientations) can have residual entropy. For example, CO could have a left or right orientation ...
Sun Feb 14, 2021 1:14 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Entropy of gases
Replies: 7
Views: 41

### Re: Entropy of gases

An easy way to think about entropy is as "molecular disorder". You can also think about it as an increase in entropy indicates a greater number of possible microstates. In a solid, the molecules do not have the freedom to move around a lot and have many different orientations, so solids ha...
Sun Feb 14, 2021 1:06 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Calculating entropy
Replies: 1
Views: 7

### Re: Calculating entropy

Entropy is a state function, so you can combine multiple equations to find out the entropy of another equation (similar to what we did with Hess's Law). If you were given the temperature change of one step and then the volume change in another step, you could find the change in entropy in each of th...
Sun Feb 07, 2021 10:51 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Sapling #11 (Week 4-5)
Replies: 5
Views: 43

### Re: Sapling #11 (Week 4-5)

For this question, set -q(iron) = q(water), because the heat lost by the iron is equal to the heat gained by the water since no heat is lost to the surroundings. So, you will have -27.6g(0.449 J/(g·°C))(Tf- 94.7 °C)= 50.0 (4.184 J/(g·°C))(Tf- 25 °C) Use this equation to solve for Tf, or the final te...
Sun Feb 07, 2021 10:42 pm
Forum: Calculating Work of Expansion
Topic: Negative Work
Replies: 5
Views: 16

### Re: Negative Work

q is used to represent heat energy, so when heat energy enters a system, the value for q is positive, and when heat leaves the system, q is negative. Similarly, when work is done by the system, the value for work is negative, and when the surroundings do work on the system, work is positive. Hope th...
Sun Feb 07, 2021 10:38 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Sapling week 3&4 #10
Replies: 3
Views: 39

### Re: Sapling week 3&4 #10

I expanded out your work so the equation is

17076 + 214.22 Tf = -1870 Tf + 84161

and I seem to be getting the right answer.

There was probably just a math error somewhere.

Hope this helps!
Sun Feb 07, 2021 10:29 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Sapling #8 Week 4
Replies: 3
Views: 40

### Re: Sapling #8 Week 4

For this question, you can use the equation q=n*C*delta T. The value of C can be found by multiplying R*4 since the problem says that the constant‑pressure molar specific heat for SO2(g) is equal to 4R. From here, just plug in the moles and the change in temperature to find q. To find delta U, you c...
Sun Feb 07, 2021 10:22 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Hess's Law
Replies: 12
Views: 57

### Re: Hess's Law

The best way to figure it out is to look at which compound appears as a product vs a reactants. Try to begin with "unique" compounds that are only in one of the equations and get those on the correct side. Once all the compounds are on the correct sides, you can figure out what needs to be...
Sun Feb 07, 2021 10:19 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Sapling #7, Week 4
Replies: 3
Views: 17

### Re: Sapling #7, Week 4

For this question, you can use the equation deltaU= q + w to find q. Then, set up the equation q=mass* specific heat * deltaT. This should lead you to find the specific heat.

Hope this helps!
Sun Feb 07, 2021 10:16 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Sapling Question 14
Replies: 2
Views: 33

### Re: Sapling Question 14

For path A, you need to use w=-nRTln(V2/V1) to find the work done. Use PV=nRT to find the moles of gas, and then plug that into the equation. For path B, the first step has no change in volume, so there is no work being done. In the second step, you can use the equation w=-Pex*deltaV to find the wor...
Sun Jan 31, 2021 11:35 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Relationship between pka and conjugate base
Replies: 2
Views: 34

### Re: Relationship between pka and conjugate base

A strong conjugate base indicates a weak acid. The pKa would be high because weak acids have low Ka values and high pKa values.

