## Search found 123 matches

Mon Mar 08, 2021 6:05 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Inert Conductor
Replies: 2
Views: 12

### Re: Inert Conductor

Hi, you would only need to use an inert electrode when you don't have a metal involved in the anode or cathode. I'm not sure what other cases you are referring to. Could you give an example?
Mon Mar 08, 2021 6:01 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook 6O.3
Replies: 1
Views: 8

### Re: Textbook 6O.3

a.png (32.86 KiB) Viewed 6 times
Mon Mar 08, 2021 5:59 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook 6L.5 Part D
Replies: 1
Views: 27

### Re: Textbook 6L.5 Part D

For this kind of question, you need to find 2 half reactions from the half reactions table that when manipulated and combined will result in the reaction asked. In this case, you need to find 2 reactions that contain Au, Au+, and Au3+. Looking at the table, you find only one set of possible reaction...
Mon Mar 08, 2021 5:54 pm
Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
Topic: Textbook 6.45
Replies: 1
Views: 27

### Re: Textbook 6.45

In general, the more negative the standard reduction potential, the more strongly reducing it is. So substances lower in the standard reduction potential table are stronger reducing agents.
Mon Mar 08, 2021 5:50 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: units of k
Replies: 7
Views: 48

### Re: units of k

Choose unit for k such that the unit for rate is $M\cdot s^{-1}$
Therefore, If the overall order is n then the unit of k is $M^{1-n}\cdot s^{-1}$
Mon Mar 01, 2021 12:34 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook Problem 6M.7
Replies: 1
Views: 14

### Re: Textbook Problem 6M.7

In general, the more negative the standard reduction potential, the more strongly reducing it is. So substances lower in the standard reduction potential table are a stronger as reducing agents.
Mon Mar 01, 2021 12:29 pm
Forum: Balancing Redox Reactions
Topic: Textbook Problem 6K.1
Replies: 2
Views: 16

### Re: Textbook Problem 6K.1

Your two half reactions would be 14H^{+}(aq)+Cr_{2}O_{7} ^{2-}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq) + 7H_2O(l) C_2H_5OH(aq)\rightarrow C_2H_4O(aq)+2H^+ (aq)+ 2e^- (times 3 to balance the e - ) then adding them together you get 8H^{+}(aq)+...
Mon Mar 01, 2021 12:17 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook Question 6L.5 part d
Replies: 1
Views: 26

### Re: Textbook Question 6L.5 part d

For this type of question, you would have to refer to the table for half reactions and try to find two reactions that when combined will result in the reaction given in the question. In this case, you would need reactions that have Au, Au + , Au 3+ which will be Au + + e - -> Au Au 3+ + 3e - -> Au Y...
Mon Mar 01, 2021 12:11 pm
Forum: Balancing Redox Reactions
Topic: 6L.1
Replies: 3
Views: 30

### Re: 6L.1

Find the two half-reactions. For example, in letter a:
2Ce4+(aq) + 2e- → 2Ce3+(aq)
3I(aq) → I3-(aq) + 2e-
so n=2
Mon Mar 01, 2021 12:06 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6N.7 part b
Replies: 1
Views: 22

### Re: 6N.7 part b

Hi, I'm not sure but maybe it's because we reduced the stoichiometric coefficients of the original reaction. 2H^+(aq,10^{-3}M)\rightarrow 2H^+(aq,10^{-4}M) Q=\frac{[H^+(10^{-3}M)]^2}{[H^+(10^{-4}M)]^2} E=-\frac{RT}{2F}\ln\frac{(10^{-3})^2}{(10^{-4})^2}...
Mon Mar 01, 2021 11:55 am
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Textbook 6N.5A
Replies: 1
Views: 17

### Re: Textbook 6N.5A

You would use the partial pressure for H2 instead of its molar concentration
$0.33=0.27-\frac{0.025693}{2}\ln\frac{[H^+]^2[Cl^-]^2}{P_{H_2}}$
$\frac{0.33-0.27}{-\frac{0.025693}{2}}=\ln\frac{[H^+]^2(1)^2}{1^2}$
$[H^+]=\sqrt{e^{-4.67}}$
$pH=-\log \sqrt{e^{-4.67}} = 1.0$
Wed Feb 24, 2021 8:40 am
Forum: Balancing Redox Reactions
Topic: Sapling week 7/8 #8
Replies: 2
Views: 22

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

ox #
Cl2: 0
Cu: 0
Cl in CuCl: -1
Cu in CuCl: +1

Cl2 -> 2Cl- (0 -> -1) reduction
Cl2 + 2e- -> 2Cl-
2Cu -> 2Cu+ (0 -> +1) oxidation
2Cu -> 2Cu+ + 2e-
Wed Feb 24, 2021 8:02 am
Forum: Balancing Redox Reactions
Topic: Sapling HW Week 7 & 8, Balancing the Formation of Iron(III)Oxide Trihydrate
Replies: 3
Views: 14

