## Search found 101 matches

Mon Mar 09, 2020 12:23 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: 7A.15
Replies: 4
Views: 200

### Re: 7A.15

Since you would be able to find the differential rate law, you should not take into account the concentration of C since it is a zero order. Because of this, the rate would become rate = k[A][B] so that you're able to determine the orders of concentration of A compared to the concentration of A in t...
Mon Mar 09, 2020 12:20 pm
Forum: General Rate Laws
Topic: 7A.15
Replies: 3
Views: 94

### Re: 7A.15

In figuring out the answer, you would see that the concentration of C would be changing, but you do not take it into account since it is a zero order. The differential rate law would instead be rate = k[A][B] instead, so this way, you are able to focus on the differences between each experiment from...
Mon Mar 09, 2020 12:17 pm
Forum: General Rate Laws
Topic: 7B3a
Replies: 3
Views: 66

### Re: 7B3a

This is from the equation ln2. The equations sheet showcases t1/2 = ln2/k which is able to determine the half life reaction time. In the solutions manual, they only use ln2 as a substitute for 0.693 in order to get to the answer. This is able to identify the reaction time in comparison to the equati...
Mon Mar 09, 2020 12:15 pm
Forum: General Rate Laws
Topic: 7A.15
Replies: 2
Views: 47

### Re: 7A.15

You should be able to compare by focusing on one initial concentration that is different, while the others stay the same. For example, to find the order of A, you should be able to compare the first and second experiment since the rate shows that the concentration of A differs form 10 to 20, while t...
Mon Mar 09, 2020 12:12 pm
Forum: General Rate Laws
Topic: 7A.1
Replies: 3
Views: 46

### Re: 7A.1

This is a comparison of the stoichiometric coefficients which relates to kinetics. For example, the first part in a, you should compare N2 and H2 to each other in the chemical equation. Since it takes 3 moles of H2 to make 1 mole of N2, then you should be able to multiply H2 by 1/3 to get 1 mole of ...
Mon Mar 02, 2020 12:25 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.3
Replies: 3
Views: 32

### Re: 6L.3

Yes, we should be able to calculate the cell potential with the equations in order to showcase the spontaneous reaction.

Specifically for part A, it should have E cell potential = E cathode - E anode.
Thus, it should be 0.80 V - (-0.23 V) = 1.03 V as a result.
Mon Mar 02, 2020 12:22 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6M.5
Replies: 4
Views: 81

### Re: 6M.5

Adding on, once you determine the half reactions, you are able to write the cell diagram for cathodes on the right and anodes on the left. Make sure to include the H+ in the cell diagram. Upon calculating the cell potential, you are able to determine that the reaction is spontaneous since it becomes...
Mon Mar 02, 2020 12:19 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.3
Replies: 2
Views: 23

### Re: 6L.3

Since the cell potential is negative, this means that the delta G would be positive, making the reaction non-spontaneous.

However, if the reaction is in reverse, the cell potential would be positive, making delta G negative to have the reaction spontaneous.
Mon Mar 02, 2020 12:17 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Textbook question 6.65
Replies: 4
Views: 117

### Re: Textbook question 6.65

Because pH ranges from 1 to 14, you need to use the E knot cell = RTlnK/nF equation, which is derived from when delta G knot is equal to each other for both equations of -nFE knot cell and -RTknK. Then, since it begins at pH = 7, you should know that E knot cell is 0. You should be able to calculate...
Mon Mar 02, 2020 12:13 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6N.3
Replies: 6
Views: 143

### Re: 6N.3

In the book, Faraday's constant is not used because the solutions manual utilizes a short cut. If you multiplied the values of the constants together and divided with temperature, R, and Faraday's constant, you will be able to get 0.025693 which is divided by the moles. This is only used when the te...
Mon Feb 24, 2020 12:24 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.5 D.
Replies: 7
Views: 126

### Re: 6L.5 D.

For this specific reaction, Au + acts in both the cathode and anode reaction, so you should see both in each reaction. For cathode, the half reaction is: Au+ + e- -> Au For anode, the half reaction is" Au3+ + 3e- -> Au+ Then, you should multiply the half reaction of cathode by 3 and flip the an...
Mon Feb 24, 2020 12:21 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.5 part b)
Replies: 2
Views: 37

### Re: 6L.5 part b)

To tell if you need an inert electrode like platinum, you should be able to tell by looking at the specific matter or state of the equations.
In this specific question, you can see that both Ce and I are aqueous, meaning that they need a solid inert electrode to stabilize the redox reaction.
Mon Feb 24, 2020 12:19 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6M.1
Replies: 4
Views: 85

### Re: 6M.1

Make sure to start with the equation of E cell = E cathode - E anode. Once you put it in the equation, you can solve for the remaining or missing variable. -0.689 V = -1.038 V - E anode E anode = -0.349 V. The negative signs can be confusing, but as long as you plug it in for specific variables and ...
Mon Feb 24, 2020 12:14 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.7
Replies: 2
Views: 83

