Search found 104 matches
- Sun Mar 15, 2020 6:59 am
- Forum: First Order Reactions
- Topic: Pseudo first order rxn
- Replies: 5
- Views: 491
Re: Pseudo first order rxn
A pseudo first-order reaction is one that is not normally first order but can be changed to a first-order reaction by significantly increasing or decreasing the concentration of one of the reactions. By doing so, you would be measuring the effect that the reactant whose concentration you are not cha...
- Sun Mar 15, 2020 6:56 am
- Forum: Zero Order Reactions
- Topic: When to use each order
- Replies: 19
- Views: 1019
Re: When to use each order
You can tell the orders based on the graphs: for zero-order the graph of [A] vs time would be linear, for first-order the graph of lnA vs time would be linear, and for second-order the graph of 1/[A] vs time would be linear. You can also determine the order based on a data table by using experiment ...
- Sun Mar 15, 2020 6:51 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: pseudo rate law
- Replies: 13
- Views: 991
Re: pseudo rate law
For the pseudo rate law, you would write the rate law in terms of the reactant that is in excess. Thus, you would substitute k' for k[reactant that is maintained at its original concentration]. For example in the reaction A + B --> C, and B's concentration were increased so that it would be in exces...
- Sun Mar 15, 2020 6:47 am
- Forum: General Rate Laws
- Topic: Elementary steps
- Replies: 3
- Views: 342
Re: Elementary steps
You can use the coefficients of reactants in an elementary step to determine the orders of each reactant fr the rate law. For the nonelementary step, you can't determine the rate law without experimental data. The combination of elementary steps can help you derive the overall reaction and the overa...
- Sun Mar 15, 2020 6:44 am
- Forum: General Rate Laws
- Topic: 7C.7
- Replies: 5
- Views: 430
Re: 7C.7
The rate will generally correspond to the slower reaction because that will govern the rate of the overall reaction. In this case, the rate of the slow reaction is rate = k[NO][Br2] which matches the answer.
- Sun Mar 08, 2020 12:45 am
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: order of reactions
- Replies: 7
- Views: 460
Re: order of reactions
Thee order can also tell you which graph [A] vs time, ln[A] vs time, or 1/[A] vs time will show a straight line.
- Sun Mar 08, 2020 12:39 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Homework 7D5
- Replies: 1
- Views: 205
Re: Homework 7D5
I think you can use the equation ln (kr2/kr1) = (Ea/R)(1/T1 - 1/T2) and plug in the value for k1 at 25 degrees C as well as the Ea, R and the calculated values of temperature in Kelvin. This should get you the correct answer.
- Sun Mar 08, 2020 12:23 am
- Forum: First Order Reactions
- Topic: how to know actual order
- Replies: 4
- Views: 366
Re: how to know actual order
You shouldn't use stoichiometric coefficients to find the order of a reaction for overall reactions. This is the case because it has not been experimentally determined that the reaction is in second order with respect to A. You can use them to write the forward and reverse reaction rates for element...
- Sun Mar 08, 2020 12:20 am
- Forum: Zero Order Reactions
- Topic: half life calculations
- Replies: 3
- Views: 375
Re: half life calculations
The integrated rate law for the zero order reaction is [A] = - k t + [A]0. When t=t1/2, we can substitute this value into the equation to get [A] at t1/2 = -k(t1/2) + [A]o. So t1/2 = [A]o / 2 k. The integrated rate law for first order reaction is ln [A] = -k t + ln [A]o. When t = t1/2 we can substit...
- Sat Mar 07, 2020 11:56 pm
- Forum: General Rate Laws
- Topic: 7A.17
- Replies: 1
- Views: 238
Re: 7A.17
I think you are forgetting to square the concentration of C. When you input the values for A, B, and C for k[A][B]^2[C]^2, I think you calculated (2.85 x 10^12 L^4/mol^4 s)(0.00301 mol/L)(0.001 mol/L)^2(0.00115 mol/L) which would get you 9865.275 L/mol s. Since the reaction is second order with resp...
