If you use -Ea/R then it is (1/T2 - 1/T1)
If you use Ea/R then it is (1/T1 - 1/T2)
Search found 102 matches
- Sat Mar 13, 2021 3:47 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Textbook Problem 7D.5
- Replies: 2
- Views: 304
- Sat Mar 13, 2021 3:39 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Textbook 4A.13
- Replies: 2
- Views: 286
Re: Textbook 4A.13
The change in temperature in Celcius is the same in Kelvin. For example (30C - 10C) = 20C, (303K - 283K) = 20K. The q of rxn is always equal to the negative q of calorimeter because we assume that all the heat released by the rxn is gained by the calorimeter and no heat enters/escapes the calorimeter.
- Sat Mar 13, 2021 3:28 pm
- Forum: Balancing Redox Reactions
- Topic: Textbook 6K3 (d)
- Replies: 2
- Views: 223
Re: Textbook 6K3 (d)
That's a typo. The correction is noted on the solution manual but this is the intended rxn: Cl2(g) → ClOH(aq) + Cl-(aq)
- Sat Mar 13, 2021 3:24 pm
- Forum: Student Social/Study Group
- Topic: Left/Right Electrode
- Replies: 9
- Views: 614
Re: Left/Right Electrode
For Ecell to be positive, Eright is the cathode/reduction half-rxn and Eleft is the anode/oxidation half-rxn. The right rxn will have a more positive/less negative standard reduction potential than the left rxn. That way, Eright/cathode - Eleft/anode results in a positive Ecell
- Sat Mar 13, 2021 3:20 pm
- Forum: First Order Reactions
- Topic: 7B.7
- Replies: 2
- Views: 288
Re: 7B.7
Since the half-life of a first-order rxn is independent of the initial concentration of A, we can determine how much time it takes by multiplying the amount of half-lives by the half-life. For example, a) [A] = (1/8)[A]o, [A]/[A]o = (1/2)^3 which means 3 half-lives have past t = 3(half-life) = 3(355s)
- Sun Mar 07, 2021 7:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Texbook 6M.5
- Replies: 2
- Views: 229
Re: Texbook 6M.5
I used the same table found in appendix 2 experimental data/standard potentials for all the problems
- Sun Mar 07, 2021 7:36 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 7A.17
- Replies: 4
- Views: 247
Re: 7A.17
If you compare rate 2 and 3, both [A] and [B] are changing, so you would only be able to figure out the order of B if you know the order of A. You can also compare rates 1 and 3 to solve for the order of B
- Sun Mar 07, 2021 7:30 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: textbook 6O.1
- Replies: 2
- Views: 177
Re: textbook 6O.1
Sorry I can't help you with this problem, but I don't think you need to worry about it because 6O.1 is not on the outline/syllabus anymore.
- Sun Mar 07, 2021 7:24 pm
- Forum: First Order Reactions
- Topic: Textbook problem 7B.1
- Replies: 2
- Views: 183
Re: Textbook problem 7B.1
[A]naught is the initial concentration of the drug given to the patient which is 20mg
- Sun Mar 07, 2021 7:21 pm
- Forum: First Order Reactions
- Topic: C is [A]nought
- Replies: 8
- Views: 350
Re: C is [A]nought
To get the integrated rate laws, we have to solve for C and we can do so by setting t=0. For example, in a 1st order rxn: ln[A]=-k(0)+C C=ln[A]naught which just means ln of the initial [A] or [A] when t=0. Same thing for a 2nd order rxn: 1/[A]=k(0)+C C=1/[A]naught which means 1/ the initial [A] or [...
- Sat Feb 27, 2021 11:05 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Sapling #9 Weeks 7/8
- Replies: 1
- Views: 188
Re: Sapling #9 Weeks 7/8
Since both solutions are 1M and the cell is at 25C, the cell is at standard conditions and we can use the values in the table. First, find the reduction half-reactions of Zn/Zn+ and Cu/Cu+. Then, compare the standard reduction potentials to determine with is the cathode and which is the anode. The c...
- Sat Feb 27, 2021 10:57 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: E naught Cell equation
- Replies: 9
- Views: 444
Re: E naught Cell equation
It is always cathode - anode when we directly use the standard reduction potentials.
If we flip one of the half-reactions, the charge of E changes. When we add two half-reactions to get the overall redox reaction, we can simply add the two Es to get E overall.
If we flip one of the half-reactions, the charge of E changes. When we add two half-reactions to get the overall redox reaction, we can simply add the two Es to get E overall.
- Sat Feb 27, 2021 10:53 am
- Forum: Balancing Redox Reactions
- Topic: Sapling #5
- Replies: 6
- Views: 315
Re: Sapling #5
H2O2 is hydrogen peroxide and peroxide (O2^2-) has an overall -2 charge, so the oxidation number of O is -1. The oxidation number of H is +1. In ClO2-, the oxidation of O is -2 and the oxidation number of Cl is +3. I hope this will help you balance out the charges and figure the correct number of el...
- Sat Feb 27, 2021 10:45 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: sapling week 7/8 #12
- Replies: 3
- Views: 205
Re: sapling week 7/8 #12
You can find the number of electrons by writing out the half-reactions and balancing them. You can also compare the oxidation numbers. For example, the oxidation number of Au3+ is (+3) and the oxidation number of Au is (0), which means each Au3+ gains 3 electrons to become Au. Since there are 2 mole...