### Re: 6L.7

For b, you would introduce the cathode half reaction as:
O2 + 4 H+ + 4e- -> 2 H2O (1.23 V)
The anode half reaction is:
O2 + 2 H2O + 4e- -> 4 OH- (0.40 V)

Thus, this would be equal to
4 H+ + 4 OH- -> 4 H2O
which would give 1.23 V - 0.40 V = 0.83 V.
Mon Feb 24, 2020 12:11 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagrams
Replies: 7
Views: 127

### Re: Cell Diagrams

Since there is already a solid on both sides of the cell diagram for an anode and a cathode, you do not need to introduce platinum into the equation. Platinum does not interfere with the equation and is inert, but in this example, you do not need it since the Au (s) acts to balance the equation for ...
Wed Feb 19, 2020 1:03 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 5.55
Replies: 4
Views: 79

### Re: 5.55

C is not included because it is a solid. Since it is a solid, it is not able to be included in the chemical equation for ICE because we do not include solids or liquids.

You should only include aqueous solutions or gases, and then, you are able to find the equilibrium concentrations for the question.
Wed Feb 19, 2020 1:01 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: delatG= -RTInK
Replies: 4
Views: 98

### Re: delatG= -RTInK

When you do not know the delta G of the chemical equation, you can use this by calculating with the R and temperature as well as the K if it is given in the equation. K is able to be easily measured when given a chemical equation for any types of reactions. Thus, if you have an example to find delta...
Wed Feb 19, 2020 12:52 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 5G.13
Replies: 8
Views: 263

### Re: 5G.13

You are able to first calculate the Q value of the equation. Then, once getting this, you can use the delta G = delta G knot + RTlnQ, but since we do not know the delta G knot, we can replace it with -RTlnK. Thus, the whole equation would be delta G = -RTlnK + RTlnQ. You are able to insert all of th...
Wed Feb 19, 2020 12:38 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: G vs G knot
Replies: 15
Views: 150

### Re: G vs G knot

G, or more specifically, delta G, is used in a spontaneous reaction with the energy released that is able to be used as work. However, delta G is the free energy in a standard state condition.To calculate this, you can use it as a state function by adding or subtracting as a Hess Type approach or fr...
Wed Feb 19, 2020 12:20 pm
Forum: Balancing Redox Reactions
Topic: 6K.1
Replies: 1
Views: 45

### Re: 6K.1

To first approach the question, you should see at which indicate which part is being oxidized and reduced. To see a compound's overall charge, you should look at the superscript. To look at the charge of the compound as a whole, you should be able to see the total net charge. Then, look at the indiv...
Mon Feb 10, 2020 12:30 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Isolated vs Insulated
Replies: 3
Views: 110

### Re: Isolated vs Insulated

Isolated systems are defined as both matter and energy are unable to enter the surroundings.

Because of this, insulated objects should be isolated since they are unable to pass both matter and energy to the surroundings. You should be able to know open, closed, and isolated.
Mon Feb 10, 2020 12:25 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: State Function
Replies: 7
Views: 101

### Re: State Function

Hess's law can be applied to Gibb's equation because Gibb's is a state function. Because they are state functions, they are able to be added or subtracted together since the pathway is not dependent on them. In this, Gibbs is able to be added or subtracted when given specific equations of formation ...
Mon Feb 10, 2020 12:23 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Writing balanced equation for formation reactions
Replies: 4
Views: 32

### Re: Writing balanced equation for formation reactions

The coefficients should match what the equation is asking for. If you were to do it without the fractions, it would greatly affect the answer and would change the values of entropy, enthalpy, and Gibbs. Since some equations only ask for 1.00 mols of a product in formation, then you should use the fr...
Mon Feb 10, 2020 12:21 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 4J.5
Replies: 1
Views: 29

### Re: 4J.5

To calculate the equation, you should be able to look at the back of the textbook in order to find specific values dealing with each part of the equation. You should have the sum of the products minus the sum of the reactants as well as having each part of the equilibrium equation be multiplied by t...
Mon Feb 10, 2020 12:18 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 4J.7
Replies: 2
Views: 37

### Re: 4J.7

In order to calculate the equation for each, you should be able to look at the values in the back of the textbook for each part of the chemical equation. To calculate, you should get the sum of the products minus the sum of the reactants. Additionally, you should be able to figure out the values for...
Mon Feb 03, 2020 12:26 pm
Forum: Calculating Work of Expansion
Topic: 4B.9
Replies: 3
Views: 66