### Re: Sapling HW Week 7 & 8, Balancing the Formation of Iron(III)Oxide Trihydrate

I don't think you need the parenthesis.
Tue Feb 23, 2021 2:03 pm
Forum: Balancing Redox Reactions
Topic: Basic and Acidic Solutions
Replies: 2
Views: 25

### Re: Basic and Acidic Solutions

Hi there's another method which is basically the same but I think really reduces the amount you need to write down since you don't have to balance the half reactions separately. Using example 6K.1 as comparison: MnO_4^-+H_2C_2O_4->Mn^{2+}+CO_2 1. Balance atoms other than O and H MnO_4^-+H_2C_2O_4->M...
Tue Feb 23, 2021 1:47 pm
Forum: Balancing Redox Reactions
Topic: Adding H+, OH-, and H2O in Redox Reactions
Replies: 7
Views: 71

### Re: Adding H+, OH-, and H2O in Redox Reactions

Hi, I don't know if this method applies to all cases but I think it really shortens the amount of things you need to write when solving. Using the example from above: Cr_2O_7^{2-}+HNO_2->Cr^{3+}+NO_3^- 1. balance atoms except H and O Cr_2O_7^{2-}+HNO_2->2Cr^{3+}+NO_3^- 2. balance change in oxidation...
Tue Feb 23, 2021 1:26 pm
Forum: Balancing Redox Reactions
Topic: Sapling Week 7/8 #5
Replies: 2
Views: 43

### Re: Sapling Week 7/8 #5

Remember that you could always check your answer by comparing the number of atoms and the sum of charges in both sides of the equation and making sure that they match.
Tue Feb 23, 2021 1:18 pm
Forum: Balancing Redox Reactions
Topic: Textbook 6L.5 (d)
Replies: 1
Views: 12

### Re: Textbook 6L.5 (d)

By looking at the half reactions table: Au + + e - -> Au 1.69 Au 3+ +3e - ->Au 1.40 Since 1.4 < 1.69, the second reaction will undergo in the reverse direction becoming an oxidation reaction. And after balancing the number of e- 3Au + + 3e - -> 3Au Au->Au 3+ +3e - Summing the two half reactions 3Au ...
Tue Feb 16, 2021 9:07 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy density
Replies: 2
Views: 38

### Re: Enthalpy density

I think the sign is just ignored when calculating enthalpy density since it is just used to mention the energy per unit volume.
The table in p. 278 also uses positive values.
Tue Feb 16, 2021 8:48 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: textbook problem 4c. 15
Replies: 2
Views: 24

### Re: textbook problem 4c. 15

Solid and Gas phase: 30J/mol Liquid phase: 60J/mol These values indicate that more energy is required for the liquid phase than solid/gas phase to increase them by the same amount of temperature. For the same amount of energy input, the temperature increase will be higher for the solid phase than th...
Tue Feb 16, 2021 8:25 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Textbook 4C.5b
Replies: 2
Views: 22

### Re: Textbook 4C.5b

C2H6 is nonlinear molecule so the contributions would be
translational: 3/2 R
rotational(nonlinear): 3/2 R
total: 3R
You can find an explanation to how these numbers are found in section 4B.3
Tue Feb 16, 2021 8:18 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Change in enthalpy for constant volume system
Replies: 1
Views: 25

### Re: Change in enthalpy for constant volume system

You can use this formula
$\Delta H=\Delta U + nR\Delta T$
At this point you have values for $\Delta T$ and $\Delta U=q+w=q+0=q=765J$
Tue Feb 16, 2021 8:09 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Textbook Problem 4.15
Replies: 5
Views: 49

### Re: Textbook Problem 4.15

You could start by writing out the balanced chemical equation representing the reaction. The question then asks for the final temperature which hints that you need to find the heat transferred during the reaction (q). In constant pressure, q=delta H and you can find delta H of reaction using the cha...
Mon Feb 08, 2021 10:53 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Relationship between constant volume heat capacity and constant pressure heat capacity.
Replies: 1
Views: 28

### Re: Relationship between constant volume heat capacity and constant pressure heat capacity.

Heat capacity is $C$ with units $J\cdot K^{-1}$
Molar heat capacity is $C_{m}$ with units $J\cdot K^{-1}\cdot mol^{-1}$
$C_m=C/n$
The V or P indicates constant volume or constant pressure

$C_P=C_V+nR$
divide everything by n and you get
$C_{P,m}=C_{V,m}+R$
Mon Feb 08, 2021 10:48 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: lg (log) in delta S (change in entropy)
Replies: 1
Views: 22

### Re: lg (log) in delta S (change in entropy)

in an isothermal process \Delta S=\frac{q}{T} Also, \Delta U=q+w=0 so q=-w applying this to the change in entropy equation and assuming the process is reversible, \Delta S=\frac{q}{T}=\frac{-w}{T}=\frac{-(-nRT\ln(V_2/V_1))}{T}=nR\ln(V_2/V_1) But since entropy is a state funct...
Mon Feb 08, 2021 10:43 am
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Textbook Problem 4F.5
Replies: 1
Views: 13