- Sat Feb 27, 2021 10:40 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Sapling week 7/8 Question 17
- Replies: 4
- Views: 230
Re: Sapling week 7/8 Question 17
If your Q = 4.2/(0.81)^2 and you still have the wrong answer maybe round to two significant figures
- Sun Feb 21, 2021 6:28 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidation States for TM
- Replies: 6
- Views: 342
Re: Oxidation States for TM
Yes, the oxidation states of TM need to be calculated in relation to the other elements in the compound. I think the only exceptions are Ag+1 and Zn+2
- Sun Feb 21, 2021 6:23 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 4.15 Solutions Manual Clarification
- Replies: 1
- Views: 199
Re: 4.15 Solutions Manual Clarification
It's negative because they used the relationship qrxn = -qsol, so
-20000 = -(4.184)(800)deltaT.
You can also use -qrxn = qsol
-(-20000) = (4.184)(800)deltaT.
-20000 = -(4.184)(800)deltaT.
You can also use -qrxn = qsol
-(-20000) = (4.184)(800)deltaT.
- Sun Feb 21, 2021 6:18 pm
- Forum: Ideal Gases
- Topic: Calculating temperature change
- Replies: 4
- Views: 286
Re: Calculating temperature change
at constant P, (delta)U = qp + w
(delta)U = (3.4 atm.L)(101.325 J/atm.L)
w = (2.26 atm.L)(101.325 J/atm.L)
qp = (delta)U - w
qp = nCp(delta)T
(delta)U - w = (0.17 mol)(5/2 R)(delta)T
(delta)U = (3.4 atm.L)(101.325 J/atm.L)
w = (2.26 atm.L)(101.325 J/atm.L)
qp = (delta)U - w
qp = nCp(delta)T
(delta)U - w = (0.17 mol)(5/2 R)(delta)T
- Sun Feb 21, 2021 6:06 pm
- Forum: Balancing Redox Reactions
- Topic: oxidizing agent
- Replies: 3
- Views: 214
Re: oxidizing agent
The oxidizing agent promotes an oxidation rxn. In a redox reaction, the substance that undergoes reduction can be said to promote the oxidation half rxn/oxidize the other substance. The best oxidizing agent would have the greatest standard reduction potential.
- Sun Feb 21, 2021 5:56 pm
- Forum: Balancing Redox Reactions
- Topic: Adding an Inert Conductor
- Replies: 3
- Views: 198
Re: Adding an Inert Conductor
In the example provided in the lecture, an inert conductor is used when one of the half-reactions don't involve a solid (e.g. Fe2+ and Fe3+). The inert conductor is used as an electrode to transfer e-.
- Mon Feb 15, 2021 12:02 am
- Forum: Calculating Work of Expansion
- Topic: 4D.7 7th Edition
- Replies: 2
- Views: 231
Re: 4D.7 7th Edition
DeltaV is related to deltan
PdeltaV=deltanRT
In the balanced chemical equation the difference but the moles of gas of products and the moles of gas of reactants is -1.5 moles
PdeltaV=deltanRT
In the balanced chemical equation the difference but the moles of gas of products and the moles of gas of reactants is -1.5 moles
- Sun Feb 14, 2021 11:57 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Favorability of a Reaction and ∆U
- Replies: 4
- Views: 234
Re: Favorability of a Reaction and ∆U
I don’t think so. I think deltaG would tell more about the favorability of a rxn
- Sat Feb 13, 2021 9:24 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Textbook Problem 4D.11 part c
- Replies: 1
- Views: 135
Re: Textbook Problem 4D.11 part c
Given the balanced chemical equation N2+O2-->2NO and deltaH(rxn) = 180.6kJ, the oxidation of 1 mol of N2 requires 180.6kJ.
If 492J were absorbed, then mol of N2 = (0.492kJ)/(180.6kJ/mol)
then multiply the mols of N2 with the molar mass of N2 to get grams of N2.
If 492J were absorbed, then mol of N2 = (0.492kJ)/(180.6kJ/mol)
then multiply the mols of N2 with the molar mass of N2 to get grams of N2.
- Fri Feb 12, 2021 6:20 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Textbook 4.15
- Replies: 2
- Views: 123
Re: Textbook 4.15
In 4.15, we assume that the density and heat capacity of HCl are the same as those of water because water is the solvent and it is in excess. The density of water is 1g/ml, so 800ml of HCl is equal to 800g of HCl
- Fri Feb 12, 2021 6:16 pm
- Forum: Calculating Work of Expansion
- Topic: Textbook 4.7a
- Replies: 1
- Views: 160
Re: Textbook 4.7a
Yes, the change in moles of gas is related to the change in volume:
(delta)n = n(final, P) - n(initial, R)
w = -P(delta)V where P(delta)V = (delta)nRT
w = -P x ((delta)nRT/P)
w = -(delta)nRT
(delta)n = n(final, P) - n(initial, R)
w = -P(delta)V where P(delta)V = (delta)nRT
w = -P x ((delta)nRT/P)
w = -(delta)nRT
- Sun Feb 07, 2021 8:26 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Sapling #10 Week 3/4
- Replies: 6
- Views: 221
Re: Sapling #10 Week 3/4
The specific heat of ice is used to calculate the overall q from the initial temperature to the final temperature if the ice starts at a temperature below 0C, which means it is still ice. In this question, the ice starts at 0C, which means it is in the solid-liquid phase change portion of the heatin...
