Search found 100 matches

by Pegah Nasseri 1K
Wed Mar 11, 2020 5:28 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: dilutions and Ecell
Replies: 8
Views: 47

Re: dilutions and Ecell

Remember that E˚(cell) = E˚(cathode) - E˚(anode). The greater the concentration difference between the cathode and anode, the greater your standard cell potential. If you dilute the cathode, you will decrease the cell potential because you are lowering the concentration of the cathode, lowering the ...
by Pegah Nasseri 1K
Wed Mar 11, 2020 5:22 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Enaught in Concentration Cells
Replies: 4
Views: 35

Re: Enaught in Concentration Cells

No, remember that to find the change in gibbs free energy, you need the cell potential. You calculate the cell potential through the equation E = E˚ - (RT/nF) lnQ. This equation incorporates your standard cell potential value but it is not the only value that determines the cell potential. Then you ...
by Pegah Nasseri 1K
Wed Mar 11, 2020 5:17 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Catalysts
Replies: 2
Views: 14

Re: Catalysts

Catalysts are enzymes that speed up the rate of reactions by lowering the activation energy (Ea) of the reaction. A smaller activation energy corresponds to a quicker reaction. Remember that the catalysts do not react with anything in the reaction and are not affected by the process.
by Pegah Nasseri 1K
Wed Mar 11, 2020 5:09 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Relationship Between Ecell and K
Replies: 9
Views: 49

Re: Relationship Between Ecell and K

Looking at the two equations E˚= (RT/nF)lnK and Ecell = E˚- (RT/nF)lnQ, you can see that if K>1 then E˚> 0. When E˚>0, then Ecell will be a greater value and will most likely be greater than zero as well. So if K>0 then E˚>0 and Ecell>0.
by Pegah Nasseri 1K
Wed Mar 11, 2020 4:59 pm
Forum: General Rate Laws
Topic: Rate of Formation vs Unique Rate
Replies: 4
Views: 98

Re: Rate of Formation vs Unique Rate

Given the reaction 2A + B -> 3C, the rate following the rate law would be rate = k[A]^{l}[B]^{m} . Using the unique rate instead of the rate law, you would calculate the rate by plugging in the values for any of these: -\frac{1}{2}\frac{d[A]}{dt}=-\frac{d[B]}{dt}=\frac{1}{3}\frac{d[C]}{dt} . The uni...
by Pegah Nasseri 1K
Sat Mar 07, 2020 3:49 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Reducing/Oxidizing agents
Replies: 5
Views: 61

Re: Reducing/Oxidizing agents

Looking at this example of half reactions: Cu2+(aq) + 2e- -> Cu(s) (reduction) Zn(s) -> Zn2+ (aq) + 2e- (oxidation) The reducing agent is the one that is being oxidized. In this case, it would be Zn(s). The oxidizing agent is the one that is being reduced. Here, it would be Cu2+(aq). Think about it ...
by Pegah Nasseri 1K
Sat Mar 07, 2020 3:33 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: How to calculate for n
Replies: 7
Views: 67

Re: How to calculate for n

Like others have mentioned, you would find n by balancing the oxidation and reduction half reactions and adding them up. For example, if you have these as your half reactions: Fe3+(aq) + e- -> Fe2+(aq) Zn(s) -> Zn2+(aq) + 2e- You would multiply the top reaction by 2 so you can have 2 electrons react...
by Pegah Nasseri 1K
Sat Mar 07, 2020 3:20 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Concentration cell
Replies: 8
Views: 61

Re: Concentration cell

In a concentration cell, both of your electrodes in your cathode and anode are the same material. The solutions in the cathode and anode are the same. This indicates that your standard cell potential is zero. So a concentration cell is a galvanic cell whose standard reduction potential is zero. In t...
by Pegah Nasseri 1K
Sat Mar 07, 2020 3:13 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: temperature and k
Replies: 5
Views: 50

Re: temperature and k

In an endothermic reaction, heat is required so it is part of the reactants of the reaction. Adding heat by increasing the temperature will produce more products, shifting the equilibrium reaction towards the products. This will cause K to increase since the numerator of the equilibrium constant wil...
by Pegah Nasseri 1K
Sat Mar 07, 2020 11:00 am
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: HW 6L.9
Replies: 4
Views: 45

Re: HW 6L.9

You must look at the reduction potentials for each reaction and the cathode will be the reaction with the highest (most positive) reduction potential. Then you can subtract the other reduction potential (the anode) from the cathode to get your standard cell potential. When drawing the cell diagram, ...
by Pegah Nasseri 1K
Fri Feb 28, 2020 7:05 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Difference between ΔG = 0 and ΔG° = 0?
Replies: 1
Views: 19

Re: Difference between ΔG = 0 and ΔG° = 0?