- Sun Mar 01, 2020 12:48 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Units for delta G
- Replies: 31
- Views: 1730
Re: Units for delta G
Since one Faraday is 96485 C/mol, it can be multiplied by n (moles) in the equation delta G = -nFE to just get kJ. What are the units for E and how do we get to Joules from coulombs? The units for E are Volts, and one Volt = 1 J/C. So, when you multiply n (moles) by F (C/mol) and E ( V = J/C) you g...
- Sat Feb 29, 2020 11:28 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: how concentration affects cell potential
- Replies: 2
- Views: 323
Re: how concentration affects cell potential
A cell reaction with a positive cell potential has a strong tendency to take place and will result in more products at equilibrium. As a result, K will be larger than 1 when E*cell > 0. When a cell has negative cell potential, the reaction is less likely to take place thus generating fewer products ...
- Sat Feb 29, 2020 11:23 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6N.9
- Replies: 2
- Views: 265
Re: 6N.9
You should use the following half-reactions:
oxidation: Sn --> Sn2+ + 2e-
reduction: 2H+ + 2e- --> H2
This will help you calculate E*. then you can combine the two to find the net which would be Sn + 2H+ --> Sn2+ + H2 writing the expression for Q should give you Q = [H2][Sn2+]/[H2]^2
oxidation: Sn --> Sn2+ + 2e-
reduction: 2H+ + 2e- --> H2
This will help you calculate E*. then you can combine the two to find the net which would be Sn + 2H+ --> Sn2+ + H2 writing the expression for Q should give you Q = [H2][Sn2+]/[H2]^2
- Sat Feb 29, 2020 11:16 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Units for delta G
- Replies: 31
- Views: 1730
Re: Units for delta G
Since one Faraday is 96485 C/mol, it can be multiplied by n (moles) in the equation delta G = -nFE to just get kJ.
- Sat Feb 29, 2020 11:09 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Maximum Work
- Replies: 1
- Views: 208
Re: Maximum Work
For maximum work calculations, the anode is in the reactants while the cathode is in the products. For example with question 6N.3 part a, Pt(s)|H2(g, 1.0 bar)|HCl(aq, 0.075 mol/L)||HCl(aq, 1.0 mol/L)|H2(g, 1.0 bar)|Pt(s), the oxidation reaction is H2 --> 2H+ + 2e- while the reduction reaction is 2H+...
- Sun Feb 23, 2020 1:38 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Use of Platinum
- Replies: 10
- Views: 563
Re: Use of Platinum
Platinum should be used when there is no metal present to conduct electrons from the anode to the cathode. You would include platinum when a solid metal isn't present to allow for the transfer of electrons.
- Sun Feb 23, 2020 1:29 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: adding/subtracting half-redox rxns
- Replies: 4
- Views: 304
Re: adding/subtracting half-redox rxns
Also, unlike in Hess' law where you could multiply the enthalpy values by the same coefficient that you multiplied the entire reaction by, you can't use the same strategy with cell potential. Standard reduction potential gives the voltage difference between two standard electrodes which is always th...
- Sun Feb 23, 2020 1:15 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: max work
- Replies: 1
- Views: 178
Re: max work
I think charge is still being transferred when you have maximum cell potential and maximum work. The relationship between E and work is E = -w/charge where charge = moles of e- transferred x F = nF. As a result, wmax = -nFE which suggests that in order for a maximum work value to result, there shoul...
- Sun Feb 23, 2020 1:07 pm
- Forum: Balancing Redox Reactions
- Topic: Steps to Balance a Basic Solution
- Replies: 2
- Views: 198
Re: Steps to Balance a Basic Solution
For an acidic solution, if there are more Oxygen atoms in the reactions you would add H3O+ to supply more H+. For a basic solution, if there are more Oxygen atoms in the reactants you would add H2O to supply more H+.
- Sun Feb 23, 2020 12:58 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidation Numbers
- Replies: 3
- Views: 278
Re: Oxidation Numbers
The formal charge is a charge on a molecule that is present when the shared electrons are equally divided between atoms in each bond. The oxidation numbers are not always equal to the actual charge on the molecules, they are used as a way to keep track of the number of electrons an atom has.