- Sun Feb 07, 2021 8:10 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Capacity of Calorimeter
- Replies: 7
- Views: 739
Re: Heat Capacity of Calorimeter
The heat capacity of a calorimeter is the amount of heat needed to raise the temperature by 1C. From the textbook and Sapling problems, heat capacity of a calorimeter can be calculated using Ccal = q/deltaT when the question states the heat input and change in temperature. It can also be calculated ...
- Thu Jan 28, 2021 8:27 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Textbook Problem 6B.11
- Replies: 3
- Views: 164
Re: Textbook Problem 6B.11
ai) given the pH of the diluted solution, you can find the pOH and [OH]. pOH = 14 - pH and [OH-] = 10^-(pOH)
aii) [OH-] = [NaOH] because NaOH is a strong base
using the [OH-] from (ai), you can calculate the initial concentration using M1V1=M2V2.
V1=5ml, M2=[OH-]from (ai), V2=500ml and solve for M1
aii) [OH-] = [NaOH] because NaOH is a strong base
using the [OH-] from (ai), you can calculate the initial concentration using M1V1=M2V2.
V1=5ml, M2=[OH-]from (ai), V2=500ml and solve for M1
- Thu Jan 28, 2021 8:17 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Textbook Problem 6.19
- Replies: 1
- Views: 182
Re: Textbook Problem 6.19
The reason why HHb increases is because there is an increase of H3O+ from the lactic acid. The H3O+ from the lactic acid reacts with HbO2- to produce more HHb. So, an increase in reactants will cause the rxn to shift right and produce more products, therefore HbO2- (reactant) will decrease.
- Thu Jan 28, 2021 8:13 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Textbook Problem 6B.3
- Replies: 3
- Views: 95
Re: Textbook Problem 6B.3
a) to solve for the desired pH, use the initial molarity of the acid. Since HCl is a strong acid, [HCl] = [H3O+], and pH = -log[H3O+] b) the newly prepared solution has a different concentration than the original solution, so you need to solve for the new molarity, where M1V1=M2V2. The new concentra...
- Thu Jan 28, 2021 8:05 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 4.31 Q (b)
- Replies: 1
- Views: 125
Re: 4.31 Q (b)
The standard enthalpy of formation is defined as the enthalpy of formation of a substance from its elements in their most stable form. In 4.31, H2(g) is the most stable form so the standard enthalpy of formation of H2 (element in its most stable form) is 0. H2(g) --> H2(g)
- Thu Jan 28, 2021 7:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook 5.35
- Replies: 1
- Views: 129
Re: Textbook 5.35
I found it helpful to fill in as much info into the ICE table: A <-> B + C I: 0.027 C: E: 0.017 0.005 0.01 Since the coefficient of the change is equal to the stoichiometric coefficient, we can fill out the Change. Assume x=0.005 (bc it's the smallest value), then C(A)= -2x, C(B)=+x, and C(C)=+2x. T...
- Wed Jan 27, 2021 8:18 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Concentration, Volume, and Pressure
- Replies: 3
- Views: 85
Re: Concentration, Volume, and Pressure
The concentration of a gas changes if the volume changes.
P=(n/V)RT
conc. = n/V
If the moles are constant and volume decreases, concentration increases. If volume increases, concentration decreases.
P=(n/V)RT
conc. = n/V
If the moles are constant and volume decreases, concentration increases. If volume increases, concentration decreases.
- Sun Jan 24, 2021 7:37 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Reversible Reaction
- Replies: 5
- Views: 253
Re: Reversible Reaction
I think they are written with a one-way arrow to show explicitly that the enthalpy applies to the forward rxn, and the reverse rxn would require manipulation of the enthalpy value.
- Sun Jan 24, 2021 7:33 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Sapling #4 Week 2-3
- Replies: 3
- Views: 245
Re: Sapling #4 Week 2-3
This is what I got:
6.3×10^−6 = x^2/(0.10-x) x<<0.10 bc Kb is small
6.3×10^−6 = x^2/0.10
x = sqrt(0.1 x 6.3×10^−6)
x = 7.94x10^-4
6.3×10^−6 = x^2/(0.10-x) x<<0.10 bc Kb is small
6.3×10^−6 = x^2/0.10
x = sqrt(0.1 x 6.3×10^−6)
x = 7.94x10^-4
- Sun Jan 24, 2021 7:25 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Week 2 #5
- Replies: 4
- Views: 172
Re: Sapling Week 2 #5
The initial concentration of B is equal to the sum of the equilibrium concentration of B and BH+ because of the law of conservation of matter. Any B that is protonated will become BH+.
- Sun Jan 24, 2021 7:19 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: approximating
- Replies: 10
- Views: 389
Re: approximating
We don't take out the x from the numerator because x does not equal to 0, therefore the x in the numerator is a significant value. We can omit the x because x is so small.