ΔG° is the change in gibbs free energy of formation, which is calculated using the equation ΔG°= -RTlnK. This is the gibbs free energy that occurs at equilibrium. ΔG takes into account ΔG°in its calculation: ΔG = ΔG°+ RTlnQ. Looking at the equation, just because ΔG°= 0, this doesn't necessarily mean...
by Pegah Nasseri 1K
Fri Feb 28, 2020 6:24 pm
Forum: General Rate Laws
Topic: Deriving the Rate Equation
Replies: 5
Views: 39

Re: Deriving the Rate Equation

Understanding how the equation is derived could help you a lot in answering all the conceptual questions in kinetics. I would spend less time trying to memorize the derivation and instead look over and understand each step taken to derive it and what this indicates about the rates. For example, it w...
by Pegah Nasseri 1K
Fri Feb 28, 2020 6:14 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: standard gibbs free energy
Replies: 2
Views: 22

Re: standard gibbs free energy

It depends on what is already given to you in the question. If the question gives you constants for Delta H, Delta S, and T, or if it provides you with constants that can help you calculate these variables, then you should use Delta G = Delta H -T(Delta S). If not much is provided to you, you can lo...
by Pegah Nasseri 1K
Fri Feb 28, 2020 5:47 pm
Forum: Van't Hoff Equation
Topic: Outline Clarification on Van't Hoff EQ
Replies: 1
Views: 30

Re: Outline Clarification on Van't Hoff EQ

The Van't Hoff Equation is derived using two equations: \Delta G= -RTlnK and \Delta G=\Delta H-T\Delta S Setting the two equations to each other gives you the Van't Hoff Equation: ln K = - \frac{\Delta H}{RT} + \frac{\Delta S}{R} In the Van't Hoff Equation, you can use the whatever constants are giv...
by Pegah Nasseri 1K
Fri Feb 28, 2020 4:57 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Spontaneity
Replies: 12
Views: 121

Re: Spontaneity

A reaction is spontaneous when \Delta G< 0 . This is because there is available energy released by the reaction that can used to perform work. Remember that if the change in gibbs free energy of a forward reaction is positive, its reverse reaction can be spontaneous because the change in gibbs free ...
by Pegah Nasseri 1K
Fri Feb 21, 2020 7:33 pm
Forum: Lewis Acids & Bases
Topic: Midterm Question 3B
Replies: 3
Views: 83

Re: Midterm Question 3B

You can think about it in terms of which reaction will be the strongest. The strongest chemical reaction would give off the most amount of heat energy when producing the products. If you allow two diluted substances to react, you wouldn't see a strong enough result as compared to if you reacted a hi...
by Pegah Nasseri 1K
Fri Feb 21, 2020 7:26 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 5G. 13
Replies: 6
Views: 92

Re: 5G. 13

You would use the equation \Delta G = \Delta G + RTlnQ = -RTlnK + RTlnQ . Substitute the 6.8 for the K value in the equation and use all the partial pressures given to you to calculate Q (partial pressure of the products over the partial pressure of the reactants). You use the Delta G=-RTlnK part be...
by Pegah Nasseri 1K
Fri Feb 21, 2020 7:12 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: midterm Q6A
Replies: 2
Views: 59

Re: midterm Q6A

I think you have to pay attention to the wording. The answer choice doesn't specifically state that the enthalpy is positive. It only states that 23.1 kJ of heat will be evolved. The key word used here is "evolved". This word indicates that the heat was released into the surroundings, maki...
by Pegah Nasseri 1K
Fri Feb 21, 2020 7:05 pm
Forum: Identifying Acidic & Basic Salts
Topic: Midterm 3C
Replies: 6
Views: 99

Re: Midterm 3C

When NH4Cl dissociates (since it is a salt), you are left with NH4+ and Cl-. NH4+ is an acid because it will donate protons to the solution, making the solution more acidic. Think of NH4+ as the conjugate acid of NH3.
by Pegah Nasseri 1K
Fri Feb 21, 2020 7:01 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Midterm 6B
Replies: 1
Views: 38

Re: Midterm 6B

There are two key concepts you need to know to be able to do this question. First of all, you have to use the equation \Delta G = \Delta H-T\Delta S . Looking at the equation, Delta G will be most similar to Delta H when Delta S has a small value near zero. Second of all, you have to understand whic...
by Pegah Nasseri 1K
Thu Feb 13, 2020 10:22 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Isothermal reversible of ideal gas
Replies: 4
Views: 44

Re: Isothermal reversible of ideal gas

An ideal gas in an isothermal, reversible reaction will have an internal energy of 0. This means that the heat will equal the negative energy of the work done: Delta U = q + w = 0 This indicates that: q = -w In addition, the total change in entropy will be zero: \Delta S_{total} = \Delta S_{system} ...
by Pegah Nasseri 1K
Thu Feb 13, 2020 10:11 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Differences
Replies: 5
Views: 76

Re: Differences

In an isolated system, the system does not interact with the surroundings at all. In terms of matter and energy, neither of them can be exchanged or transferred between the system and its environment. In a closed system, only energy can be transferred between the system and its surroundings. Matter,...
by Pegah Nasseri 1K
Thu Feb 13, 2020 9:59 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bond Enthalpies with Molecules
Replies: 4
Views: 52

Re: Bond Enthalpies with Molecules

You can draw the Lewis structures of each molecule to look at all the bonds present between the atoms of each molecule. From this you can deduce which bonds are broken and which ones are formed to produce the products of the reaction. Use the values of the bond enthalpies given to you for each bond ...
by Pegah Nasseri 1K
Thu Feb 13, 2020 9:51 pm
Forum: Ideal Gases
Topic: initial values
Replies: 6
Views: 61

Re: initial values

It depends on the values you've been given. If you've been given the initial pressure and the initial temperature, then use the initial volume to find the number of moles. Usually, you will only have been given all the initial values and will thus calculate the number of moles using the initial volu...
by Pegah Nasseri 1K
Thu Feb 13, 2020 9:48 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Delta S
Replies: 8
Views: 67