### Re: Textbook Problem 4F.5

isothermal expansion means constant T, change in volume
$\Delta S=nR\ln(V_2/V_1)$
Mon Feb 08, 2021 10:36 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: week 5/6 sapling #8
Replies: 1
Views: 19

### Re: week 5/6 sapling #8

You would use the formula for $\Delta S$ from T1 -> T2
$\Delta S=nC\ln(T_2/T_1)$
$n=\frac{PV}{RT}$
Mon Feb 08, 2021 10:18 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: How is ∆U affected by V and P?
Replies: 4
Views: 50

### Re: How is ∆U affected by V and P?

\Delta U = q+w We are only considering expansion work for now. At constant volume, w=0 since no expansion work occurs. \Delta U = q + 0 = q At constant pressure, \Delta H = q , \Delta U=\Delta H-P\Delta V In an adiabatic process, there is no heat transfer (q=0). Therefore, \Delta U=w See exercise 4...
Mon Feb 08, 2021 10:05 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Assuming 298 K
Replies: 7
Views: 26

### Re: Assuming 298 K

"All reaction enthalpies used in this text are for 298.15 K unless another temperature is indicated." ^ this can be found in the first paragraph in page 277 of the textbook So since you would need to use \Delta H and the temperature is not given, you can just assume that T=298.15 K. I'm no...
Mon Feb 08, 2021 9:57 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Textbook 4D.3
Replies: 2
Views: 22

### Re: Textbook 4D.3

Reaction of 1.40g CO causes the temperature of the calorimeter to rise from 22.113 C to 22.799 C. Therefore the heat absorbed by the calorimeter from the reaction of 1.40g CO: q=C\Delta T=(3.00\ kJ/C)(0.686\ C)=2.058\ kJ q of reaction is -2.058 kJ per 1.40g CO You are asked to find \...
Mon Feb 01, 2021 5:18 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Sapling Week 3/4 #19
Replies: 1
Views: 35

### Re: Sapling Week 3/4 #19

Bomb calorimeter also means constant volume. This is from an example problem in page 253 of the textbook. Screen Shot 2021-02-02 at 09.12.02.png In your case, the calorimeter was calibrated using 0.700L of solution which is the same volume as the one in the experiment (0.350L + 0.350L = 0.700L). Als...
Mon Feb 01, 2021 4:42 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Positive or negative work
Replies: 3
Views: 27

### Re: Positive or negative work

\Delta U = q+w This formula does not change. What changes is whether w is positive or negative. For example if 50J of work is done on the system, w = 50J and you would plug 50J into w in the formula. If 50J of work is done by the system, w = -50J but the formula is still \Delta U = q+w . You would ...
Mon Feb 01, 2021 4:34 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Textbook 4B.3b
Replies: 2
Views: 18

### Re: Textbook 4B.3b

I think the answer key at the back of the book has typos.
^This is from the answer key you can find in the resources page of your sapling.
Mon Feb 01, 2021 1:12 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Change in Internal Energy Formula
Replies: 3
Views: 37

### Re: Change in Internal Energy Formula

Yes, I believe you would use the appropriate formula for w.
In the case of a reversible expansion, you would use the one with the integral. Also, the pressure is not constant for reversible expansion.
Mon Feb 01, 2021 1:11 pm
Forum: Calculating Work of Expansion
Topic: Textbook 4A.3 [ENDORSED]
Replies: 4
Views: 58

### Re: Textbook 4A.3[ENDORSED]

It's in the sapling course page under resources.
Mon Feb 01, 2021 11:29 am
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: External Pressure
Replies: 32
Views: 105

### Re: External Pressure

I think it is the pressure that acts on the system. So for in the gas with piston example its the pressure pushing on the piston trying to compress the gas.
Mon Feb 01, 2021 11:25 am
Forum: Calculating Work of Expansion
Replies: 5
Views: 39

You always use w=-PdeltaV if your point of reference is the systen. A positive change in volume ( \Delta V ) will result in negative work because work is done by the system. On the other hand a negative change in volume will result in positive work because work is done on the system (gases being com...
Mon Feb 01, 2021 10:55 am
Forum: Calculating Work of Expansion
Topic: Textbook 4A.3 [ENDORSED]
Replies: 4
Views: 58

### Re: Textbook 4A.3[ENDORSED]

You are right and my answer key says 28J instead of 8J like you said.
Mon Jan 25, 2021 2:05 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Sapling Week 3-4 #10
Replies: 2
Views: 45

### Re: Sapling Week 3-4 #10

You have : ice cube at 0C and water at 45C To reach the same temperature, ice cube will absorb the heat (increase in temp) which is released by the water (drop in temp) But remember that the ice cub has to melt first into the liquid phase before its temperature can increase so Heat to melt ice cube ...
Mon Jan 25, 2021 1:39 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: textbook question 5J #5
Replies: 3
Views: 28