### Re: 4B.9

For adiabatic processes, you should know that it is a closed system so there is fixed matter but energy going into the surroundings. Thus for a closed system, you are able to figure out when q = 0 since there is no heat being generated and delta U or internal energy is always going to equal work for...
Mon Feb 03, 2020 12:24 pm
Forum: Calculating Work of Expansion
Topic: 4A.13
Replies: 2
Views: 47

### Re: 4A.13

Because q = CdeltaT, you are able to solve this equation by first calculating for the C. C = q/delta T, and thus, it is -3.50 kJ/7.32 K = 0.478 kJ/K^-1. You know that it is negative since the q of a calorimeter is negative since energy is leaving. Then, you can calculate it again from q(Cal) = Cdelt...
Mon Feb 03, 2020 12:21 pm
Forum: Calculating Work of Expansion
Topic: 4C.13
Replies: 2
Views: 33

### Re: 4C.13

In order to get this equation, you need to calculate the enthalpy of heating the ice cube by converting it to the kJ. Once you do this, you are able to add it to the equation to calculating heat through q = m*C*delta T, leaving delta T a variable as (Tfinal - 0). Then, you should do the same for cal...
Mon Feb 03, 2020 12:17 pm
Forum: Calculating Work of Expansion
Topic: 4A.5
Replies: 2
Views: 57

### Re: 4A.5

To calculate the reverse and isothermal reaction, this basically means that it is when there is constant temperature. In lecture today, we learned how to derive the equation for work, and the equation you would have to use is w = -n*R*T*ln(V2/V1) in order to get the work. This is able to calculate f...
Mon Feb 03, 2020 12:14 pm
Forum: Calculating Work of Expansion
Topic: 4A.3 and 4A.5
Replies: 1
Views: 27

### Re: 4A.3 and 4A.5

I am not certain if you are talking about part a or part b in both questions. However, for both part A questions, the work was in Liters, so you needed to convert the Liters into Joules by (101.325 J/ 1 L) in order to get the answer since work is always calculated in Joules. In 4A.5 for part B, you ...
Mon Jan 27, 2020 2:20 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Homework Question 4A.3
Replies: 2
Views: 95

### Re: Homework Question 4A.3

I think this is an error in the textbook.
The answer that I got for part C for question 4A.3 was 28 Joules which is the answer found in part A of the question since there is no change other than the change in volume.
Mon Jan 27, 2020 2:18 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: work
Replies: 5
Views: 68

### Re: work

Work is used by the equation of: w = opposing force x distance moved. There are many different equations of work that is used depending on the situation, but this is the overall equation for work. For example, to calculate the work with pressure as well as a change in volume, you can use the equatio...
Mon Jan 27, 2020 2:15 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Standard enthalpy of formation
Replies: 6
Views: 64

### Re: Standard enthalpy of formation

The standard enthalpy of formation for a stable element is zero because since it is diatomic, it requires no energy in order to form. Diatomic gases such as O2, I2, H2, and so forth are natural and form without needing energy, making it extremely stable. To form an O2 for example, O2 -> O2, and ther...
Mon Jan 27, 2020 2:11 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Delta U
Replies: 3
Views: 44

### Re: Delta U

Delta h is the change in the enthalpy, while the delta u is the change in the internal energy. With internal energy, it is the amount of the energy that a system contains.

Delta h is equal to the delta u plus the pressure times change in volume.
AKA:
ΔH = ΔU + PΔV.
Mon Jan 27, 2020 2:06 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Standard enthalpy of formation of O2
Replies: 4
Views: 39

### Re: Standard enthalpy of formation of O2

O2 is already in it's most stable form, so this is why the enthalpy is only 0. We stated in the additional comments that the bond enthalpies of diatomic molecules are accurate and measured for those molecules.

Because the reaction goes from O2 to O2, it is already stable, making it zero.
Mon Jan 20, 2020 1:14 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I 33
Replies: 2
Views: 26

### Re: 5I 33

For this specific equation, solids and liquids are not part of the K constant equilibrium equation. This is shown here because the ammonium carbamate of NH4(NH2CO2) is not included since it is a solid for the reactant side. Because of this, the ICE table would only include NH3 and CO2. Since the gra...
Mon Jan 20, 2020 1:05 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5.I.23
Replies: 3
Views: 56

### Re: 5.I.23

For this specific equation, you should have the H20 as a part of the ICE table because it is part of the product made from the chemical equilibrium. H20 would only be disregarded if there was a solid or a liquid, and it may be also disregarded as reactant in the equation since the solvent is insigni...
Mon Jan 20, 2020 1:00 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6D.7
Replies: 3
Views: 48

### Re: 6D.7

To do this, you must find the initial HClO molar concentration. Since you are given the pH, you can find out the equilibrium molar concentration of the H30+ by using the pH through 10^-4.60. You will then get 2.51 x 10^-5. Because the equilibrium molar concentration of H30+ is x due to the ICE table...
Mon Jan 20, 2020 12:55 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.13
Replies: 4
Views: 56