Re: Delta S

The total change in entropy is simply the sum of the change in entropy of the system and the change in entropy of the surroundings. For example, in a reversible isothermal reaction, the change in entropy of the system is Delta S= nRln(V2/V1) and the change in entropy of the surrounding would be the ...
by Pegah Nasseri 1K
Sat Feb 08, 2020 1:12 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Homework Problem 4F.9
Replies: 3
Views: 33

Re: Homework Problem 4F.9

You can use the ideal gas equation PV=nRT to create a proportion: P1V1 = nRT , P2V2 = nRT Since the reaction is isothermal, nRT will be the same value in both equations and you can set the two equations equal to each other: P1V1=P2V2 Then you can create the proportion P1/P2 = V2/V1 and plug in the c...
by Pegah Nasseri 1K
Sat Feb 08, 2020 12:56 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 4A.11
Replies: 5
Views: 39

Re: 4A.11

Heat capacity (C) is used in the equation Q = C*delta T and represents the heat energy needed to raise the temperature of something by one degree. Molar heat capacity is used in the equation Q = n*C*delta T and is the amount of heat energy it takes to raise 1 mol of something by 1 Kelvin. Specific h...
by Pegah Nasseri 1K
Sat Feb 08, 2020 12:41 am
Forum: Phase Changes & Related Calculations
Topic: 4B.5
Replies: 7
Views: 85

Re: 4B.5

You would use the equation delta U = q + w. You know that the gas gained 5.50 kJ of heat energy so q will be positive (q = 5500J). To find the work, you must use the equation W = - (P*delta V). However make sure to convert the volumes given to you from milliliters to liters and to convert the pressu...
by Pegah Nasseri 1K
Sat Feb 08, 2020 12:30 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Irreversible v Reversible
Replies: 4
Views: 44

Re: Irreversible v Reversible

An irreversible reaction occurs when work is done against a constant external pressure. In a reversible reaction, work is done against changes in external pressure that occur in small increments. Reactions that happen in nature are irreversible. Reversible reactions are slower and perform the most w...
by Pegah Nasseri 1K
Sat Feb 08, 2020 12:18 am
Forum: Calculating Work of Expansion
Topic: Equation for Work
Replies: 3
Views: 23

Re: Equation for Work

Yes, when calculating the work always use the equation W = - (P*deltaV) and make sure to include the negative sign. If work is done on the system, then you know a compression will occur which means the volume will decrease. If the volume decreases due to compression, then the change in volume will b...
by Pegah Nasseri 1K
Fri Jan 31, 2020 12:05 am
Forum: Phase Changes & Related Calculations
Topic: Work Equation
Replies: 3
Views: 50

Re: Work Equation

When work is done by the system, the system will lose energy to its surroundings. The negative sign in the equation w = - (P * delta V) accounts for this loss of energy by the system when it performs work. Negative work is performed by the system when the volume increases or expands at constant pres...
by Pegah Nasseri 1K
Thu Jan 30, 2020 11:48 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Ka and percent ionization problem
Replies: 2
Views: 24

Re: Ka and percent ionization problem

Yes, everything is correct up until the percent ionization calculation. For the percent ionization calculation, you will divide the concentration of HF that was ionized over the initial concentration of HF you started with. The concentration of the HF that ionized is equal to the concentration of F-...
by Pegah Nasseri 1K
Thu Jan 30, 2020 6:19 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 4a.9
Replies: 3
Views: 20

Re: 4a.9

You would use the equation q = m*c*delta T where q is the heat energy, m is the mass of the copper, c is the specific heat of the copper and delta T is the change in temperature. You know that since no energy is lost to the surroundings, the heat lost by copper will equal the negative value of the h...
by Pegah Nasseri 1K
Thu Jan 30, 2020 6:07 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: delta V & delta U
Replies: 1
Views: 15

Re: delta V & delta U

Delta V = 0 indicates that there was no change in volume, which means that the system was neither compressed or expanded. This indicates that there was no work done on the system. The change in internal energy of a closed system is composed of energy transferred through work or compression (w) and e...
by Pegah Nasseri 1K
Thu Jan 30, 2020 6:00 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bond enthalpies
Replies: 4
Views: 35

Re: Bond enthalpies

No, when calculating enthalpies of a reaction using bond enthalpies you would add up all the enthalpies of the bonds broken along with all the enthalpies of the bonds formed. If the sum of the enthalpies of the bonds broken are greater than the sum of the bonds formed, then the enthalpy of the react...
by Pegah Nasseri 1K
Thu Jan 23, 2020 6:45 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6B.3
Replies: 2
Views: 17

Re: 6B.3

The actual volume is 250 ml so you have to determine what the new concentration is because the volume changed. You would use the equation M1V1=M2V2 to find the new molar concentration which would be found as follows: M2 = (200 ml X 0.025M) / 250 ml. Then take the negative log of this new molar conce...
by Pegah Nasseri 1K
Thu Jan 23, 2020 6:39 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: K value
Replies: 7
Views: 37

Re: K value

If K<10^-3, then you can leave out the x in the denominator when solving for the quadratic. After you obtain a value for x, check if the x value is less than 5% of the initial concentration. If x is less than 5% of the initial concentration then the approximation is valid and you were correct in lea...
by Pegah Nasseri 1K
Thu Jan 23, 2020 6:31 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Hw problem 5.35 help
Replies: 3
Views: 38