Hope this helps!
Sun Jan 31, 2021 11:21 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Homework Problem
Replies: 3
Views: 27

### Re: Homework Problem

The enthalpy of formation is defined as the energy when 1 mole of a substance is made from the elements in their standard states. Therefore, CO2 and H2O cannot both be products if the delta H was supposed to be the enthalpy of formation. Additionally, the reactants would have to be the elements in t...
Sun Jan 31, 2021 11:07 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 6B.11 Textbook Question
Replies: 1
Views: 11

### Re: 6B.11 Textbook Question

Since every OH- ion comes from one NaOH molecule, the ratio of OH- to NaOH is 1:1, so the molar concentration of hydroxide ions in the original solution is equal to the molar concentration of NaOH in the original solution. You can take the molarity of the original solution which you found in part A ...
Sun Jan 31, 2021 10:57 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: pka vs ka
Replies: 28
Views: 139

### Re: pka vs ka

I believe it is because of the product property of logarithms which states that
log(MN) = log(M) + log(N). Multiplying Ka and Kb will give you Kw, but if you want to find the pKw, you would be doing log(Kw)=log(Ka*Kb), which equals log(Ka)+log(Kb) or pKa+pKb.

Hope this helps!
Sun Jan 31, 2021 10:41 pm
Forum: Phase Changes & Related Calculations
Topic: X2 vs 2X
Replies: 14
Views: 85

### Re: X2 vs 2X

In general, bonds require energy/heat to be broken and release energy/heat when they are formed. In this case, one bond is broken, which requires heat, but none are formed, so no heat is released. This means that the reaction must be endothermic as heat is taken up by the system and used to break th...
Sun Jan 31, 2021 10:31 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Acids and Bases
Replies: 15
Views: 86

### Re: Acids and Bases

The best way to know the strong acids is just to memorize them, unfortunately. Luckily, there are only 7 (HClO4, HBr, HI, HCl, HNO3, H2SO4, HClO3). Strong bases are any alkali metal or alkaline earth metal cation with the hydroxide ion (ex. NaOH, Ca(OH)2, KOH), as well as metal oxides. Hope this hel...
Sun Jan 24, 2021 5:33 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: percent ionization
Replies: 4
Views: 42

### Re: percent ionization

To find percent ionization for an acid, it is [H+]/[HA] x 100. This is the amount of acid that has been deprotonated, divided by the initial amount of the acid, and then multiplied by 100 to find the percent ionization. For equilibrium questions, the [H+] will typically be the value at equilibrium a...
Sun Jan 24, 2021 4:47 pm
Forum: Calculating the pH of Salt Solutions
Topic: pH sig figs
Replies: 4
Views: 40

### Re: pH sig figs

You want the least number of sig figs in the problem to match the number of sig figs in your answer. The only thing to look out for is that if you take the log of anything, the answer of the log should have the same number of decimal places as the H+ or OH- concentration does. For example, a H+ conc...
Sun Jan 24, 2021 4:35 pm
Forum: Phase Changes & Related Calculations
Topic: sapling weeks 3-4 #4
Replies: 6
Views: 51

### Re: sapling weeks 3-4 #4

Yes, any formation of a bond will always be exothermic, and the breaking of a bond will be endothermic. To figure out which individual bond is stronger, you can analyze the intermolecular forces.
Sat Jan 23, 2021 11:39 pm
Forum: Phase Changes & Related Calculations
Topic: sapling weeks 3-4 #4
Replies: 6
Views: 51

### Re: sapling weeks 3-4 #4

An endothermic reaction requires energy, while an exothermic reaction releases energy. Breaking bonds requires energy (endothermic) and forming bonds releases energy (exothermic). Since all of the reactions in this problem have some bonds that are broken and some bonds that are formed, you need to f...
Sat Jan 23, 2021 11:25 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Calculating the pH of a Salt Solution
Replies: 2
Views: 14

### Re: Calculating the pH of a Salt Solution

You are right! Since the initial concentration of NH4Cl was 0.15 M, this means that there are 0.15 moles of NH4Cl per 1 liter of solution. Since 1 mole of NH4Cl is made up of 1 mole NH4 and 1 mole Cl (according to the ratio given by the formula of the compound), 0.15 moles of NH4Cl per liter in the ...
Sat Jan 23, 2021 6:35 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling 2 #5
Replies: 2
Views: 40