### Re: 5I.13

For this specific part C, I compared the two K values to each other with F2 and Cl2. With this comparison, I saw that Cl1 had a much smaller K value than F2. Because of this, it is evident that Cl2 is much more stable because they are less likely to react with a smaller K value in comparison.
Mon Jan 20, 2020 12:51 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE
Replies: 3
Views: 41

### Re: ICE

ICE tables are often used when you are given some initial molar concentrations of specific molecules. If the question is asking for the equilibrium molar concentration, you can use ICE to find it. Additionally, you can use it on order to find the composition of equilibrium mixtures. For later method...
Mon Jan 13, 2020 12:26 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.27
Replies: 8
Views: 99

### Re: 5I.27

Equilibrium composition means the equilibrium concentrations of all the reactants and the products.
To use this for the question, make sure to divide each of the moles divided by the volume in order to find the molarity of each.
Specifically in 27, you should be finding the Q of the equation.
Mon Jan 13, 2020 12:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.13 part c
Replies: 2
Views: 24

### Re: 5I.13 part c

Cl2 is more stable because in the question, it shows the K calculated for both a and b parts. Because the K for Cl2 was smaller than the K for F2, this ultimately shows that the CL2 equation is less likely to react in response. Because it is least likely to react, this means that Cl2 is much more st...
Mon Jan 13, 2020 12:22 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I. 19
Replies: 3
Views: 42

### Re: 5I. 19

For this specific problem, you know that 60% must be left as the equilibrium. Thus, when you are calculating the equilibrium, all you have to do to find x is getting 0.133 - x = .4(0.133). 40% of the actual product must be left for H2. Then, you are able to find the x and you can plug it in to find ...
Mon Jan 13, 2020 12:18 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework 5I.13
Replies: 2
Views: 41

### Re: Homework 5I.13

For this specific equation, I used the ICE table from above in order to find the answer. However, instead of using the quadratic equation, it is evident that the values would be too small and the x from the denominator would be too small to affect the K. It would be 1.2 x 10^-7 = (2x)^2 / (0.001-x)....
Mon Jan 13, 2020 12:13 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5H.3
Replies: 2
Views: 36

### Re: 5H.3

For this reaction, you will not see the equation on the table. Instead, because the equation requires both BrCl and HCl, you will actually need to utilize the K from each reaction found in the table. To do this, use the K from 2 BrCl -> Br2 + Cl2, which is 377. Use the K from H2 + Cl2 -> 2 HCl which...
Mon Jan 06, 2020 6:01 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Effect on the equilibrium constant
Replies: 2
Views: 38

### Re: Effect on the equilibrium constant

To add on, besides the equilibrium constant, there is an effect on the position of the equilibrium. For change in concentration, the k does not change, but when more reactants are added, more products will form until the original product / reactant ratio is attained. The chemical reaction remains at...
Mon Jan 06, 2020 5:56 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Writing Equilibrium Constant Labels
Replies: 4
Views: 74

### Re: Writing Equilibrium Constant Labels

P does means partial pressure because you are using a gas throughout the entire equilibrium. The equilibrium constant would also be denoted as Kp. You can often use the equation, PV = nRT in order to get the concentration of the whole equilibrium if you were asked to find it. Both Kp and Hc have the...
Mon Jan 06, 2020 5:53 pm
Forum: Ideal Gases
Topic: ICE Table Variables
Replies: 6
Views: 105

### Re: ICE Table Variables

Using variables should be essential in establishing the change in molar concentration. However, you should only not use when there is a solid involved or water leftover as a liquid. When you are given equilibrium molar concentrations, you can use that number in order to sometimes determine the chang...
Mon Jan 06, 2020 3:37 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Question 5H.3
Replies: 3
Views: 61

### Question 5H.3

I have a question on 5H.3, which states, "Use the information in Table 5G.2 to determine the value of K at 300 K for the reaction 2 BrCl(g) + H2(g) ->/<- Br2(g) + 2 HCl(g)." I know that with the previous exercises with 5H, we had to use the K values listed on the table, but I am confused w...
Mon Jan 06, 2020 2:54 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Post Assessment Problem 18
Replies: 2
Views: 62

### Re: Post Assessment Problem 18

To solve this problem, you should make sure that the equation is first balanced for the equilibrium constant. The equation balanced should be 2 NH3 (g) ⇌ N2 (g) + 3 H2 (g). Then, after this, you should put the products over the reactants in order to calculate the equilibrium constant. For this, usin...
Mon Dec 02, 2019 12:01 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: pH vs pOH
Replies: 4
Views: 63

### Re: pH vs pOH

You should be able to look at the actual molecule to determine if it's an acid or a base. If it is an acid, you know that you are solving for the pH with the -log. If it is a base, you know that you are solving for the pOH with the log. Determining the opposite would just have pH + pOH = 14 as a tot...
Mon Dec 02, 2019 11:58 am
Forum: Conjugate Acids & Bases
Topic: Naming
Replies: 4
Views: 93