Re: Hw problem 5.35 help

You don't need to know the volume because you will use the partial pressures given to you in the graph for your calculation of the equilibrium constant. The equilibrium constant is calculated by dividing the partial pressure of the products over the partial pressure of the reactant: K= (Pb)(Pc)^2 / ...
by Pegah Nasseri 1K
Thu Jan 23, 2020 6:23 pm
Forum: Ideal Gases
Topic: Homework problem 5I.11
Replies: 3
Views: 42

Re: Homework problem 5I.11

For the first part of this question, you are calculating Q which is the concentration of the products over the concentration of the reactants (not at equilibrium). Use the moles of the reactants and products given to you in the question and divide each one by the volume (0.500 L) to obtain a concent...
by Pegah Nasseri 1K
Thu Jan 23, 2020 6:17 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 5.61f
Replies: 5
Views: 34

Re: 5.61f

Water would not have any effect on the reactants and products no matter whether the reactants or products are gases or aqueous. This is because the water is a liquid and acts as a solvent in this reaction. As a solvent, its change in concentration is not significant and will not affect the concentra...
by Pegah Nasseri 1K
Fri Jan 17, 2020 7:08 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: ICE tables
Replies: 3
Views: 29

Re: ICE tables

When solving the quadratic, the x in the denominator can be left out because its effect on the quantity will negligible. This can only be done if you obtain a value of x that is less than 5% of the initial concentration which allows you to approximate. For example, if your quadratic is K = x^2/(0.15...
by Pegah Nasseri 1K
Fri Jan 17, 2020 6:44 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.13 Equilibrium Composition
Replies: 2
Views: 41

Re: 5I.13 Equilibrium Composition

Use the Kc value provided in table 5G.2 because the concentrations of the gases are given to you in the question. The Kc value would be 1.2 X 10^7. The molarity of Cl2 is 0.0010 M (2.0 X 10^-3 mol / 2.0 L). Using the ICE table, you should get the quadratic Kc = [Cl]^2/[Cl2] = (2x)^2/(0.0010-x) = 1.2...
by Pegah Nasseri 1K
Fri Jan 17, 2020 5:41 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Homework question 5J.5
Replies: 3
Views: 30

Re: Homework question 5J.5

When pressure increases, the volume of the container decreases. The molar concentrations of the reactants and products change due to this change in volume. It is helpful to remember that if pressure increases, the reaction shifts to the side with fewer moles. For part (d) of this question, there are...
by Pegah Nasseri 1K
Thu Jan 16, 2020 7:36 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Inert gas
Replies: 4
Views: 22

Re: Inert gas

An inert gas doesn't react. Noble gases can be inert gases because they are not apart of chemical reactions. However, some noble gases such as Xenon can react to form compounds. Most noble gases are inert and do not chemically react to form compunds.
by Pegah Nasseri 1K
Thu Jan 16, 2020 7:27 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Acids and Bases, understanding the Ka value
Replies: 3
Views: 45

Re: Acids and Bases, understanding the Ka value

A high Ka value corresponds to a low pKa value. If the Ka value is high, it indicates that there are a lot of products being produced in this reaction, which are the H3O+ ions that affect the pH of the solution. The lower the pKa, the stronger the acid. The stronger the acid, the more stable its con...
by Pegah Nasseri 1K
Fri Jan 10, 2020 3:55 pm
Forum: Ideal Gases
Topic: converting Kc to Kp
Replies: 13
Views: 111

Re: converting Kc to Kp

Kc is calculated using molar concentration while Kp is calculated using partial pressures. To convert from Kc to Kp, use the equation PV=nRT. We know that the molar concentration is calculated by n/V so this equation can be manipulated to find the concentration which is concentration = n/V= P/RT. Fo...
by Pegah Nasseri 1K
Fri Jan 10, 2020 3:45 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: K vs Q
Replies: 9
Views: 85

Re: K vs Q

Knowing that both K and Q are the ratios of products to reactants will help you determine which value has more products or more reactants than the other. If Q<K then you would know that Q has less products than K, meaning that the reaction has more reactants than what it would have at equilibrium, s...
by Pegah Nasseri 1K
Fri Jan 10, 2020 3:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Example 5I.4
Replies: 4
Views: 33

Re: Example 5I.4

Always when calculating K for gases, partial pressure is used. However, if the question specifically asks you to find K_{c} then you would use the concentration of the gases. You would convert the partial pressure of the gases to molar concentration using the equation PV=nRT (n/V (concentration) = P...
by Pegah Nasseri 1K
Fri Jan 10, 2020 3:32 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Change in k
Replies: 11
Views: 102

Re: Change in k

Lavelle mentioned in lecture that if more reactant is added to a chemical reaction at equilibrium more product will form until the original product to reactant ratio is maintained. The same is true for when more product is added at equilibrium which causes more reactant to form until the product to ...
by Pegah Nasseri 1K
Fri Jan 10, 2020 3:23 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.5
Replies: 4
Views: 50

Re: 5I.5

For this problem you know that K=25 so using the chemical equation PCl_{5} (g) \rightleftharpoons PCl_{3} (g) + Cl_{2} (g) you know that: K = \frac{(P_{PCl_{3}})(P_{Cl_{2})}}{P_{PCl_{5}}} = 25 Plug in the values of the partial pressures that have already been ...
by Pegah Nasseri 1K
Tue Dec 03, 2019 4:23 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Calculating pH for weak Acids
Replies: 3
Views: 28