- Sat Jan 23, 2021 7:45 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Problem 5.35
- Replies: 2
- Views: 125
Re: Textbook Problem 5.35
I found it helpful to fill in as much info into the ICE table: A <-> B + C I: 0.027 C: E: 0.017 0.005 0.01 Since the coefficient of the change is equal to the stoichiometric coefficient, we can fill out the Change. Assume x=0.005 (bc it's the smallest value), then C(A)= -2x, C(B)=+x, and C(C)=+2x. T...
- Sat Jan 16, 2021 9:37 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling HW #2
- Replies: 6
- Views: 283
Re: Sapling HW #2
Yes, fill in the information for the ICE table. You have the initial concentration of SO3 and the equilibrium concentration of O2. Using the equilibrium concentration of O2, you can figure out the change and fill out the rest of the table and use the equilibrium values to calculate Kc.
- Sat Jan 16, 2021 9:32 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Textbook 6D.13 part b
- Replies: 2
- Views: 128
Re: Textbook 6D.13 part b
Since CH3NH3+ is the conjugate acid of a weak base (CH3NH2), it will affect the pH.
CH3NH3+ + H2O <-> CH3NH2 + H3O+
[CH3NH3+] = [CH3NH3Cl] = 0.20 because salts dissociate in water
the Kb of CH3NH2 is given in table 6C.2, but Ka is needed (Ka = Kw/Kb)
CH3NH3+ + H2O <-> CH3NH2 + H3O+
[CH3NH3+] = [CH3NH3Cl] = 0.20 because salts dissociate in water
the Kb of CH3NH2 is given in table 6C.2, but Ka is needed (Ka = Kw/Kb)
- Sat Jan 16, 2021 9:20 pm
- Forum: Calculating the pH of Salt Solutions
- Topic: Textbook Problem 6D.13
- Replies: 2
- Views: 135
Re: Textbook Problem 6D.13
the pH of d is larger than b. The value you calculated for d is the pOH, so you just need to convert it to pH.
- Sat Jan 16, 2021 9:17 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Question 6D.15 (b)
- Replies: 1
- Views: 114
Re: Textbook Question 6D.15 (b)
Al3+ can interact with water to form Al(H2O)6 3+. Table 6D.1 includes that example and the Ka value. Figure 6D.2 next to the table also explains why some metal ions are acidic.
- Sat Jan 16, 2021 9:12 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Question 6E.1
- Replies: 2
- Views: 119
Re: Textbook Question 6E.1
I think we have to calculate the [H3O+] from the second deprotonation because Ka2 is still relatively large and will affect the pH.
- Sun Jan 10, 2021 10:39 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chemical Equilibrium Part 2 Question 29
- Replies: 2
- Views: 156
Re: Chemical Equilibrium Part 2 Question 29
2BrCl(g) ⇌ Br2 (g) + Cl2(g)
1.84 x 10-4
-2x +x +x
3.37x10-5 x x
E: [BrCl] = 0.183(1.84 x 10-4) = 3.37x10-5
[Br2] = [Cl2] = x and 2x = 3.37x10-5
1.84 x 10-4
-2x +x +x
3.37x10-5 x x
E: [BrCl] = 0.183(1.84 x 10-4) = 3.37x10-5
[Br2] = [Cl2] = x and 2x = 3.37x10-5
- Sun Jan 10, 2021 10:33 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chemical Equilibrium Part 2 Question 27
- Replies: 3
- Views: 196
Re: Chemical Equilibrium Part 2 Question 27
C(s) + H2O (g) ⇌ CO (g) + H2 (g) 0.05 M 0 0 -x +x +x 0.05-x x 0.040 I: [H2O] = 2.5mol/50L = 0.05 M; pure solids (C) don't affect K and you can disregard the amount of C C: the change is x and all the coefficients are 1 E: the equilibrium concentration of H2 is given, so you can figure out x and equi...
- Thu Jan 07, 2021 5:25 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium Partial Pressure (Problem 5I.3)
- Replies: 1
- Views: 101
Re: Equilibrium Partial Pressure (Problem 5I.3)
Your setup is correct, but the Kc value you have identified from the table corresponds to 298K while the problem states that the temperature is 500K. That is probably why your answer doesn't match the solution manual
- Thu Jan 07, 2021 3:48 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5.35
- Replies: 2
- Views: 93
Re: 5.35
From the graph, the partial pressures at equilibrium are A=17kPa, B=5kPa, and C=10kPa. The change from B initial to B equilibrium is 5. Assume x = 5, then the change from C initial to C equilibrium is 2x and the change from A initial to A final is also 2x. After checking, x = 5 makes sense. The stoi...
- Wed Jan 06, 2021 8:56 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Difference between "sits" and "shifts"
- Replies: 5
- Views: 280
Re: Difference between "sits" and "shifts"
I think it would be more appropriate to use "sits" when the system is at equilibrium and you need to identify whether products (R) or reactants (L) are favored. This implies no change to the system and equilibrium doesn't change/move. "Shift" would be more appropriate when a chan...