### Re: Naming

In order to begin, look at the name of the ligands first for coordination compounds. Using the number of ligands, you should use mono-, di-, tri-, tetra-, penta-, or hexa-. Add an o to the end of an anion name, like cyanide s cyan. You should also look at the metal to be placed at the end of the nam...
Mon Dec 02, 2019 11:55 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Water as a acid or base
Replies: 4
Views: 92

### Re: Water as a acid or base

Depending on the chemical reaction, water can either act as an acid or a base. For example, if paired with an acid, water would then act as a base and accept an H+ from the acid. Water is an amphoteric molecule that can either act as an acid or a base. You should look at what the other molecule is i...
Mon Dec 02, 2019 11:52 am
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Final Study Methods
Replies: 11
Views: 232

### Re: Final Study Methods

For the exam, the textbook has a lot of information in order to study since it reinforces a lot of information! Also, I would go back and study the problems that he assigned within the textbook because there is a possibility of them being in the final. The notes do hold information, but I think doin...
Mon Dec 02, 2019 11:37 am
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Resonance Hybrid
Replies: 4
Views: 58

### Re: Resonance Hybrid

Resonance is able to make an acid stronger because they are able to stabilize the conjugate base which will increase the acidity. It depends on the structure with delocalization for the anion. This will in turn make the acid be able to give off more H+ ions due to its stability with the conjugate ba...
Mon Dec 02, 2019 11:30 am
Forum: Hybridization
Topic: 2F.5
Replies: 4
Views: 73

### Re: 2F.5

For hybridization, you should be able to know it by drawing out the Lewis Structure. Then, you can look at the number of electron densities in order to figure out the hybridization. For rules of hybridization, 2 electron densities means sp, 3 electron densities means sp2, 4 electron densities means ...
Mon Nov 25, 2019 11:49 am
Forum: Naming
Topic: 9C #9c/d
Replies: 3
Views: 47

### 9C #9c/d

For Question #9 on parts c and d, with Problems 9C, I'm confused on what en and edta mean. Can someone explain these to me?
c) [PtCl2(en)2]^2+
d) [Cr(edta)]^-
Mon Nov 25, 2019 11:45 am
Forum: Naming
Topic: 9C.1
Replies: 3
Views: 50

### Re: 9C.1

To start, you should look at the name of ligands first in naming coordination compounds. Looking at the number of ligands, you should use mono-, di-, tri-, tetra-, penta-, or hexa- to name the ligand. Make sure to add an o to the end of an anion name, for example, cyanide should be cyano or cyanido....
Mon Nov 25, 2019 11:39 am
Forum: Hybridization
Topic: 2F.5
Replies: 3
Views: 52

### Re: 2F.5

In order to find the answer, you should draw out the Lewis structure of each molecule. Then, you are able to see the type of hybridization for their specific atoms. For example, BeCl2 should be drawn out with a Lewis structure. In this, you can see that the shape of the molecule is linear according ...
Mon Nov 25, 2019 11:35 am
Forum: Naming
Topic: 9C #5
Replies: 3
Views: 69

### 9C #5

On Question #5 of problems 9C, I'm confused on what it's asking in regards to a polydentate.
Can someone give me a brief definition of a polydentate and how the question is being asked?
Thank you so much!
Mon Nov 25, 2019 11:32 am
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 Bonding Sites
Replies: 2
Views: 44

### Re: Hydrogen Bonding Sites

Hydrogen bonding sites are determined by hydrogen bonding, where it is present in molecules with N, O, and F atoms. In a molecule, when you see a hydrogen bonded directly to a nitrogen, oxygen, or fluorine, this is a hydrogen bonding site. However, it also considers the lone pairs of electrons as we...
Mon Nov 18, 2019 11:57 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Why is SF4 Polar?
Replies: 6
Views: 76

### Re: Why is SF4 Polar?

SF4 is polar because being polar means that it has an unequal distribution of electrons in it's structure. Because of the seesaw shape, it indicates that there are 4 bonded pairs as well as 1 lone pair of electrons. Because of the 1 lone pair of electrons around the central atom, it is characterized...
Mon Nov 18, 2019 11:54 am
Forum: Determining Molecular Shape (VSEPR)
Topic: VSEPR notation
Replies: 6
Views: 131

### Re: VSEPR notation

Yes, in CH2Cl2, the VSEPR notation would be AX4, corresponding to tetrahedral shape as well as bond angles of 109.5 degrees. For PCl3, the VSEPR notation would be AX3E, corresponding to trigonal pyramidal as well as bond angles of less than 109.5 degrees. It is helpful to memorize the notations in o...
Mon Nov 18, 2019 11:49 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Molecular Shape
Replies: 3
Views: 46