### Re: textbook question 5J #5

Increasing the pressure by compression will shift the equilibrium with the least number of moles of gases.
For part b there is only 1 mole of gases on the reactant compared to 2 moles on the product side so reactant side is favored.
The same for e, 1 on reactant and 2 on product.
Mon Jan 25, 2021 1:35 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4D.15
Replies: 1
Views: 25

### Re: 4D.15

We are asked to find the reaction enthalpy of C2H2(g) + 2H2(g) -> C2H6(g) using the values for enthalpy change of combustion that were given combustion is usually : Fuel + O2 -> CO2 + H2O C2H2 + 5/2 O2 -> 2CO2 + H2O \Delta H= -1300. kJ C2H6 + 7/2 O2 -> 2CO2 + 3H2O \Delta H= -1560. kJ H2 + 1/2 O2 -> ...
Mon Jan 25, 2021 1:08 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Textbook problem 4D.3
Replies: 1
Views: 35

### Re: Textbook problem 4D.3

CO + H2O -> CO2 + H2 \Delta T = 22.799 - 22.113 = 0.686\degree C q =-C\Delta T = 3.00\ kJ\cdot\degree C^-1\times 0.686\degree C = -2.06 kJ this is the heat released for 1.4g for 1 mol: q=\frac{ -2.06 kJ}{1.4g}\times\frac{28g}{1mol}=-41.2kJ\cdot mol^{-1} since volume is constant in a calorimeter, w=0...
Mon Jan 25, 2021 12:38 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Clarification on the sign of q
Replies: 3
Views: 31

### Re: Clarification on the sign of q

Like the answer above it is used for problems regarding calorimetry. For example, if you know the heat capacity of the calorimeter and the temperature change of the calorimeter solution then you can calculate the heat lost or gained by the reaction using q sys = -q surrounding (in this case calorime...
Tue Jan 19, 2021 12:33 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: textbook 6.19
Replies: 2
Views: 25

### Re: textbook 6.19

You can use Le Chatelier's principle here. In part a, lactic acid is produced meaning [H 3 O + ] will increase. As you can see from the equation, the forward reaction will be more favored to reduce the effect of the change. HbO 2 - will decrease In part b, there will be more O 2 making the equilibri...
Tue Jan 19, 2021 8:01 am
Forum: Phase Changes & Related Calculations
Topic: 4e.3 sample problem bond enthalpy
Replies: 2
Views: 44

### Re: 4e.3 sample problem bond enthalpy

Looking at the chemical equation, bromine has to be vaporized first from its liquid to gas phase before it can react with gaseous propene. Energy is added into the system to vaporize the bromine (just like how you put in heat to boil water). C 3 H 6 Br 2 will then be formed but will be in the gaseou...
Tue Jan 19, 2021 7:49 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Textbook Problem 6B.5 part e
Replies: 3
Views: 31

### Re: Textbook Problem 6B.5 part e

$13.6\ mg\ NaOH\times\frac{1\ g}{1000\ mg}\times\frac{1\ mol\ NaOH}{40\ g\ NaOH}\times\frac{1}{0.350\ L}=9.714\times 10^{-4}\ M\ NaOH$
$[OH^-]=9.714\times 10^{-4}\ M$
$pOH=-\log{9.714\times 10^{-4}}=3.013$
$pH=pKw-pOH=14-3.013=10.987$
Mon Jan 18, 2021 2:45 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: #10 Sapling HW
Replies: 2
Views: 19

### Re: #10 Sapling HW

pKa of BH + = 8.46 -> pKb of B = 14-8.46 = 5.54 pH = 9.71 -> pOH = 14-9.71 = 4.29 \begin{aligned} K_b &= \frac{[BH^+][OH^-]}{[B]}\\ pK_b&=-log\frac{[BH^+][OH^-]}{[B]}=-log[OH^-]-log\frac{[BH^+]}{[B]}\\ pK_b&=pOH-log\frac{[BH^+]}{[B]}\\ pOH &= pKb + log\frac{[BH^+]}{[B]}\\ 4.29&=5...
Mon Jan 18, 2021 11:27 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Problem 6D5
Replies: 1
Views: 25

### Re: Textbook Problem 6D5

The K values would probably be given unless that is what you are asked to find. In this case you can refer to this table and others like it form the textbook.
Mon Jan 18, 2021 11:15 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook 6D.15
Replies: 2
Views: 22

### Re: Textbook 6D.15

You would use this table found in page 478 of the textbook.