### Re: Sapling 2 #5

You need to divide by the initial concentration of B, but using the Kb, you can only find the equilibrium concentration of B. By adding the concentration of the protonated species, you can find the initial concentration of B, which you need to find the % protonation which is the concentration of the...
Sun Jan 17, 2021 6:55 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Sapling Week 2 #2
Replies: 4
Views: 42

### Re: Sapling Week 2 #2

A good rule to follow is when the K value is less than 10^-4. This means that there is a lot of the reactant present, so any small change will not make a significant impact on the concentration of the reactant. However, in this question the K value is much larger than that, so you cannot use the app...
Sun Jan 17, 2021 6:46 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling #6
Replies: 3
Views: 38

### Re: Sapling #6

If the salt has the conjugate acid of a weak base, the solution will be acidic because the conjugate acid will attract the OH- from water and leave H+ ions behind, making the solution acidic. It is also important to note that the other ion (in this case the anion) must be a spectator ion in this sit...
Sun Jan 17, 2021 5:29 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling Week 2 Question 5
Replies: 2
Views: 54

### Sapling Week 2 Question 5

For #5 on the week 2 Sapling hw, the question is as follows: The Kb for an amine is 5.266×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.572 ? Assume that all OH− came from the reaction of B with H2O. I was wondering how to tell that this solution is at eq...
Sun Jan 17, 2021 5:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling HW Q7
Replies: 3
Views: 44

### Re: Sapling HW Q7

First, convert the Ka value to Kb using the equation Kw=Ka x Kb where Kw is 1.0 x 10^-14 since the salt acts as a weak base. Then set up an ICE table for the equation ClO- + H2O <--> HClO + OH-, using 0.056 as the initial value. Set the Kb value equal to [HClO][OH-]/[ClO-], and solve for the OH- con...
Sun Jan 17, 2021 5:15 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling HW #5
Replies: 2
Views: 37

### Re: Sapling HW #5

For this problem, I recommend converting the pH to a pOH since you are working with a base. Once you have a pOH, you can use it to figure out the OH- concentration. Since for every one mole of OH- there is also one mole of BH+, you now know the concentration of OH- and BH+. Set up the equation for K...
Sun Jan 17, 2021 5:10 pm
Forum: Identifying Acidic & Basic Salts
Topic: Sapling HW Q6
Replies: 3
Views: 72

### Re: Sapling HW Q6

For this question, it is important to know that the cation of a strong base and the anion of a strong acid create a neutral salt, the cation of a strong base and the anion of a weak acid create a basic salt, and the cation of a weak base and the anion of a strong acid create an acidic salt. HClO4, H...
Sun Jan 17, 2021 5:00 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling HW Q5
Replies: 1
Views: 41

### Re: Sapling HW Q5

For this problem, I recommend converting the pH to a pOH since you are working with a base. Once you have a pOH, you can use it to figure out the OH- concentration. Since for every one mole of OH- there is also one mole of BH+, you know can figure out the numerator of the equation for Kb. Set the Kb...
Sun Jan 10, 2021 1:05 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: When to use ice table
Replies: 4
Views: 33

### Re: When to use ice table

You typically use an ICE table when you are given initial concentrations. The question should state if the concentration is at equilibrium, or say that the initial values are given. If the concentrations of all the species in the reaction are given and you are asked to find the K value, those values...
Sun Jan 10, 2021 1:00 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling #5 on finding k
Replies: 2
Views: 38

### Re: Sapling #5 on finding k

You basically just combine equations to get the equation you want. Equations can be reversed, in which case the K value would now be equal to 1/K, and the equation can be multiplied by a number, in which case the K value would be raised to the power of that number. For this question, multiplying the...
Sun Jan 10, 2021 12:55 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Kc calc with Cubic Equations
Replies: 1
Views: 20