### Re: Molecular Shape

Yes, you should primarily focus on the central atom when taking in the shape of the molecule and solving for VSEPR. In this way, you can see the central atom's regions of electron density to perform the right molecular shape. You need to see the number of bonded pairs as well as lone pairs in order ...
Mon Nov 18, 2019 11:47 am
Forum: Hybridization
Topic: Hybridization
Replies: 3
Views: 32

### Hybridization

For the examples used in class today, I was a little confused on the very last example for hybridization. When we used C2H4 for ethylene, I am confused on why the 2sp^2 hybridization goes to 2p orbital as well. Can someone explain to me for these hybrid orbitals?
Mon Nov 18, 2019 11:45 am
Forum: Determining Molecular Shape (VSEPR)
Topic: VSPER
Replies: 7
Views: 221

### Re: VSPER

VSEPR stands for valence shell electron pair repulsion, and it is used in order to model for different structures of molecules. It is used in order to predict the shape for a molecule as well as the bond angles. It takes in both the amount of electron density and the number of lone pairs and bonded ...
Wed Nov 13, 2019 12:32 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Single/Double Bonds in Resonance
Replies: 6
Views: 77

### Re: Single/Double Bonds in Resonance

Each bond for the single, double, and triple are essentially the same as looking at all single bonds. Thus, the resonance structures do not change what type of structure it is because each bond acts as a single region of electron density, which remains true for these resonance structures.
Wed Nov 13, 2019 12:17 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: Intermolecular forces
Replies: 10
Views: 188

### Re: Intermolecular forces

Van der Waals and London forces are used as the same term. For the definition of London forces, it is essentially the weakest type of intermolecular forces, and this is usually between two dipoles. They are also known as induced dipole- induced dipole interactions as well.
Wed Nov 13, 2019 12:12 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge vs Octet Rule
Replies: 12
Views: 226

### Re: Formal Charge vs Octet Rule

Yes, primarily you should start with the octet rule first in order to check the lewis structure that you drew. Only afterwards should you concern yourself with the formal charge in drawing a stable structure because you have already finished the octet rule.
Wed Nov 13, 2019 12:09 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: Ion-Ion
Replies: 2
Views: 36

### Re: Ion-Ion

Yes, both are essentially the same thing as each other. For ionic bonding or ion-ion bonding, it is defined as a chemical bond that shows the interactions between two opposite charged ions. This can be shown through the negative charged anion and a positive charged cation.
Tue Nov 12, 2019 11:48 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: 3F13
Replies: 6
Views: 224

### 3F13

For 3F13, I am confused on how to identify the actual arrangement of molecules for the strongest intermolecular attraction. How would you justify your answer and how would you know which one is the strongest? Thank you!
Mon Nov 04, 2019 2:22 pm
Forum: Photoelectric Effect
Topic: Photoelectric Effect Word Problems
Replies: 1
Views: 73

### Photoelectric Effect Word Problems

For the test, I am a bit confused on what the question is asking in regards to energy of photon, threshold energy, and kinetic energy. Do any of you have tricks on what the question is asking for in regards to electrons being emitted, etc? Sometimes, I get confused on whether it is asking for certai...
Mon Nov 04, 2019 2:18 pm
Forum: Lewis Structures
Topic: hw problem 2c.3
Replies: 1
Views: 43

### Re: hw problem 2c.3

Professor Lavelle did say that we needed to know certain common molecules such as H20 as well as nitrate of NO3- for the test. However, I do not think he is going to put difficult molecules and expect us to know what it is. If there were difficult molecules, he would probably put the formula and the...
Mon Nov 04, 2019 2:10 pm
Forum: Lewis Structures
Topic: Lewis Structures
Replies: 3
Views: 57

### Re: Lewis Structures

First, you have to count the number of valence electrons in the molecule. For example, for CO, it would have 10 total electrons. Then, since it's CO, it would have a shared single bond totaling to 8 e-. Because it needs to have the octet rule for each atom, this is not enough to satisfy the octet ru...
Mon Nov 04, 2019 12:15 pm
Forum: Lewis Structures
Topic: Tips and Tricks for Lewis structures
Replies: 2
Views: 34

### Re: Tips and Tricks for Lewis structures

For Lewis structures, you should be able to follow the octet rule first before moving around the electrons with the formal charge. Once all of the atoms for the Lewis structure follows the octet rule, then you should use the formal charge. You should memorize the equation so that you can quickly see...
Mon Nov 04, 2019 12:13 pm
Forum: Octet Exceptions
Replies: 8
Views: 113