Re: Calculating pH for weak Acids

No, we will only be tested on calculating pH for strong acids and bases. To find pH of strong acids, you know the acid will dissociate completely and so you can calculate the number of moles of H+ the acid produces based on the weight and molar mass of the acid. Divide the number of moles of H+ by t...
by Pegah Nasseri 1K
Tue Dec 03, 2019 4:18 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Determining binding sites
Replies: 3
Views: 40

Re: Determining binding sites

You count the number of atoms with lone pairs, not the number of lone pairs. So in the water molecule, only the oxygen atom has lone pairs. So the central metal atom will only bind to the ligand once by binding once to the one oxygen atom.
by Pegah Nasseri 1K
Tue Dec 03, 2019 4:16 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: 9C.5
Replies: 4
Views: 65

Re: 9C.5

You have to look at how many atoms in the ligand have lone pairs, because the lone pairs are potential binding sites to the central metal. The ligand in part (a) has three nitrogen atoms that have lone pairs. Thus there are three binding sites so the ligand is tridentate. The ligand in part (b) has ...
by Pegah Nasseri 1K
Tue Dec 03, 2019 4:03 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: pH, pKa, pKb
Replies: 5
Views: 43

Re: pH, pKa, pKb

Ka is the concentration of products over the concentration of reactants. If Ka is high, then you know a lot of products are being produced. A higher Ka corresponds to a stronger acid because the acid is generating more H+ products by dissociating more. A high Ka corresponds to a lower pKa (pKa=-log ...
by Pegah Nasseri 1K
Tue Dec 03, 2019 3:56 pm
Forum: Biological Examples
Topic: Biological Examples
Replies: 4
Views: 56

Re: Biological Examples

Myoglobin and hemoglobin are both proteins that transport oxygen in the red blood and muscle cells of vertebrates. They both possess metals with different oxidation states which make them useful as electron carriers. Essentially, what you need to know is that they are important in electron transport...
by Pegah Nasseri 1K
Tue Nov 26, 2019 10:23 am
Forum: Naming
Topic: Chelating Complex
Replies: 4
Views: 40

Re: Chelating Complex

Chelates are coordination compounds that form a structure in the form of a ring. The ligand forms a bond with the central metal atom at 2 or more different points.
by Pegah Nasseri 1K
Tue Nov 26, 2019 10:16 am
Forum: Sigma & Pi Bonds
Topic: Unhybridized Orbital vs Spin Pair?
Replies: 8
Views: 42

Re: Unhybridized Orbital vs Spin Pair?

An electron will prefer to occupy the unhybridized orbital because it takes less energy for the electron to be in the higher p orbital since the electron repulsion as a spin pair in a hybridized orbital would be too unstable.
by Pegah Nasseri 1K
Tue Nov 26, 2019 10:09 am
Forum: Hybridization
Topic: How to write
Replies: 3
Views: 39

Re: How to write

To figure out the hybridization, you have to count the number of regions of electron density surrounding the atom. A single bond, double bond, triple bond, and lone pair each count as one region of electron density. Number of regions of electron density: 2 : sp 3 : sp^2 4 : sp^3 5 : sp^3d 6 : sp^3d^...
by Pegah Nasseri 1K
Tue Nov 26, 2019 10:00 am
Forum: Dipole Moments
Topic: Intermolecular force
Replies: 5
Views: 348

Re: Intermolecular force

Yes, if the intermolecular forces between two molecules is stronger, it will be harder to break the bonds between them. The strengths of the different types of intermolecular forces in increasing order are as follows: induced dipole-induced dipole (London forces) < dipole-dipole < ion-dipole < ion-ion
by Pegah Nasseri 1K
Tue Nov 26, 2019 9:41 am
Forum: Sigma & Pi Bonds
Topic: Strength of the Bonds
Replies: 2
Views: 30

Re: Strength of the Bonds

As mentioned in the other post, a sigma bond is stronger than a pi bond. Then, the reason why a double bond is stronger than a single bond is because it consists of both a sigma and pi bond while the single bond consists of only a sigma bond.
by Pegah Nasseri 1K
Tue Nov 19, 2019 11:58 pm
Forum: Dipole Moments
Topic: dipole moments vs dipole-dipole
Replies: 4
Views: 109

Re: dipole moments vs dipole-dipole

please reply to this by quoting my question so i get a notification. thanks in advance! what is the difference between dipole moments and induced dipoles and dipole-dipole interactions? From what I understand dipole moments are permanent dipoles within a molecule, between atoms (like C-F). Induced ...
by Pegah Nasseri 1K
Tue Nov 19, 2019 11:44 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: SO2 bond sigma and pi bond
Replies: 4
Views: 48

Re: SO2 bond sigma and pi bond

In SO2, the central atom S is double bonded to each O. There are two double bonds in the structure. You know that there can only be one sigma bond in every bond. So in each double bond, there will be a sigma bond and a pi bond. In total, there will be two sigma bonds and two pi bonds. Another thing ...
by Pegah Nasseri 1K
Tue Nov 19, 2019 11:34 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: 3F.5 b
Replies: 2
Views: 34

Re: 3F.5 b

Hydrogens are most often bonded to carbons when drawing lewis structures. Other than this, you can figure it out by looking at the formal charge. If the 2 hydrogens are bonded to the oxygen, the oxygen would have a formal charge of 2 and the carbons at the end of the structure would have formal char...
by Pegah Nasseri 1K
Tue Nov 19, 2019 11:27 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: hydrogen bond
Replies: 3
Views: 34