AlCl3 dissociates into Al3+ and Cl- ions in water and Al3+ is hydrated to Al(H2O)63+ which acts as an acid.
Wed Jan 13, 2021 8:58 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Textbook question 6B 11
Replies: 2
Views: 34

### Re: Textbook question 6B 11

Working backwards: pH of dilute -> concentration of dilute -> concentration of original -> moles of Na2O -> mass of Na2O a.i.Dilute solution : pOH = 14 - pH = 14 - 13.25 = 0.75 [OH-] = 10 -pOH = 10 -0.75 = 0.1778M (final answer in 2 sig figs) a.ii. use 5mL for volume of original because thats the vo...
Tue Jan 12, 2021 7:11 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Affect of Changing Pressure on K
Replies: 5
Views: 71

### Re: Affect of Changing Pressure on K

To add on to the answer above, you might think that adding more gases into the system/container will increase the pressure inside the container. You then will be right but remember that we are only concerned about the Partial pressures of the substances involved in the equilibrium reaction. So while...
Tue Jan 12, 2021 7:01 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Table 5G.2
Replies: 3
Views: 49

### Re: Textbook Table 5G.2

Kp is the equilibrium constant found using partial pressures while Kc is the equilibrium constant found using molar concentrations. The book refers to Kp with K for gaseous reactions because a.png You can convert between Kc and K using K=K_c(RT)^{\Delta n_r} where \Delta n_r is the differenc...
Mon Jan 11, 2021 4:09 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: textbook question 5.35
Replies: 4
Views: 34

### Re: textbook question 5.35

You can make an ICE table for partial pressures and fill out the initial and equilibrium pressures based on the graph. Screen Shot 2021-01-12 at 08.05.21.png Since we didn't start out with any B and C, their values for the "Change" row would be the same with Equilibrium. For A, 27.5-x=17.5...
Mon Jan 11, 2021 11:31 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook 5I.3
Replies: 3
Views: 35

### Re: Textbook 5I.3

The equilibrium constant you used is for T=298K not 500K.
Mon Jan 11, 2021 11:30 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling 5.57 b)
Replies: 3
Views: 43

### Re: Sapling 5.57 b)

Your ICE table will start like this since you initially have 0.245 M SO3 and you know that at equilibrium you will have 0.24 M SO2 (Volume is 1L so 1 moles -> 1 M). x is the amount of NO you add at the beginning of the reaction 3.png Since you do not have any SO2 at the beginning and at equilibrium ...
Tue Jan 05, 2021 10:36 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Audio Visual Assessment Question
Replies: 3
Views: 36

### Re: Audio Visual Assessment Question

Rearranging $PV=nRT$
$P=\frac{n}{V}RT=cRT$
for example,
$P_{SO_2}=c_{SO_2}RT=(0.075)(8.206x10^-2)(312)=1.92$
Tue Jan 05, 2021 10:30 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Audio Visual Assessment Question 1B
Replies: 2
Views: 27

### Re: Audio Visual Assessment Question 1B

I believe the answer is A. Kc represents the equilibrium constant using the equilibrium molar concentrations for the reactants and products. Kp represents the equilibrium constant using the equilibrium partial pressures for the reactants and products. Values for K are always values when system is in...
Tue Jan 05, 2021 9:18 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Reactant in Excess
Replies: 9
Views: 92

### Re: Reactant in Excess

I'm assuming you are asking what would happen if you added more reactant to the system after it has achieved equilibrium. In this case, the reactant will be consumed and products will form until Q = K at which the system has achieved equilibrium again.
Tue Jan 05, 2021 5:09 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Table 5G.2
Replies: 1
Views: 27

### Re: Textbook Table 5G.2

K p is the equilibrium constant found using partial pressures while K c is the equilibrium constant found using molar concentrations. The book refers to K p with K for gaseous reactions because Screen Shot 2021-01-05 at 21.02.15.png You can convert between K c and K using K=K_c(RT)^{\Delta n...
Tue Jan 05, 2021 4:57 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Module Part 2 #29
Replies: 2
Views: 23

### Re: Module Part 2 #29

You maybe had a calculation error because I answered using molar concentrations and got the right answer (4.98).
Concentrations at equilibrium:
[BrCl] = 3.367x10-5 M
[Br2]=[Cl2] = 7.515x10-5 M
$K_c=\frac{[Br_2][Cl_2]}{[BrCl]^2}=4.98$
Tue Dec 08, 2020 3:40 am
Forum: Calculating the pH of Salt Solutions
Topic: Sampling week10 Q13
Replies: 2
Views: 51

### Re: Sampling week10 Q13

pKa = -log(Ka) just like how pH = -log([H+]) and pOH = -log([OH-]) From the equilibrium equation, \begin{aligned} K_a&=\frac{[A^-][H_3O^+]}{[HA]}\\-\log(K_a)&=-\log\frac{[A^-][H_3O^+]}{[HA]}\\pK_a&=pH-\log\frac{[A^-]}{[HA]}\\pH-pK_a&=\log\frac{[A^-]}{[HA]}\end{aligned} From t...
Mon Dec 07, 2020 10:37 am
Forum: Quantum Numbers and The H-Atom
Topic: probabilty of finding e- in a location
Replies: 6
Views: 94

### Re: probabilty of finding e- in a location

I don't think that will be included since we only covered the topic conceptually and did not do any calculations about wavefunctions.
Mon Dec 07, 2020 10:10 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: coordination numbers
Replies: 2
Views: 29