- Wed Jan 06, 2021 8:49 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Stability of reactants vs products
- Replies: 5
- Views: 521
Re: Stability of reactants vs products
for reactants <-> products if K is large, then there are more products and the forward rxn is favored. That means reactants are less likely to stay as reactants and become whatever the products are, so the products would be more stable than the reactants. if K is small, then there are more reactants...
- Tue Jan 05, 2021 9:08 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Audio-Visual Module Question #15
- Replies: 3
- Views: 114
Re: Audio-Visual Module Question #15
a) water is added since H2O (reactant) is added, more products would be formed to use up the extra H2O. Another way of approaching this is by comparing the equilibrium constant and the reaction quotient. If there is an increase in H2O, then Q would be smaller than K. To adjust and reach equilibrium,...
- Sat Dec 12, 2020 8:48 pm
- Forum: Bronsted Acids & Bases
- Topic: Why is H2SO3 an acid when the oxygens have lone pairs?
- Replies: 2
- Views: 472
Re: Why is H2SO3 an acid when the oxygens have lone pairs?
SO3^2- has 2 oxygens with a formal charge of -1 and those are the sites where H+ from H2O will form HSO3- or H2SO3. That makes SO3^2- a base.
- Sat Dec 12, 2020 8:38 pm
- Forum: Naming
- Topic: Textbook Question 9C.1
- Replies: 5
- Views: 457
Re: Textbook Question 9C.1
OH2: aqua, (CN)5: pentacyano, Co: cobaltate because the complex has a negative charge
aquapentacyanocobaltate(III) ion
aquapentacyanocobaltate(III) ion
- Sat Dec 12, 2020 8:34 pm
- Forum: Conjugate Acids & Bases
- Topic: Sapling 14
- Replies: 6
- Views: 350
Re: Sapling 14
Yes, it will be neutral. This is how I like to think of it. When pH>pKa, that means the surrounding is more basic and there are less H+ and more OH- present. Since there are more OH- present, the equilibrium will shift left and produce more B, which is neutral.
B + H2O <--> BH+ + OH-
B + H2O <--> BH+ + OH-
- Sat Dec 12, 2020 8:25 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Prioritizing Polarizing Power of Polarizabilty
- Replies: 3
- Views: 247
Re: Prioritizing Polarizing Power of Polarizabilty
It may be easier to approach this problem by comparing electronegativities where the smaller the difference the greater the covalent character. If the difference of electronegativities is high, that means that electrons are more likely to be attracted to the more electronegative atom and display cha...
- Sat Dec 12, 2020 8:14 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: HBrO2 vs HClO2
- Replies: 7
- Views: 992
Re: HBrO2 vs HClO2
In HBrO2 and HClO2, the H is bonded to the O, so the strength of the acid depends on the stability of the anions (BrO2- and ClO2-). ClO2- is more stable because Cl is more electronegative than Br which means it can stabilize the negative charge on the O by withdrawing e-
- Sun Dec 06, 2020 7:44 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Textbook 2E 5
- Replies: 2
- Views: 85
Re: Textbook 2E 5
a) what is the shape of a ClO2 + ion
b) what is the expected OClO bond angle
b) what is the expected OClO bond angle
- Sun Dec 06, 2020 7:25 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Figuring Out monodentate, bidentate, etc
- Replies: 5
- Views: 232
Re: Figuring Out monodentate, bidentate, etc
A ligand is a species bound to the metal ion which means it can be an atom or a molecule. If a ligand is a molecule that has multiple sites with lone pairs and single bonds that allow it to rotate in a way where multiple sites can bind to the metal ion simultaneously, then it is polydentate.
- Sun Dec 06, 2020 7:18 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Chelates
- Replies: 3
- Views: 152
Re: Chelates
A chelate is formed when the ligand is polydentate.
- Sun Dec 06, 2020 7:17 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Sapling #6
- Replies: 3
- Views: 328
Re: Sapling #6
The coordination sphere has an overall charge of +1 since the oxidation state of Cl (outside) is -1 and the entire unit is neutral. The oxidation state of NH3 is 0 and the oxidation number of Cl (inside) is -1 and there is only one Cl within the coordination sphere so the oxidation state of Co must ...
- Fri Dec 04, 2020 5:25 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Calculations involving strong acids
- Replies: 3
- Views: 200
Re: Calculations involving strong acids
The 100% ionization of HCl means that all 0.1 M of HCl will become H+ and Cl- HCl(aq) --> H+ (aq) + Cl-(aq) or HCl(aq) + H2O(l) --> H3O+(aq) + Cl-(aq) 0.1M of HCl yields 0.1M of H+ and 0.1M of Cl- because of 1:1 between HCl and H+/Cl- In this scenario, [H+]=0.1M so calculating the pH would be simple...
- Sun Nov 29, 2020 2:30 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Sapling #3
- Replies: 5
- Views: 299
Re: Sapling #3
In an octahedral arrangement, it doesn't matter which position (equatorial or axial) is made the lone pair. If the equatorial position is made the lone pair, there will be four 90 angles. Two between the two adjacent equatorial atoms; two between the two axial atoms. If the axial position is made th...