### Re: Kc calc with Cubic Equations

If the K value is less than 10^-4, that means the reactants are strongly favored and x will not significantly change the concentration of the reactants, so the approximation will work. I believe for this class most cubic equations will have small K values where the approximation can be applied, but ...
Sun Jan 10, 2021 12:49 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Endothermic and Exothermic Reactions
Replies: 6
Views: 72

### Re: Endothermic and Exothermic Reactions

I think the main concept to understand with endothermic and exothermic reactions is that endothermic reactions require heat to form the products and exothermic reactions release heat. If heat was added to an endothermic reaction, then more products would be formed (product formation would be favored...
Sun Jan 10, 2021 12:44 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Reaction Quotient Definition
Replies: 7
Views: 62

### Re: Reaction Quotient Definition

The reaction quotient is used when the reaction is not at equilibrium. At equilibrium, the ratio or products to reactants is the K value, but when the reaction is not at equilibrium, there can be more products or reactants, so the ratio would no longer equal the K value. When a product or a reactant...
Sat Jan 09, 2021 11:13 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Endothermic and Exothermic Concepts
Replies: 12
Views: 71

### Re: Endothermic and Exothermic Concepts

I think Dr. Lavelle meant that since there is now more heat being added, that additional heat will be used by the reaction, which will result in more products being formed since the reaction is endothermic and needs heat to convert the reactants to products.

Hope this helps!
Sat Jan 09, 2021 6:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: solving concentration from chem. equilibrium (sapling #3)
Replies: 2
Views: 39

### Re: solving concentration from chem. equilibrium (sapling #3)

For this problem, you can take the square root of both sides after setting up the ratio, so you do not need to solve a quadratic equation. Typically, if you have two positive concentrations, the one that is larger than the initial concentrations is incorrect.

Hope this helps!
Wed Jan 06, 2021 4:28 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Q compared to K
Replies: 6
Views: 27

### Re: Q compared to K

You are correct in that Q would be less than K, but K would not change. The equilibrium constant, K, only changes for a reaction due to a change in temperature. Therefore, not having reached equilibrium does not change the equilibrium constant. Calculating Q in this situation would let us know that ...
Wed Jan 06, 2021 12:14 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Partial Pressure
Replies: 4
Views: 39

### Re: Partial Pressure

Partial pressure is the pressure of one of the gases in a mixture. Essentially, the pressure of one gas if it was placed alone in the same volume as the mixture, is its partial pressure in the mixture. Since there is a mixture of gases, each gas has its own partial pressure, which can be added to fi...
Wed Jan 06, 2021 12:00 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Solvents absence in equilibrium constant eq
Replies: 3
Views: 42

### Re: Solvents absence in equilibrium constant eq

Solvents aren't written in the equilibrium constant equations because there tends to be such an excess of solvent, that even if there was a small change, it is insignificant. For example, if there are 1 million bees and 10 more are added, we still say there are about 1 million bees because the chang...
Wed Jan 06, 2021 11:46 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Q and K
Replies: 17
Views: 178

### Re: Q and K

K only changes due to a change in temperature. In this situation, adding more reactants or products would change Q because the concentrations are no longer at equilibrium. You could then make an ICE table and find the new equilibrium concentrations, which would still have the same ratio and therefor...
Wed Jan 06, 2021 11:31 am
Forum: Ideal Gases
Topic: K and PV=nRT
Replies: 9
Views: 96

### Re: K and PV=nRT

The ideal gas law, PV=nRT is only used to convert from pressure to concentration, or vice versa. You always use K=[P]/[R] to find the equilibrium constant, but if you were given some values of the products or reactants as concentrations and other values as pressures, you can use the ideal gas law to...
Wed Jan 06, 2021 11:23 am
Forum: Ideal Gases
Topic: Difference between real and ideal gas
Replies: 10
Views: 114