Re: hydrogen bond

There are two rules for hydrogen bonding:
1. The hydrogen has to be covalently bonded to a fluorine, oxygen, or nitrogen ( a highly electronegative atom)
2. The hydrogen must be near a molecule containing a fluorine, oxygen, or nitrogen with a lone pair
by Pegah Nasseri 1K
Tue Nov 19, 2019 11:21 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond angles
Replies: 3
Views: 43

Re: Bond angles

In the CH2CHCN lewis structure, you would have two trigonal planar structures and one linear structure. The two trigonal planar structures have bond angles of 120 degrees between the carbons and hydrogens. The C-C-N bond is the linear structure because the central atom C has two bonding pairs. The l...
by Pegah Nasseri 1K
Sat Nov 16, 2019 12:53 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Distortion and Polarizability
Replies: 2
Views: 31

Re: Distortion and Polarizability

Anions that are larger with lower charges are more polarizable, meaning their electron cloud can be more easily distorted. Anions that are smaller with a higher charge have more polarizing power, meaning they are more able to distort an electron cloud.
by Pegah Nasseri 1K
Sat Nov 16, 2019 12:44 pm
Forum: General Science Questions
Topic: Test 2
Replies: 7
Views: 67

Re: Test 2

In discussion section, my TA mentioned that it would be more conceptual and there would be less calculations compared to test 1 and the midterm. The test will cover types of intermolecular interactions such as hydrogen bonding, VSEPR formulas and shapes, determining if a molecule is polar or non-pol...
by Pegah Nasseri 1K
Sat Nov 16, 2019 12:38 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: T-shape
Replies: 6
Views: 40

Re: T-shape

An example of a t-shaped structure is iodine trichloride (ICl3). The central atom iodine has 3 bonding pairs and 2 lone pairs so its formula is AX3E2. Imagine the shape as being tetrahedral but there are 2 lone pairs so there are three bonding pairs that are being repelled by the lone pairs, forming...
by Pegah Nasseri 1K
Sat Nov 16, 2019 12:28 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Shape
Replies: 6
Views: 72

Re: Shape

Some common formulas for general shapes are:
AX2 - Linear
AX3 - Trigonal Planar
AX4 - Tetrahedral
AX4E - Trigonal Bipyramidal
AX4E2 - Square Planar
AX3E - Trigonal Pyramidal
AX2E2 or AXE - Bent or Angular
AX6 - Octahedral
by Pegah Nasseri 1K
Sat Nov 16, 2019 12:21 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Shape of ClO2+
Replies: 6
Views: 64

Re: Shape of ClO2+

ClO2+ will be bent because the central atom chlorine has 2 bonding pairs and one lone pair so its VSEPR formula would be AX2E which corresponds to the bent shape. If the central atom had 3 bonding pairs, then it would trigonal planar but in this case there is a lone pair.
by Pegah Nasseri 1K
Tue Nov 05, 2019 5:44 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Probability
Replies: 3
Views: 74

Re: Probability

An orbital is just the probability of the electron's location. It represents a 3D space in the form of a shape where you could find the electron within the atom and around the nucleus. Orbitals can be referred to as probability maps and the shape of the region of 3D space where you could find the el...
by Pegah Nasseri 1K
Tue Nov 05, 2019 5:38 pm
Forum: DeBroglie Equation
Topic: Dino Nuggets 8C
Replies: 2
Views: 54

Re: Dino Nuggets 8C

6.78 X 10^14 Hz is the frequency of the light. Use the equation c=wavelength*frequency to solve for the wavelength of the light (wavelength= c/frequency). You should get around 442 nm which would be within the visible range of 400-700 nm.
by Pegah Nasseri 1K
Tue Nov 05, 2019 5:28 pm
Forum: General Science Questions
Topic: Coordinate covalent bonds/ Lewis acids and bases
Replies: 1
Views: 33

Re: Coordinate covalent bonds/ Lewis acids and bases

No, Lavelle only briefly introduced us to Lewis acids and bases to show that some central atoms in a lewis structure don't need an octet because they can later accept electrons (which is what makes them an acid). You don't have to know anything about Lewis acids and bases. Just know that the first 4...
by Pegah Nasseri 1K
Tue Nov 05, 2019 5:21 pm
Forum: Lewis Structures
Topic: CN-
Replies: 2
Views: 43

Re: CN-

Yes, CN- has a triple bond. There are 10 valence electrons in total. Using a double bond would not fulfill the octet rule for carbon. Nitrogen would have a formal charge of -1 if it had 2 lone pairs. However, in this case the Carbon has to take the formal charge of -1 in order for it to have an octe...
by Pegah Nasseri 1K
Tue Nov 05, 2019 5:13 pm
Forum: Quantum Numbers and The H-Atom
Topic: Midterm Content
Replies: 2
Views: 63

Re: Midterm Content

Yes, the Lyman series is for UV light and it starts at n=1. The Balmer series is used for visible light and starts at n=2. We have to know the range of wavelengths in the visible light region (400-700 nm). Aside from that, you should know where the other ranges of light would be located relative to ...
by Pegah Nasseri 1K
Thu Oct 31, 2019 4:23 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge equation
Replies: 9
Views: 59