- Sun Feb 16, 2020 7:23 pm
- Forum: Balancing Redox Reactions
- Topic: In Class Example
- Replies: 4
- Views: 269
Re: In Class Example
The reaction of the permanganate ion with Fe2+ in water is written as: 8H+ + MnO4- + 5Fe2+ --> Mn2+ + 5Fe3+ + 4H2O This reaction can show that MnO4- is the oxidizing agent because it is reduced to Mn2+ and Fe2+ is the oxidizing agent because it is oxidized to Fe3+. This can be seen in the half-react...
- Sun Feb 16, 2020 7:15 pm
- Forum: Van't Hoff Equation
- Topic: delta s
- Replies: 9
- Views: 529
Re: delta s
Delta S is constant in the Vant Hoff equation because the change in entropy of the system at equilibrium is constant. Delta S doesn't increase because it has reached its maximum.
- Sat Feb 15, 2020 4:46 pm
- Forum: Van't Hoff Equation
- Topic: Equation
- Replies: 5
- Views: 304
Re: Equation
You can generally only use lnK = (-delta H/RT) + (delta S/T) when the reaction is at equilibrium because you are setting Q = K.
- Sat Feb 15, 2020 4:38 pm
- Forum: Balancing Redox Reactions
- Topic: Homework 6K.3 part c
- Replies: 2
- Views: 193
Re: Homework 6K.3 part c
H2S(aq) + Cl2(g) --> S(s) + Cl-(aq)
In this reaction, Cl2 gains electrons to form Cl- thus Cl2 is being reduced. An oxidizing agent is the electron acceptor in a reaction, and in this case, Cl2 gains electrons while H2S loses electrons.
In this reaction, Cl2 gains electrons to form Cl- thus Cl2 is being reduced. An oxidizing agent is the electron acceptor in a reaction, and in this case, Cl2 gains electrons while H2S loses electrons.
- Sat Feb 15, 2020 4:18 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: finding enthalpy of non-isobaric process
- Replies: 3
- Views: 302
Re: finding enthalpy of non-isobaric process
Delta H only = q at constant pressure, so if the pressure is not constant you won't be able to calculate a value for Delta H
- Sun Feb 09, 2020 9:32 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Difference between standard reaction enthalpy and enthalpy
- Replies: 2
- Views: 193
Re: Difference between standard reaction enthalpy and enthalpy
The reaction enthalpy is the heat released or absorbed for the reaction; it is the difference in enthalpy for the products and reactants. The enthalpy of formation for a mol of substance in its standard state is the change in enthalpy that occurs when that substance is formed from its elements. The ...
- Sun Feb 09, 2020 9:26 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Change in concentration
- Replies: 2
- Views: 100
Re: Change in concentration
Yes, if you decrease the concentration of NO, the concentration of H2O will increase because the reaction will shift to the right to generate more products. If you increase the concentration of NO, the concentration of H2O will decrease as the reaction will shift toward the left to generate more rea...
- Sun Feb 09, 2020 9:12 pm
- Forum: Calculating Work of Expansion
- Topic: Constant pressure
- Replies: 2
- Views: 75
Re: Constant pressure
Constant pressure situations could also be referred to as isobaric, which would lead you to use the equation w = -PdeltaV. however, if the external pressure for a system is zero even though a value greater then zero is provided for the internal pressure, the work done would be zero.
- Sun Feb 09, 2020 9:10 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat capacity
- Replies: 2
- Views: 73
Re: Heat capacity
Heat capacity is a state function because its value doesn't depend on the way by which the temperature was changed or the path taken to add or subtract heat. The heat capacity is determined by the overall change in these two factors.
- Sun Feb 09, 2020 9:05 pm
- Forum: Phase Changes & Related Calculations
- Topic: Rotational/Translational Contributions?
- Replies: 2
- Views: 135
Re: Rotational/Translational Contributions?
The values given indicate what the heat capacity of an ideal gas could be at constant volume and constant pressure. the main concept to know here is that the change in internal energy = (3/2)Rdelta T. You can find the heat capacity by dividing delta U by delta T. You can use the "How Do We Know...
- Fri Jan 31, 2020 8:17 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 4A.13
- Replies: 4
- Views: 116
Re: 4A.13
Since diluted aqueous solutions have similar heat capacities as water, you can assume that the heat capacity of the calorimeter is the same as that of the solution. This would allow the qreaction to equal -qcalorimeter.