### Re: Difference between real and ideal gas

An ideal gas is based on certain assumptions of the Kinetic Molecular Theory. For an ideal gas, it is assumed that the molecules are constantly in motion, they have negligible volume, they do not exert any force on each other, and that they have perfectly elastic collisions. These assumptions allow ...
Wed Jan 06, 2021 11:11 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Endothermic vs Exothermic Stabilities
Replies: 8
Views: 127

### Re: Endothermic vs Exothermic Stabilities

I believe you are correct. The way I think of it is that anything with lower energy is more stable, so in an endothermic reaction the products would have more energy than the reactants and therefore, be less stable. On the other hand, in an exothermic reaction, the products would have a lower energy...
Sat Dec 12, 2020 6:11 pm
Forum: *Titrations & Titration Calculations
Topic: Titration Formulas With Conjugate Acids/Bases
Replies: 2
Views: 68

### Re: Titration Formulas With Conjugate Acids/Bases

I believe the conjugate acid would just be water in this situation, or 2 water molecules since the equation has (OH)2. Your second equation with the conjugate base of the weak acid is correct as that would be the reason as to why the pH increases to above 7.

Hope this helps!
Sat Dec 12, 2020 4:19 pm
Forum: Sigma & Pi Bonds
Topic: Valence Bonding Descriptions for Triple Bonds
Replies: 1
Views: 45

### Valence Bonding Descriptions for Triple Bonds

How do you use valence bonding descriptions when it comes to triple bonds? For example, you use valence bonding descriptions such as (C2sp^2, C2sp^2) to describe a bond between two carbon atoms with trigonal planar geometries, but in HCN, how would you describe the triple bond between the C and the ...
Sat Dec 12, 2020 3:51 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Textbook 6b1
Replies: 4
Views: 77

### Re: Textbook 6b1

The easiest way to do this problem is to simply choose a molar concentration of HCl and find the pH. Then, take 12% of that molar concentration and find the pH value. Subtract the values to find the difference in pH. For example, I used a 0.5 M solution of HCl, and then a 0.06 M solution (which is 1...
Thu Dec 10, 2020 10:28 pm
Forum: Calculating the pH of Salt Solutions
Topic: Why is Cu(NO3)2 Acidic?
Replies: 2
Views: 54

### Re: Why is Cu(NO3)2 Acidic?

In Cu(NO3)2, the NO3- is the conjugate base of a strong acid (HNO3), so it will not affect the pH of the aqueous solution. However, Cu2+ is not the conjugate acid of a strong base, so it will affect the pH. Since Cu2+ has a positive charge, it will attract the oxygen atom from the water molecule. Th...
Thu Dec 10, 2020 7:13 pm
Forum: Identifying Acidic & Basic Salts
Topic: textbook problem 6D #11
Replies: 1
Views: 32

### Re: textbook problem 6D #11

In this problem, you want to see what the effect of the salt would be on the aqueous solution, so you want to understand how the salt will interact with the water molecules. Within the salt, the conjugate base of any strong acid and the conjugate acid of any strong base will have no effect on the pH...
Thu Dec 10, 2020 7:02 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: AX3E2
Replies: 7
Views: 92

### Re: AX3E2

T-shaped does not have lone pairs in the axial position (top and bottom) but instead, has them on the equatorial position. This is because when there is only 1 lone pair, making the atom on the top a lone pair would cause the more repulsive lone pair to interact with three atoms (all the atoms in th...
Thu Dec 10, 2020 6:32 pm
Forum: Identifying Acidic & Basic Salts
Topic: 6D.11 Chemical Equations for salts
Replies: 4
Views: 53

### Re: 6D.11 Chemical Equations for salts

You want to see what the effect of the salt would be on the aqueous solution, so you want to understand how the salt will interact with the water molecules. Within the salt, the conjugate base of any strong acid and the conjugate acid of any strong base will have no effect on the pH, so that ion can...
Mon Dec 07, 2020 10:37 am
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Acid Deprotonation
Replies: 1
Views: 37