Re: Formal Charge equation

It is easier to use the equation that Lavelle gave us in class for formal charge. Another quick way for calculating formal charge without using the equation is by simply knowing the number of valence electrons an element has and subtracting that by the amount of valence electrons it has in the lewis...
by Pegah Nasseri 1K
Thu Oct 31, 2019 4:01 pm
Forum: Octet Exceptions
Topic: En
Replies: 3
Views: 39

Re: En

As stated by the previous posts, En stands for electronegativity and indicates how much an element attracts a shared pair of electrons. Lavelle mentioned to us to not worry about an electronegativity difference between 1.5 and 2 for now. We will not be asked about that yet.
by Pegah Nasseri 1K
Thu Oct 31, 2019 2:51 pm
Forum: Lewis Structures
Topic: lewis structures
Replies: 4
Views: 39

Re: lewis structures

If there are resonance structures, we will probably be asked to draw more than one possible structure of a molecule. If there are no resonance structures, you would have to draw the structure with the lowest formal charge which would be the best structure since it is the most stable. The best struct...
by Pegah Nasseri 1K
Thu Oct 31, 2019 2:41 pm
Forum: Octet Exceptions
Topic: expanded valence shell confusion
Replies: 3
Views: 44

Re: expanded valence shell confusion

For lewis structures with a central element that is located in row 3 of the periodic table or beyond, the element is allowed to have more than an octet of valence electrons. So in your lewis structure, the central element could have more bonds than the amount required for it to have an octet. This i...
by Pegah Nasseri 1K
Thu Oct 31, 2019 2:36 pm
Forum: Electronegativity
Topic: Electronegativity difference
Replies: 2
Views: 21

Re: Electronegativity difference

You are not expected to know the values of electronegativity for all the elements. They would be provided to you if you were asked to determine the type of bond. If the difference of the two electronegativity values is greater than 2, you would know that the bond is ionic. If the difference is less ...
by Pegah Nasseri 1K
Tue Oct 22, 2019 11:47 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: KeV
Replies: 2
Views: 40

Re: KeV

This is just another unit for expressing energy. When given a quantity with this unit, make sure to convert the unit to joules. 1 KeV = 1000 eV and 1 eV = 1.602 X 10^-19 J. So when converting KeV to joules you would use dimensional analysis: \large KeV (\frac{1000 eV}{1 KeV})(\frac{1.602...
by Pegah Nasseri 1K
Tue Oct 22, 2019 11:21 pm
Forum: Einstein Equation
Topic: Usage of Einstein Equation
Replies: 2
Views: 97

Re: Usage of Einstein Equation

The equation E=hv gives you the energy of a single photon. If you have the total energy of a source of light, you can figure out the number of photons by dividing the total energy of the light by the energy of a single photon of the light.
by Pegah Nasseri 1K
Tue Oct 22, 2019 11:12 pm
Forum: Einstein Equation
Topic: 1B.9
Replies: 3
Views: 49

Re: 1B.9

You have two equation E=hv (the energy of a single photon) and c=lambda*frequency: \large E=hv \large c=\lambda v You can isolate v, the variable for frequency in the equation c=lambda*frequency: \large c=\lambda v \large v =\frac{c}{\lambda } Plug this into the equation E=hv for v. \large E=hv, v=\...
by Pegah Nasseri 1K
Tue Oct 22, 2019 10:57 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: s, p, d, f orbitals
Replies: 15
Views: 138

Re: s, p, d, f orbitals

The s-orbital is filled up first but when the s-orbital is occupied and electrons enter the d-orbital, the d-orbital will have a lower energy than the s-orbital. For example, 3d has a lower energy than 4s when electrons are in the 3d orbital and 4s is occupied. So when writing out the electron confi...
by Pegah Nasseri 1K
Mon Oct 21, 2019 5:16 pm
Forum: Limiting Reactant Calculations
Topic: Test Question
Replies: 1
Views: 75

Re: Test Question

Use the 50.0 mg of glucose to determine how much of each reactant is needed to produce that much glucose: \large (50.0*10^{-3} g C_{6}H_{12}O_{6})(\frac{1}{180.156 g*mol^{-1}})(\frac{6 mol CO_{2}}{1 mol C_{6}H_{12}O_{6}}) = 1.67 * 10^{-3} mol CO_{2} \large (50.0*10^{-3} g...
by Pegah Nasseri 1K
Tue Oct 15, 2019 7:06 pm
Forum: Photoelectric Effect
Topic: Wavelength Emitted
Replies: 4
Views: 54

Re: Wavelength Emitted

The wavelength of the light emitted will always be positive and the energy of the photon will be positive as well. However, the change in energy of the electron as it changes states can be negative. The energy is negative when the electron emits energy when moving down energy levels. The absolute va...
by Pegah Nasseri 1K
Tue Oct 15, 2019 6:40 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Uncertainty Measurement Clarification
Replies: 2
Views: 32

Re: Uncertainty Measurement Clarification

You would use the full range of values. So in your example you would use 10 as the value of uncertainty for your equation. Another example would be if you were given a measurement of 13.5 +/- 0.5 m the range of values you would have would be between 13.0 m - 14.0 m and so you would have an uncertain...
by Pegah Nasseri 1K
Tue Oct 15, 2019 6:25 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Principle Quantum Level
Replies: 2
Views: 39