### Re: Radicals: Homework Problem #2C1

For this question, it is evident that only B and C are radicals because they have one electron leftover that isn't paired. The answer key must be a mistake since NO2- does not have any radicals in it. It is important to note that when drawing lewis structures, you should be able to see an odd number...
Mon Oct 28, 2019 11:57 am
Forum: Lewis Structures
Topic: 2C.7
Replies: 2
Views: 40

### Re: 2C.7

To determine the number of electron pairs with both lone and shared, you should draw out the Lewis structure for each molecule. When you see electrons that aren't paired, you should take them in account for lone pairs, while electrons that are paired through bonding should be accounted for.
Mon Oct 28, 2019 11:51 am
Forum: Lewis Structures
Topic: lewis structures
Replies: 5
Views: 77

### Re: lewis structures

To determine what element goes into the center, it depends on the ionization energy, which is the energy needed to remove an electron from an atom (gas phase). The trend for ionization energy shows that it increases as you go right and up in the periodic table. Thus, to determine the center, you nee...
Mon Oct 28, 2019 11:43 am
Forum: Lewis Structures
Topic: 2C.1
Replies: 2
Views: 45

### Re: 2C.1

Radicals are defined as molecules where there is at least one unpaired electron. To determine whether or not a molecule is radical, you should draw out the Lewis structures in order to see whether or not there are lone pairs of electrons. In this question, both B and C are the radicals because they'...
Mon Oct 28, 2019 11:34 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Multi-electron atoms
Replies: 6
Views: 86

### Re: Multi-electron atoms

Nodal planes are where there are no electrons found. This ultimately means that there is no electron density distribution in these parts for a certain shape. There are s, p, d, and f atomic orbitals that exist, but s is the only atomic orbital where there's no nodal plane. However, p contains 1 noda...
Mon Oct 28, 2019 11:27 am
Forum: Resonance Structures
Topic: Resonance Clarification
Replies: 8
Views: 110

### Re: Resonance Clarification

Ultimately, resonance means that lewis structures have multiple bonds in different equivalent locations. It means that sometimes, different chemical compounds have more than one correct Lewis structure. The chemical level is the same, but electrons are distributed differently around the Lewis struct...
Mon Oct 21, 2019 4:46 pm
Forum: Trends in The Periodic Table
Topic: Electron Affinity
Replies: 4
Views: 58

### Re: Electron Affinity

For the trend in electron affinity, it shows that energy is released when e- are added to gas phase atoms. For example, X(g) + e- --> X-(g) would be defined as Electron affinity = E(X)(g) - E(X-)(g). This is different because it uses initial - final instead of final - initial like most of other chem...
Mon Oct 21, 2019 4:43 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron Configuration
Replies: 4
Views: 55

### Electron Configuration

For the lecture, I noticed that for certain elements, we would put x, y, z after each orbital/state. For example, there was 1s2 2s2 2px1 2py1 2pz1. I wanted to know what the reason is for putting these and if we have to put this format down for problems on the homework assignment. Thank you!
Mon Oct 21, 2019 4:37 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: orbitals
Replies: 5
Views: 78

### Re: orbitals

We do not need to know the orbital shapes / planes to draw them out. You would only need to know the descriptions of each rather than actual pictures planned out. For s, it contains a spherical shape, while p has two lobes on each side of the nucleus with a nodal plane. The d orbital has 3 with 4 lo...
Mon Oct 21, 2019 3:47 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Are electrons always removed from the 4s orbital before the 3d orbital?
Replies: 6
Views: 84

### Re: Are electrons always removed from the 4s orbital before the 3d orbital?

Yes, electrons are removed first from 4s rather than 3d orbital. This is due to the fact that 4s always have lower ionization energy, while 3d has higher ionization energy. Thus, the electrons are always being removed from the 4s orbital easily compared to 3d, which contains filled shells.
Mon Oct 21, 2019 2:31 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 1F.19
Replies: 3
Views: 55

### Re: 1F.19

S blocks are usually more reactive due to the fact that they have lower ionization energies, and because of this, they are able to give away electrons easily. These are more likely to make a reaction, and for noble gas electron configuration, elements only need to lose 1 or 2 electrons in their s-bl...
Tue Oct 15, 2019 12:46 pm
Forum: DeBroglie Equation
Topic: 1B.5
Replies: 3
Views: 76

### 1B.5

In this question, it ask. "The gamma-ray photons emitted by the nuclear decay of a techne- tium-99 atom used in radiopharmaceuticals have an energy of 140.511 keV. Calculate the wavelength of these gamma-rays." For 140.511 keV, how would this be converted back to Joules? Would we be expect...
Tue Oct 15, 2019 12:03 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: 1B.25
Replies: 4
Views: 79

### 1B.25

For this question, it asks, "What is the minimum uncertainty in the speed of an electron confined within a lead atom of diameter 350. pm? Model the atom as a one-dimensional box with a length equal to the diameter of the actual atom." I got the answer of 1.65 x 10^5 m*s^-1. However, I'm co...
Tue Oct 15, 2019 11:33 am
Forum: DeBroglie Equation
Topic: HW Question 1.B.15
Replies: 3
Views: 47