### Re: coordination numbers

No, you don't. Just determine at how many points do the ligands attach to the metal atom. For example in [Fe(CN) 6 ] 4- , there are six CN - ligands and each of them attaches to Fe at one site. Therefore, the coordination number is 6. In [Cr(OH 2 ) 2 (ox) 2 ] 2- , there are 2 OH 2 which each attache...
Mon Dec 07, 2020 10:00 am
Forum: General Science Questions
Topic: final exam
Replies: 9
Views: 168

### Re: final exam

"Final covers all material (approximately weighted to the amount of time covered in the syllabus). Quantum was the biggest section. Questions will come from the textbook homework listed in the syllabus." https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Final_Review_Session...
Mon Dec 07, 2020 9:50 am
Forum: Hybridization
Topic: Book Problem F1
Replies: 3
Views: 40

### Re: Book Problem F1

It's how the 4 sp3 are arranged.
The relative orientations would be the same as the original electron arrangement you used to determine the hybridization.
Mon Dec 07, 2020 9:46 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Sapling Homework W10 Problem#8
Replies: 2
Views: 44

### Re: Sapling Homework W10 Problem#8

To add to the above post, you can calculate pH from [OH-] by finding pOH first
pOH = -log([OH-])
Then finding pH from it
pH = 14 - pOH
Tue Dec 01, 2020 6:00 pm
Forum: Naming
Topic: Ba[FeBr4]2
Replies: 3
Views: 42

### Re: Ba[FeBr4]2

First, recognize that the cation is Ba 2+ and the anion [FeBr 4 ] - . Since the total charge on the complex ion is -1 and we know that Br is a halogen and therefore has an oxidation number of -1, the oxidation number of Fe would be +3. Now to name the complex anion you start by naming the ligands wh...
Tue Dec 01, 2020 9:22 am
Forum: Naming
Topic: Charge (oxidation state) of atoms/molecules
Replies: 6
Views: 34

### Re: Charge (oxidation state) of atoms/molecules

You can refer to this table to determine the charges of the ions.
In the case of [Ni(CN)4]2-, charge of CN is -1 and there are 4 of it so charge of Ni + 4*-1 = -2. Therefore, charge of Ni = +2.
Tue Dec 01, 2020 9:19 am
Forum: Naming
Topic: order of ligands [ENDORSED]
Replies: 6
Views: 120

### order of ligands[ENDORSED]

Here, the dichlorido is placed before bis(oxalato) which I believe follows the rule where you name the ligands in alphabetical order, ignoring the prefix. Screen Shot 2020-12-02 at 01.09.10.png However, this one places bisoxalato before diaqua even though a precedes o alphabetically. Screen Shot 202...
Tue Dec 01, 2020 3:40 am
Forum: Naming
Topic: Identifying Anions
Replies: 4
Views: 72

### Re: Identifying Anions

Keep in mind that complex ions in a coordination compound are not always cations.

In the example above, [PtCl3(NH3)]- is an anion while [Cr(NH3)4(OH)2]+ is a cation.
Tue Dec 01, 2020 3:28 am
Forum: Hybridization
Topic: unhybridized orbitals
Replies: 4
Views: 52

### Re: unhybridized orbitals

I don't think it is necessary for the orbital involved to be hybridized. For example, the sigma bond in H2 is between two 1s orbitals, and the one in HF is $\sigma$(H1s,F2p)
Mon Nov 30, 2020 8:38 am
Forum: Naming
Topic: Textbook 9C.1
Replies: 6
Views: 75

### Re: Textbook 9C.1

"Cyano is an older naming convention while cyanido is the IUPAC preferred name"
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=2351
Wed Nov 25, 2020 12:04 am
Forum: Hybridization
Topic: Pi orbitals overlapping
Replies: 4
Views: 26

### Re: Pi orbitals overlapping

Remember that there are three kinds p-orbitals (p x , p y , and p z ). p orbitals.png The middle carbon atom will have a hybridization of sp which means that 1 s-orbital and 1 p-orbital will "mix" to form 2 sp-hybrid orbitals. The remaining 2 p-orbitals will always then be perpendicular to...
Tue Nov 24, 2020 7:43 pm
Forum: Hybridization
Topic: Pi orbitals overlapping
Replies: 4
Views: 26

### Re: Pi orbitals overlapping

Screen Shot 2020-11-25 at 11.41.41.png The 2 p-orbitals in the central C atom will form pi bonds with the two outer C atoms. These 2 p-orbitals are perpendicular to each other which means the pi bonds will also be perpendicular. This means that the H atoms will not be in one plane. Hopefully the di...
Tue Nov 24, 2020 12:31 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Focuse 2E.7
Replies: 2
Views: 18

### Re: Focuse 2E.7

The question asks for how many different O-S-Cl bond angles are there in the molecule. Like the answer above, there is only one, which is a little less than 109.5º since the lone pair pushes the bonding pair equally. An example where there would be different bond angles can be found in SF4. Its shap...
Tue Nov 24, 2020 9:45 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Section 2E Question 5b
Replies: 2
Views: 33