### Acid Deprotonation

Why does the acid deprotonate when the pka<pH?
Sun Dec 06, 2020 10:15 pm
Forum: Biological Examples
Topic: naming [Co(NH3)3Cl3]
Replies: 5
Views: 91

### Re: naming [Co(NH3)3Cl3]

You want to begin by naming the ligands in alphabetical order. Since you have ammonia and chloride, ammonia will come first. You need to specify the number of that ligand, so it would be triammine. You also need to specifiy the number of the chloride, so it would be trichloro. Finally you add the na...
Sun Dec 06, 2020 9:25 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Coordination Number
Replies: 5
Views: 60

### Re: Coordination Number

The coordination number is the number of bonds formed to the transition metal, so in this case, there are 5 NH3 molecules and 1 SO4 molecule that are bonded to the cobalt, resulting in a coordination number of 6. I recommend first counting how many ligands there are (so anything that is within the c...
Sat Dec 05, 2020 2:29 pm
Forum: Naming
Topic: Textbook Exercise 9C.3
Replies: 4
Views: 64

### Re: Textbook Exercise 9C.3

The potassium should go at the beginning (before the brackets) because it is a cation. Anions, such as Cl- or F- go after the rest of the name. This is the same idea as you would see in any salt, where the convention is to write the cation first and then the anion (ex. NaCl). In the case of potassiu...
Thu Dec 03, 2020 7:51 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: strong vs weak
Replies: 10
Views: 84

### Re: strong vs weak

The weaker the bond in an acid, the stronger the acid is because it can dissociate easily. So, if the molecules are particularly large (have a large atomic radius), then it is more likely that it is a strong acid. There are not a lot of strong acids, so it seems likely that anything not included on ...
Thu Dec 03, 2020 7:40 pm
Forum: Calculating the pH of Salt Solutions
Topic: OH calculation
Replies: 2
Views: 39

### Re: OH calculation

Using the same set up as the H+ will give you the pOH. The equation is pOH=-log[OH]. To find pH from this, you can use pH+pOH=14. Otherwise, you can use the equation [H+]=1.0 x 10^-14/[OH-] to find the H+ concentration, and then find the pH from there. This diagram is helpful to see how to convert b...
Thu Dec 03, 2020 7:33 pm
Forum: *Molecular Orbital Theory Applied To Transition Metals
Topic: planes
Replies: 4
Views: 86

### Re: planes

To figure out if compounds are in the same plane or not, you have to look at the shape of the compound. Compounds that are linear, trigonal planar, bent, t-shaped, and square planar are all coplanar, which can be seen by their molecular geometries which have all the atoms in the same plane. Hope thi...
Thu Dec 03, 2020 7:25 pm
Forum: Coordinate Covalent Bonds
Topic: cyanide
Replies: 3
Views: 56

### Re: cyanide

Cyanide does have 2 lone pairs, but since it has a triple bond and is linear, the molecule cannot orient itself in a way the both lone pairs bond to the same transition metal. To have a polydentate ligand, you need at least two lone pairs and at least one spacer between the two lone pairs. Cyanide d...
Thu Dec 03, 2020 7:17 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: how many orbitals in n=2
Replies: 4
Views: 78

### Re: how many orbitals in n=2

n=2 only has s and p orbitals, since the first quantum number possible for a d orbital is 3. There is 1 s orbital and there are 3 p orbitals, so n=2 has 4 orbitals.