- Wed Nov 25, 2020 1:19 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Chelate Lecture Example
- Replies: 1
- Views: 102
Re: Chelate Lecture Example
Yes, the ligand that forms a chelate is considered one ligand. It just has multiple bonding sites. Yes, NH2-CH2-CH2-NH2 is bidentate because there are two sites bonded to Co (i.e. the two N from NH2)
- Wed Nov 25, 2020 1:14 pm
- Forum: Hybridization
- Topic: Chelate example from the lecture
- Replies: 1
- Views: 38
Re: Chelate example from the lecture
The ligand is NH2-CH2-CH2-NH2 and the Ns from NH2 are bonded to Co. That particular ligand is bidentate that has two regions bonded to Co. A ring is formed between Co, NH2, CH2, CH2, and NH2. The Cl is a different ligand that's monodentate because it can only donate 1 e- pair at a time.
- Wed Nov 25, 2020 1:05 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: EDTA example in 11/25 lecture
- Replies: 5
- Views: 266
Re: EDTA example in 11/25 lecture
From the structure shown in lecture, EDTA4- has six atoms with single bonds and lone pairs, which means when it rotates into a certain position, EDTA4- can bond with the metal ion at those six sites
- Wed Nov 25, 2020 1:00 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: How to tell whether a ligand is polydentate
- Replies: 2
- Views: 156
Re: How to tell whether a ligand is polydentate
From the lecture, I think the ligand needs to have single bonds so it can rotate and have more than one site with lone pairs to form a chelate
- Sun Nov 22, 2020 8:40 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: sapling #4
- Replies: 3
- Views: 159
Re: sapling #4
From the textbook (2E), square pyramidal has 5 bonding pairs and 1 lone pair.
- Sun Nov 22, 2020 8:37 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: polarity from lewis structure
- Replies: 7
- Views: 273
Re: polarity from lewis structure
The shape of CO2 is linear, so it is nonpolar because the dipoles cancel.
- Fri Nov 20, 2020 8:32 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Textbook 2E.1b
- Replies: 2
- Views: 59
Textbook 2E.1b
Below are ball-and-stick models of two molecules. In each case, indicate whether there must be, may be, or cannot be one or more lone pairs of electrons on the central atom. b) O--O--O and the angle is 180 The answer from the book is "may have lone pairs" but I thought there couldn't be lo...
- Tue Nov 17, 2020 10:14 pm
- Forum: Lewis Structures
- Topic: midterm question re: sapling #3
- Replies: 3
- Views: 223
Re: midterm question re: sapling #3
My TA said we don't need to memorize the formulas. If the names are given, the formulas would also be provided.
- Tue Nov 17, 2020 10:08 pm
- Forum: Dipole Moments
- Topic: Textbook Question 3F.13
- Replies: 1
- Views: 89
Re: Textbook Question 3F.13
For II, the dipole moments are aligned in a way where the slightly positive end of the molecule is closest to the slightly negative end of the molecule +-> +-> +-> For III, the dipole moments are pointing up and down while the molecules are adjacent to each other so the oppositely charged ends of th...
- Sun Nov 15, 2020 6:13 pm
- Forum: Lewis Structures
- Topic: Number of Bonds for Elements
- Replies: 3
- Views: 88
Re: Number of Bonds for Elements
For atoms in period 3 and beyond, electrons can be in the d state because n=3 and l=0, 1, 2. This allows S or Cl to have more than 4 bonds and break the octet rule. For nitrogen, it follows the octet rule and can have up to 4 bonds, but it often forms 3 bonds and has 1 lone pair because its formal c...
- Sun Nov 15, 2020 5:54 pm
- Forum: Ionic & Covalent Bonds
- Topic: Sapling Q: Hydrogen Bonds
- Replies: 7
- Views: 349
Re: Sapling Q: Hydrogen Bonds
Regarding the second requirement that you mentioned, the "highly electronegative atom" that is bonded to a hydrogen atom can only be N, O, or F. That means a hydrogen bond can only form between a hydrogen atom that is bonded to N, O, or F and an electronegative atom (N, O, or F) with a lon...
- Thu Nov 12, 2020 6:16 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: textbook problem 2A #9,11
- Replies: 2
- Views: 65
Re: textbook problem 2A #9,11
I'll use 9A as an example I believe you added 2 electrons to 3d, which led you to believe that there were 9 electrons in the 3d state. For the transition metals, 4s needs to be filled before 3d is filled (except for Cr and Cu). So, the 2 electrons should have been removed from the 4s state and not 3...
- Thu Nov 12, 2020 6:02 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: e- p
- Replies: 1
- Views: 26
Re: e- p
the "p" is rho (Greek) and it means density
- Thu Nov 12, 2020 1:01 pm
- Forum: Trends in The Periodic Table
- Topic: Ionization energy of Copper?
- Replies: 2
- Views: 516
Re: Ionization energy of Copper?
The e- configuration of Cu is [Ar]3d^104s^1, so the 4s electron would be removed if it was ionized. The electron in the 4s subshell is further from the nucleus, so the effective nuclear charge is lower due to the "shielding" of electrons in lower energy levels. It would take less energy to...