Re: Principle Quantum Level

You know that the energy per photon needed to eject an electron is equal to the energy absorbed or emitted by the electron changing states. You also know that the final energy state of the electron is at n=4: \large E_{photon}=E_{f} - E_{i} = E_{4} - E _{i} Use the equation \large E_{n} = -hR/n^{2} ...
by Pegah Nasseri 1K
Tue Oct 15, 2019 5:47 pm
Forum: DeBroglie Equation
Topic: symbol
Replies: 4
Views: 43

Re: symbol

The symbol just indicates that the value of one measured quantity is proportional to the value of another measured quantity. For example, in the equation E=hv you can say that E∝v which means that E, the energy of one photon, is proportional to v, the frequency of the wavelength of the light.
by Pegah Nasseri 1K
Tue Oct 15, 2019 5:36 pm
Forum: Photoelectric Effect
Topic: Speed of Light
Replies: 29
Views: 251

Re: Speed of Light

The speed of light will always be a constant and is labeled in equations as the variable "c". It is somewhere around 3.0 X 10^8 m/s. On exams, make sure to use the exact number for the speed of light that is given to you on the equation sheet of the exam. This will help you with getting th...
by Pegah Nasseri 1K
Tue Oct 08, 2019 11:15 pm
Forum: SI Units, Unit Conversions
Topic: Are there going to be any questions on this week's test regarding Quanta and Photons?
Replies: 5
Views: 96

Re: Are there going to be any questions on this week's test regarding Quanta and Photons?

As mentioned in the other posts, we are only being tested on the review of high school chemistry that we did. This is the fundamentals chapter in the textbook. So you can look at the problems in the fundamentals chapter to get a good sense of what we are being tested on. Quanta and photons will not ...
by Pegah Nasseri 1K
Tue Oct 08, 2019 10:46 pm
Forum: Balancing Chemical Reactions
Topic: Chemical States
Replies: 3
Views: 81

Re: Chemical States

Try to include the states of the reactants and products if they are known. Sometimes it is helpful when determining the solute and the solvent of a solution or to figure out what solute is produced by a chemical reaction. This will help you determine different properties of the reactants and product...
by Pegah Nasseri 1K
Tue Oct 08, 2019 10:38 pm
Forum: Balancing Chemical Reactions
Topic: state of acids
Replies: 3
Views: 60

Re: state of acids

It depends on the acid and the temperature of the acid. Acids can be solids, gases, and liquids. They can also change states depending on the temperature. However, most of the common acids you come in contact with everyday are aqueous or liquid (like acetic acid aka vinegar).
by Pegah Nasseri 1K
Tue Oct 08, 2019 10:20 pm
Forum: SI Units, Unit Conversions
Topic: Accuracy vs Precision
Replies: 11
Views: 173

Re: Accuracy vs Precision

Accuracy is how close your measured values are to the actual, correct value. Precision is how close your measured values are to each other. Your results in a lab are accurate if your data matches the actual value of whatever you measured. Your results are precise if the measurements in your differen...
by Pegah Nasseri 1K
Tue Oct 08, 2019 10:13 pm
Forum: General Science Questions
Topic: Test Question
Replies: 8
Views: 138

Re: Test Question

You should be able to write out the molecular formulas of common compounds like aluminum oxide or sodium chloride using the number of valence electrons each atom has. In an ionic compound, you can figure out how many valence electrons each element needs to fill their valence electron shell and balan...
by Pegah Nasseri 1K
Thu Oct 03, 2019 2:28 pm
Forum: Significant Figures
Topic: Sig Figs
Replies: 4
Views: 55

Re: Sig Figs

You would take into account the number of significant figures of the molar masses/atomic mass numbers but usually these numbers have more significant digits than the given numbers in the word problems. Often times you would just use the least number of significant figures given to you by a number in...
by Pegah Nasseri 1K
Thu Oct 03, 2019 2:23 pm
Forum: Empirical & Molecular Formulas
Topic: Molecular formula
Replies: 2
Views: 47

Re: Molecular formula

When you are asked to find the molecular formula using the empirical formula, you would divide the molar mass of the actual molecule by the molar mass of the empirical formula to find a number. You would use this number as a multiplier and would multiply the subscripts of the empirical formula by th...
by Pegah Nasseri 1K
Thu Oct 03, 2019 2:18 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Fundamentals E.3
Replies: 7
Views: 110

Re: Fundamentals E.3

There are 9 gallium atoms on the left pan. You have to figure out how many astatine atoms need to be placed on the right pan for the masses on the two sides to be equal. Since it is known that astatine is three times as heavier like you mentioned, you would only need 3 astatine atoms to balance 9 ga...
by Pegah Nasseri 1K
Thu Oct 03, 2019 2:07 pm
Forum: Balancing Chemical Reactions
Topic: "Combustion" Term
Replies: 8
Views: 163

Re: "Combustion" Term

Professor Lavelle referred to combustion as burning or oxidation of organic molecules. He also mentioned metabolism which occurs when energy is released, such as during cellular respiration. Metabolic processes such as cellular respiration can be classified as combustion reactions because carbon dio...
by Pegah Nasseri 1K
Thu Oct 03, 2019 1:59 pm
Forum: Administrative Questions and Class Announcements
Topic: Where do we turn in the homework?
Replies: 5
Views: 74

Re: Where do we turn in the homework?

We are turning in our first homework assignment in lecture on friday. All other homework assignments after that are turned into our discussion sections.

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