### Re: HW Question 1.B.15

For this question, you have the velocity, but the mass will be given as information, so you do not need to do anything else except plug in for mass. The mass of an electron is always 9.10939 x 10^-28 g, but since we need it to be converted to kg for this equation, you need to divide by 1000 to get 9...
Tue Oct 15, 2019 11:26 am
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: HW Question 1.B.27
Replies: 2
Views: 33

### Re: HW Question 1.B.27

In the lecture, we used Heisenberg's Indeterminacy Equation as (indeterminacy in momentum)(indeterminacy in position) >/= h/4pi. However, in the book, they use the equation with the use of the h bar, which is ℏ equaling to h/2pi as 1.05457 x 10^-34 J*s. They both are the same because in the book, 1/...
Tue Oct 15, 2019 11:15 am
Forum: Einstein Equation
Topic: 1B.7 Part c
Replies: 3
Views: 77

### Re: 1B.7 Part c

For part C, you need to use the answer that you got in part A in order to find the answer. This asks to find the energy emitted by 1.00 mol of sodium atoms at this wavelength, so it is simply asking for the conversion of the atom to mols. Thus, when using this conversion, you should use Avogadro's n...
Tue Oct 08, 2019 8:47 pm
Forum: Properties of Light
Topic: Problem 1B, #15
Replies: 1
Views: 42

### Problem 1B, #15

In part b of this question, it asks, "No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 3 1016 Hz. How much energy is required to remove the electron from the metal surface?" How would you able to solve the equation for the amount of e...
Tue Oct 08, 2019 8:34 pm
Forum: Significant Figures
Replies: 8
Views: 361

For this, make sure you check with the other parts of your equation as well. For example, the book uses 6.022 x 10^23 as Avogadro's constant, and many students may only use 6.02 x 10^23. Or, in another example, the book also uses more values for elements under the periodic table, whereas Professor L...
Tue Oct 08, 2019 8:25 pm
Forum: Photoelectric Effect
Topic: Post Assessment Problem 16
Replies: 3
Views: 79

### Re: Post Assessment Problem 16

Threshold energy is the answer. This is because this is the energy that has the ability to remove the electrons from the metal. In the photoelectric experiment, it is demonstrated by shining light on metal surfaces to measure the threshold energy from these metals. It can be explained through the eq...
Mon Oct 07, 2019 4:55 pm
Forum: Properties of Light
Topic: HW help
Replies: 2
Views: 75

### Re: HW help

In 1A 7, we know the equation c = λV for speed of light (3.00 x 10^8 m*s^-1) = wavelength times frequency. This must be changed to λ = c/V in order to find the wavelength, and this equals to 3.00 x 10^8 m*s^-1 / 7.1 x 10^14 s^-1. This would in turn equal to 4.2 x 10^-7 m or converted to nm as 420 nm...
Mon Oct 07, 2019 4:38 pm
Forum: Properties of Light
Topic: Problem 1A, #15
Replies: 3
Views: 52

### Problem 1A, #15

For this question in Section 1A, it asks, "In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line." So far, I calculated...
Sun Sep 29, 2019 7:20 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Accuracy vs Precision
Replies: 10
Views: 203

### Re: Accuracy vs Precision

In lecture, precision refers to how close to one another measurements are. For accuracy, it refers to the closeness to the true value. For example, a bullseye target is able to show the differences between accuracy and precision. If a bullseye target has dots close together in the center, this means...
Sun Sep 29, 2019 6:51 pm
Forum: Significant Figures
Topic: Rounding the elements
Replies: 12
Views: 204

### Re: Rounding the elements

For me, I use Lavelle's IUPAC Periodic Table of the Elements that contains very specific constants. For example, Sodium (Na) has 22.98976928 g. Then, at the end, I round off depending on the smallest amount of significant figures.
Sun Sep 29, 2019 5:44 pm
Forum: Empirical & Molecular Formulas
Topic: Question F.13
Replies: 3
Views: 74

### Re: Question F.13

For this question, it does not need percent mass composition because it already provides that there are 4.14 g of phosphorus and 27.8 g of the white solid compound as a product. So, in order to solve this question, you first subtract with 27.8 g - 4.14 g P = 23.66 g Cl to find the amount of g in Cl....
Sun Sep 29, 2019 5:32 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: How was a mole originally determined?
Replies: 3
Views: 74

### Re: How was a mole originally determined?

1 mole means the number of atoms in exactly 12 grams of carbon-12. However, in simplier terms, it refers to 6.022 x 10^23 "things". Moles are the standard for atoms, molecules, and specific particles. Avogadro refers to the Italian physicist that proposed this hypothesis in 1811. Additiona...