Hope this helps!
Sat Nov 28, 2020 12:13 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sapling #17
Replies: 5
Views: 68

### Re: Sapling #17

C3H4 has three resonance structures. C3H4.png In the first structure given in this image, the carbon on the left has two regions of electron density and is therefore, linear and sp hybridized. This has a 180 degree bond angle. The carbon atom in the center has a triple bond and a single bond, which ...
Wed Nov 25, 2020 7:39 pm
Forum: Hybridization
Topic: Hybridization and Resonance Structures
Replies: 1
Views: 57

### Re: Hybridization and Resonance Structures

To answer the first part of your question, yes, all molecules with resonance have unhybridized molecules. Since resonance typically implies that there is a double bond, there must be an unhybridized molecule to create the pi bond. In terms of hybridized orbitals, this site has a lot of diagrams that...
Wed Nov 25, 2020 4:05 pm
Forum: Hybridization
Topic: Geometry and Hybridization
Replies: 1
Views: 44

### Re: Geometry and Hybridization

Hybridization does correlate to specific electron geometries, since it is based on the number of regions of electron density around the atom. So, for example, all molecules with linear electron geometry would have an sp hybridization, all molecules with trigonal planar electron geometry would have a...
Mon Nov 23, 2020 12:39 pm
Forum: Hybridization
Topic: Unhybridized Orbitals in pi Bonds
Replies: 1
Views: 26

### Unhybridized Orbitals in pi Bonds

From my understanding, there has to be an unhybridized orbital in order for a \pi bond to form. I was wondering if this had to be a unhybridized p orbital or if it could be an unhybridized d orbital as well. Additionally, in sulfuric acid, H2SO4, the central S atom has an sp3 hybridization. However,...
Mon Nov 23, 2020 12:27 pm
Forum: Hybridization
Topic: Lack of Spin Pairing in Hybridized Orbitals
Replies: 2
Views: 54

### Lack of Spin Pairing in Hybridized Orbitals

In today's lecture, Dr. Lavelle said that in a sp2 hybridization (in the molecule C2H2 for example) the fourth electron of the carbon would have to be in the unhybridized 2p orbital because the energy difference between the sp2 orbitals and unhybridized p orbital is very small. I know it has somethi...
Mon Nov 23, 2020 11:18 am
Forum: Hybridization
Topic: Orientation of Hybrid Orbitals
Replies: 1
Views: 32

### Orientation of Hybrid Orbitals

In the first example of today's lecture regarding the hybridization of ammonia, Dr. Lavelle says that we need to have sp3 hybridization because a bonding model with just the normal s and p orbitals would mean that the three unpaired electrons in the p orbital would have 90 degree bond angles, which ...
Sun Nov 22, 2020 6:43 pm
Forum: Hybridization
Topic: Location of Hybridized Atoms
Replies: 2
Views: 13

### Location of Hybridized Atoms

Are central atoms the only atoms in a molecule that can undergo hybridization? Why or why not?

Sun Nov 22, 2020 6:39 pm
Forum: Hybridization
Topic: hybridization
Replies: 2
Views: 57

### Re: hybridization

Hybridization occurs to create hybrid orbitals that can better accommodate the pairing of electrons. The numbers of regions of electron density can tell you the hybridization. For example, a molecule with only 2 regions of electron density would have an sp hybridization because it would only involve...
Sun Nov 22, 2020 6:33 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Shape
Replies: 10
Views: 107

### Re: Shape

It is linear because if there was one lone pair, and four bonding pairs, the lone pair would be in the equatorial plane so the more repulsive lone pair only reacts with 2 bonds at 90 degrees. Using the same logic, the next two lone pairs would also be removed from the equatorial plane, leaving two b...
Tue Nov 17, 2020 11:02 pm
Forum: Quantum Numbers and The H-Atom
Topic: n=6,l=5
Replies: 6
Views: 109

### Re: n=6,l=5

Yes, l=5 is a valid answer. There are other orbitals past the s, p, d, and f orbitals (for example there is a g orbital that exists, but there are no elements that have been found that can occupy this orbital). Any orbital that satisfies the 0, ..., n-1 requirement is a possible value for l. Hope th...
Tue Nov 17, 2020 6:48 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: shielding and penetrating
Replies: 5
Views: 111

### Re: shielding and penetrating

Since the s orbital is around the nucleus, s electrons are able to shield outer electrons well. On the other hand, p orbitals have a node where the nucleus is, so they are not able to shield outer electrons well. I also believe that d orbital electrons are not able to shield electrons very well eith...