- Sun Nov 08, 2020 12:32 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Textbook Exercise 2A.3
- Replies: 2
- Views: 152
Re: Textbook Exercise 2A.3
You wrote it in the correct order. 3d comes before 4s because n=3 is a lower energy state than n=4
You can also check your answers with the answer key on Sapling
You can also check your answers with the answer key on Sapling
- Sun Nov 08, 2020 12:11 pm
- Forum: Properties of Light
- Topic: Equation clarifications
- Replies: 4
- Views: 274
Re: Equation clarifications
If you are solving the energy of radiation (no mass), E=hc/λ
If you are solving the energy of e- (has mass) and is given the de Broglie wavelength, Ek = 0.5mv^2 and v = h/mλ where v is velocity
If you are solving the energy of e- (has mass) and is given the de Broglie wavelength, Ek = 0.5mv^2 and v = h/mλ where v is velocity
- Sun Nov 08, 2020 11:49 am
- Forum: Lewis Structures
- Topic: Lone Pairs Question
- Replies: 22
- Views: 1627
Re: Lone Pairs Question
Lone pairs are pairs of valence electrons that are not shared or not bonded. In Lewis structures, lone pairs are represented as two dots (i.e. 2 non-bonding e-)
- Fri Nov 06, 2020 2:25 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polarizability and Polarizing Power trends
- Replies: 2
- Views: 77
Re: Polarizability and Polarizing Power trends
It is true that the number of electrons increase across the period, but the number of protons also increases, which results in a higher nuclear charge that would pull the electrons in. So, the atomic radius decreases across the period.
- Thu Nov 05, 2020 6:54 pm
- Forum: Lewis Structures
- Topic: Formal Charge
- Replies: 5
- Views: 140
Re: Formal Charge
The structure is more stable if the formal charge is 0. The sum of the formal charge must always add up to the net charge, which helps you check your work.
- Sun Nov 01, 2020 6:42 pm
- Forum: Student Social/Study Group
- Topic: Sappling #25
- Replies: 4
- Views: 229
Re: Sappling #25
E(photon): size of bacterium = λ of photon = 2.3 μm To convert μm to m, (2.3 μm)(10^-6) E(photon) = hc/λ E(e-): it's the kinetic energy de Broglie λ = 2.3 μm; Ek = 0.5mv^2 convert μm to m we need the velocity of e- to solve for the kinetic energy of the electron λ=h/mv --> v = h/mλ solve of Ek with ...
- Sun Nov 01, 2020 6:33 pm
- Forum: Student Social/Study Group
- Topic: Sampling HW #7
- Replies: 2
- Views: 145
Re: Sampling HW #7
First, you can determine the energy of the radiation with λ=751nm using E = hc/λ. Remember the energy calculated is joules per photon. Then decide how much energy is needed to melt 417g of ice by multiplying the 417 by the enthalpy of fusion (found in the table). To determine how many photons are ne...
- Sun Nov 01, 2020 10:41 am
- Forum: Properties of Light
- Topic: Sapling Hw #5
- Replies: 5
- Views: 567
Re: Sapling Hw #5
With the wavelength, you can calculate the energy in joules/photon. When you multiplied the mass of ice with the enthalpy of fusion, you get the total energy in joules to melt the ice. In order to find out how many photons are needed to melt the ice, divide (333.6 x 423) by the energy of the radiati...
- Sat Oct 31, 2020 7:33 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Sapling Homework
- Replies: 8
- Views: 285
Re: Sapling Homework
Also, the subshells (s, p, d) need to be italicized
- Wed Oct 28, 2020 7:45 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Dilution module
- Replies: 6
- Views: 278
Re: Dilution module
Not necessarily. When you solve for the volume, you would get liters, but the options are in milliliters, so you need to convert your answer in L to mL.
- Thu Oct 22, 2020 8:23 pm
- Forum: Einstein Equation
- Topic: Sapling Homework - electron affinity
- Replies: 5
- Views: 292
Re: Sapling Homework - electron affinity
To convert 1.003eV/atom to kJ/mol, I used dimensional analysis
multiply (1.003eV/atom) by (6.022x10^23 atoms/mol) --> get eV/mol
then multiply by (1.602x10^-19 J/1 eV) --> get J/mol
then multiply by (1 kJ/1000 J) --> get kJ/mol
multiply (1.003eV/atom) by (6.022x10^23 atoms/mol) --> get eV/mol
then multiply by (1.602x10^-19 J/1 eV) --> get J/mol
then multiply by (1 kJ/1000 J) --> get kJ/mol
- Thu Oct 22, 2020 8:13 pm
- Forum: Photoelectric Effect
- Topic: Sapling HW Weeks 2, 3, 4 Question #9
- Replies: 10
- Views: 910
Re: Sapling HW Weeks 2, 3, 4 Question #9
The maximum wavelength of the radiation that will eject electrons occurs when the energy of the incident light equals the work function. When you are given the work function, you can calculate the wavelength of the radiation.
work function = E(photon) = hc/(wavelength)
work function = E(photon) = hc/(wavelength)
- Thu Oct 22, 2020 8:07 pm
- Forum: DeBroglie Equation
- Topic: Sapling Homework - electron affinity
- Replies: 9
- Views: 421
Re: Sapling Homework - electron affinity
Yes, the length of the bacterium will be the wavelengths of the photon and electron. To find the energy of the photon, use E = h(frequency) and c = λ(frequency) and solve for E The energy of the electron is its kinetic energy, which involves mass and velocity. To find the velocity of e-, use de Brog...