### Re: Section 2E Question 5b

I think you forgot to account for the +1 charge.
The molecule in question is ClO2+
Valence e- = 7 + 2*6 - 1 = 18. Since it is even, there should be no unpaired e-.
Tue Nov 24, 2020 1:16 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Determining Molecular Polarity using VSEPR
Replies: 8
Views: 77

### Re: Determining Molecular Polarity using VSEPR

It really helps me to think of the dipole moments in bonds as vectors represented by arrows.
So when determining the polarity of CO2, the arrows point in completely opposite directions with the same magnitude. They cancel out. Therefore, the molecule is nonpolar.
Tue Nov 24, 2020 1:11 am
Forum: Hybridization
Topic: 2F.1
Replies: 2
Views: 37

### Re: 2F.1

It is asking for how the orbitals are oriented just like how the p orbitals are oriented towards the three (x,y,z) axis. For example, the 4 sp3-orbitals are oriented in a way in which the orbitals point towards the corners of a tetrahedron. The way they are arranged is the same as the electronic arr...
Mon Nov 23, 2020 11:40 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 2E #7
Replies: 3
Views: 61

### Re: 2E #7

The question asks for how many different O-S-Cl bond angles are there in the molecule. Like the answer above, there is only one which is a little less than 109.5º since the lone pair pushes the bonding pair equally. An example where there would be different bond angles can be found in SF 4 . Its sha...
Thu Nov 19, 2020 1:02 am
Forum: Lewis Structures
Topic: Difference between Lewis acids and bases and cations and anions
Replies: 2
Views: 14

### Re: Difference between Lewis acids and bases and cations and anions

All cations are lewis acids but not all lewis acids are cations.
For example, BF3 is a lewis acid but is not a cation.
The same can applied to lewis base and anions.
Tue Nov 17, 2020 12:41 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: Vapor Pressure
Replies: 4
Views: 56

### Re: Vapor Pressure

Vapor pressure is the pressure exerted by the gas phase of a liquid in a container. So the stronger the molecules are held together, the lower the vapor pressure because fewer molecules can escape from the liquid surface to become a gas. In 3F.19, water has lower vapor pressure because its molecules...
Tue Nov 17, 2020 12:34 am
Forum: Lewis Structures
Topic: Textbook Question Number 2B 11 part C
Replies: 2
Views: 29

### Re: Textbook Question Number 2B 11 part C

Yes, there is.
However, it will not be favorable since the formal charge of the O bonded to H will become +1 while the other O will become -1.
Tue Nov 17, 2020 12:20 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: electron spin
Replies: 4
Views: 101

### Re: electron spin

I think it can be either +½ or -½. The only thing you know is that if two electrons occupy the same orbital, their spins must be paired (one +½, one -½).
Tue Nov 17, 2020 12:16 am
Forum: Lewis Structures
Topic: lewis structure
Replies: 2
Views: 38

### Re: lewis structure

Compounds in 2B.9 are ionic compounds while the ones on 2B.11 are molecular compounds (covalent bonds).
You draw the lewis structure of ions in ionic compounds separately with parentheses and charges.
Tue Nov 17, 2020 12:11 am
Forum: Lewis Structures
Topic: Structure Exceptions
Replies: 1
Views: 32

### Re: Structure Exceptions

The central atom is usually the atom with the lowest electronegativity. In this case, this atom is N.
Screen Shot 2020-11-17 at 16.12.22.png (10.99 KiB) Viewed 28 times

Also, after accounting for formal charge, you can see that NNO is more stable.
Tue Nov 10, 2020 8:28 pm
Forum: Electronegativity
Topic: Sapling #15
Replies: 9
Views: 94

### Re: Sapling #15

Hydrogen bonding also only happens when the hydrogen is bonded with a highly electronegative atom in this case N, O, F. This is so that the hydrogen will have enough $\delta ^+$ and can interact with the lone pair on N, O, or F of another molecule.
Tue Nov 10, 2020 7:54 pm
Forum: Lewis Structures
Topic: Resonance and Formal Charge
Replies: 3
Views: 45

### Re: Resonance and Formal Charge

You could start by putting the negative formal charge on the more electronegative atom since a negative formal charge means the atom has more e- than it usually has.
Tue Nov 10, 2020 7:37 pm
Forum: Quantum Numbers and The H-Atom
Topic: Quantum number ms and wave function
Replies: 2
Views: 31

### Re: Quantum number ms and wave function

You only need to know the first three.
ms is to differentiate the 2 electrons in an orbital.
Tue Nov 10, 2020 3:46 am
Forum: Ionic & Covalent Bonds
Topic: Oxidation numbers
Replies: 3
Views: 53

### Re: Oxidation numbers

Yes, when finding the oxidation number you assume that the bonds are ionic and that the electrons in a bond are assigned to the more electronegative atom.
Tue Nov 10, 2020 3:34 am
Forum: Lewis Structures
Topic: Central Atom
Replies: 20
Views: 107

### Re: Central Atom

Some exceptions exist to arranging the atoms symmetrically around the central atom.
For example, the structure for N2O is going to be NNO instead of NON.
Wed Nov 04, 2020 10:55 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: D-Orbital Electron Configuration
Replies: 3
Views: 36

### Re: D-Orbital Electron Configuration

Yes, the elements in the 4th period starting from Sc have electrons in the 3d state.
Mon Nov 02, 2020 9:23 pm
Forum: Trends in The Periodic Table
Topic: Covalent Character of Compounds
Replies: 2
Views: 44

### Re: Covalent Character of Compounds

I think the electronegativity difference between the two atoms determines this.
Electronegativity difference for AgF is 2.1 (F is 4.0 and Ag is 1.9) while electronegativity difference for BeCl2 is 1.6 (Be is 1.6 and Cl is 3.2).
This makes AgF more ionic.
Mon Nov 02, 2020 9:15 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: What Orbitals are Valence Electrons?
Replies: 3
Views: 50

### Re: What Orbitals are Valence Electrons?

Valence electrons are electrons that are outside of the noble-gas core of an atom. Since the electronic configuration of Mn4+ is [Ar]3d3, it has 3 (valence) electrons outside of the noble-gas (Ar) core.
Mon Nov 02, 2020 9:01 pm
Forum: Quantum Numbers and The H-Atom
Topic: Energy Between Subshells
Replies: 2
Views: 44

### Re: Energy Between Subshells

I think the energy of the subshell increases as you go from s to p to d when they are in the same shell. I say this because if I am not wrong, the s orbital is closer to the nucleus than the p orbital which is also closer than the d orbital. And the closer the orbital is to the nucleus, the stronger...
Mon Nov 02, 2020 8:54 pm
Forum: Trends in The Periodic Table
Topic: Sizes of ions vs elements
Replies: 4
Views: 42

### Re: Sizes of ions vs elements

Na and Cl is in the same period with Na to the left of Cl. This makes Na bigger than Cl.
But when Na becomes Na+, it loses its outermost electron that is in the 3s1 state. This means that the outermost occupied shell of Na+ is now the n=2 shell making its radius significantly smaller.
Mon Nov 02, 2020 8:51 pm
Forum: Lewis Structures
Topic: Adding electrons from the charge
Replies: 4
Views: 50

### Re: Adding electrons from the charge

I think the extra electrons in the anions come from the cations.
For example in KNO3, K gives one electron to NO3. Therefore, K+ and NO3-.
Wed Oct 28, 2020 11:11 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Bohr's Equation
Replies: 3
Views: 65

### Re: Bohr's Equation

E_n=-\frac{hR}{n^2} gives the energy of an electron of an H atom in the n energy level. \nu=R(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}) tells us the frequency of a photon that causes an electron to get excited from energy level n 1 to n 2 (frequency of photon absorbed) You can derive the Rydberg...
Wed Oct 28, 2020 11:06 pm
Forum: Photoelectric Effect
Topic: Rydberg Equation question
Replies: 3
Views: 30

### Re: Rydberg Equation question

You can derive the Rydberg equation from E_n=-\frac{hR}{n^2} When an electron gets excited from energy level n 1 to n 2 \begin{aligned} \Delta E&=E_f-E_i \\ h\nu&=(-\frac{hR}{{n_2}^2})-(-\frac{hR}{{n_1}^2})\\ h\nu&=\frac{hR}{{n_1}^2}-\frac{hR}{{n_2}^2}\\ h\nu&=hR(...
Mon Oct 26, 2020 10:27 pm
Forum: Empirical & Molecular Formulas
Topic: Limiting Reactant Calculations Module #22
Replies: 2
Views: 52

### Re: Limiting Reactant Calculations Module #22

Your chemical equation is not balanced.
It should be:
C6H9Cl3 + 3AgNO3 ---> 3AgCl + C6H9(NO3)3.
Mon Oct 26, 2020 10:25 pm
Forum: General Science Questions
Topic: help with work function hw question
Replies: 2
Views: 54

### Re: help with work function hw question

I think u mistyped the value of the frequency. E(photon) = Work function + Kinetic energy h \nu = \Phi + E k \Phi = h \nu - E k For the second part, Total energy of photons = 3.14 X 10^-7 J, and the energy required to eject an electron is \Phi you found in the first part. So number of electrons ejec...
Mon Oct 26, 2020 10:19 pm
Forum: Properties of Electrons
Topic: Atomic Spectra Post-Module Assessment #28
Replies: 3
Views: 45

### Re: Atomic Spectra Post-Module Assessment #28

1 meter = 1,650,763.73 wavelengths
Therefore, 1/1,650,763.73 meter = 1 wavelength
1 wavelength = 6.058 x 10-7 m = 605.8 nm
So it's in the visible light region.
E = hc/$\lambda$ = 3.28 x 10-19 J