- Thu Oct 22, 2020 8:00 pm
- Forum: SI Units, Unit Conversions
- Topic: SI Conversions
- Replies: 11
- Views: 342
Re: SI Conversions
I don't know if we will be given easy units that don't have to be converted, but it is best to know how to convert units since we already have to do that in Sapling and textbook problems
- Thu Oct 22, 2020 7:57 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Indeterminacy Problem
- Replies: 3
- Views: 153
Re: Indeterminacy Problem
use ∆p = m∆v, where m=8.00 kg and ∆v=10 m/s
then use ∆p∆x ≥ h/4π to find ∆x
btw, the answer in the back of the textbook is incorrect because they used ∆v=5 m/s, so that may have caused you some trouble
then use ∆p∆x ≥ h/4π to find ∆x
btw, the answer in the back of the textbook is incorrect because they used ∆v=5 m/s, so that may have caused you some trouble
- Sat Oct 17, 2020 7:28 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Photon Frequency and Energy Between Ground States
- Replies: 3
- Views: 99
Re: Photon Frequency and Energy Between Ground States
The electron will not absorb photons that doesn't match the energy differences between energy levels. The specificity results in unique spectral lines that can be used to identify the atom (or molecule).
- Sat Oct 17, 2020 7:24 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Midterm #1
- Replies: 3
- Views: 127
Re: Midterm #1
If you are referring to the Heisenberg uncertainty principle, it is part of 1B
- Sat Oct 17, 2020 6:59 pm
- Forum: Properties of Light
- Topic: Textbook Problems (Topic 1B)
- Replies: 1
- Views: 107
Re: Textbook Problems (Topic 1B)
This is how I solved the problem:
Convert 140.511keV to joules
Used E = hv, v=c/(wavelength) -> E = hc/(wavelength) to solve for wavelength
convert m to pm using 1m = 1x10^12 pm
Convert 140.511keV to joules
Used E = hv, v=c/(wavelength) -> E = hc/(wavelength) to solve for wavelength
convert m to pm using 1m = 1x10^12 pm
- Fri Oct 16, 2020 1:57 pm
- Forum: Properties of Electrons
- Topic: Using empirical equation for h-atom
- Replies: 4
- Views: 114
Re: Using empirical equation for h-atom
The -1/16 comes from -1/n^2 of En = -hR/n^2, where n=4
- Fri Oct 16, 2020 1:52 pm
- Forum: Properties of Light
- Topic: rydberg equation
- Replies: 7
- Views: 316
Re: rydberg equation
Hi
It's not necessary to know the Rydberg equation, but you do need to know how to solve difference in energy using the professors way
It's not necessary to know the Rydberg equation, but you do need to know how to solve difference in energy using the professors way
- Sun Oct 11, 2020 7:18 pm
- Forum: Limiting Reactant Calculations
- Topic: Help finding limiting reactant
- Replies: 4
- Views: 316
Re: Help finding limiting reactant
One way of finding the limiting reactant is to determine how much product each reactant yields. In this scenario, you also need to find the theoretical yield of H2O, so you can calculate how much H2O is produced from 0.0625 moles of O2, and how much H2O is produced from 0.006795 moles of aspartame. ...
- Thu Oct 08, 2020 5:50 pm
- Forum: DeBroglie Equation
- Topic: Post-module assessment question #35
- Replies: 3
- Views: 129
Re: Post-module assessment question #35
The wavelength of the electric car is so small that it is unnoticeable and insignificant so you can say large objects like an electric car don't have wavelike properties.
- Thu Oct 08, 2020 5:41 pm
- Forum: SI Units, Unit Conversions
- Topic: Review HW E23
- Replies: 2
- Views: 139
Re: Review HW E23
Your process is not wrong. I tried your method and I got the correct answer, so you may have messed up the calculations somewhere which gave you a wrong answer.
mass % of Cu2+ = 28.45%
3.00(0.2845) = 0.8535g Cu2+
0.8535g Cu2+ /(63.546g/mol Cu) = 0.0134 mol Cu2+
mass % of Cu2+ = 28.45%
3.00(0.2845) = 0.8535g Cu2+
0.8535g Cu2+ /(63.546g/mol Cu) = 0.0134 mol Cu2+
- Thu Oct 08, 2020 10:37 am
- Forum: Empirical & Molecular Formulas
- Topic: Finding the Empirical and Molecular Formulas
- Replies: 3
- Views: 500
Re: Finding the Empirical and Molecular Formulas
From the mass of CO2, you can determine how many moles and grams of C were present in the caproic acid. From the mass of H20, you can determine how many moles and grams of H were present in the caproic acid. The remaining mass of the caproic acid sample would be the grams of O present in the caproic...
- Thu Oct 08, 2020 10:23 am
- Forum: Limiting Reactant Calculations
- Topic: Sapling HW Q10
- Replies: 2
- Views: 148
Re: Sapling HW Q10
Hello, You do have to write out the formulas so you calculate for the molar masses, but it isn't necessary to figure out the limiting reactant because it states in the prompt that 2-butanone is the limiting reactant and the other reactants are in excess. When you convert milliliters of 2-butanone to...