- Fri Jan 31, 2020 8:06 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Max Heat Capacity
- Replies: 4
- Views: 102
Re: Max Heat Capacity
I don't believe that there is a maximum heat capacity for a substance. The heat capacity given for that particular substance is the amount of heat energy necessary to increase the temperature of that substance by one unit per mass. This suggests that there is a set value for the heat capacity for a ...
- Fri Jan 31, 2020 8:06 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Max Heat Capacity
- Replies: 4
- Views: 102
Re: Max Heat Capacity
I don't believe that there is a maximum heat capacity for a substance. The heat capacity given for that particular substance is the amount of heat energy necessary to increase the temperature of that substance by one unit per mass. This suggests that there is a set value for the heat capacity for a ...
- Fri Jan 31, 2020 7:56 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Reversible Process
- Replies: 4
- Views: 148
Re: Reversible Process
A reversible process can occur in either direction; it can be reverted back to its original position through very small changes made to the surroundings.
- Fri Jan 31, 2020 7:49 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: q=nCT
- Replies: 2
- Views: 1507
Re: q=nCT
The formula for heat can be either q=nCT where n stands for the number of moles, or it can be q=mCT where m stands for the mass in grams. The formula you use will depend on whether the specific heat capacity (C) is given as J/gK where you would use q=mCT or if it is given as J/molK where you would u...
- Fri Jan 31, 2020 7:45 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: When to leave out reactants/products from enthalpy calculations
- Replies: 4
- Views: 138
Re: When to leave out reactants/products from enthalpy calculations
A compound is in its purest form when it is in its most stable form. An example of this is O2 which has an enthalpy of formation of zero because it is the most stable form of oxygen, whereas O3 would have an enthalpy of formation that was not equal to zero. Generally, gases are in their most stable ...
- Sat Jan 25, 2020 11:34 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: state functions
- Replies: 4
- Views: 1030
Re: state functions
Mass, volume, and density are also state functions. As state property is not dependent on the path taken to obtain that state, it is only dependent on its current state.
- Sat Jan 25, 2020 11:29 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Reversible Expansion
- Replies: 2
- Views: 57
Re: Reversible Expansion
A reversible process is one that can return to its original state by making very small changes to the system by means of its surroundings. Reversible processes can occur in either direction.
- Sat Jan 25, 2020 11:21 am
- Forum: Phase Changes & Related Calculations
- Topic: Steam Burn
- Replies: 6
- Views: 191
Re: Steam Burn
Dr. Lavelle was saying that if you look at the heating curve of water, you will notice a transition between the phase changes of water. It starts off as solid ice and as it is heated, the temperature of the solid rises to form water. The transition between ice and water is called the melting point a...
- Thu Jan 23, 2020 2:18 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Endothermic Reaction.
- Replies: 7
- Views: 222
Re: Endothermic Reaction.
Generally, if you see smaller molecules on the reactants side forming a larger molecule on the products side, it is endothermic. If you see a molecule dissociating into smaller molecules, it is exothermic.
- Thu Jan 23, 2020 2:14 pm
- Forum: Phase Changes & Related Calculations
- Topic: - and + H values
- Replies: 5
- Views: 131
Re: - and + H values
Exothermic reactions have a negative overall enthalpy because energy is released when bonds are formed. Thus the products require less chemical potential energy than reactants, to hold atoms together. For endothermic reactions, the overall enthalpy of the reaction is positive because energy is requi...
- Sun Jan 19, 2020 6:22 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: X was ignored
- Replies: 27
- Views: 948
Re: X was ignored
The -x was ignored because the value would have been so small that 0.1-x would be rounded to 0.1. If K is less than 10^-3, you generally ignore the -x and use the initial concentration without explicitly factoring in the change.
- Sun Jan 19, 2020 3:33 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6B.11 part a ii
- Replies: 1
- Views: 88
Re: 6B.11 part a ii
You can use M1V1 = M2V2 where M1 is the concentration of the original solution, M2 is 0.178 mol/L from part i, V2 is 500 mL, and V1 is 5 mL. You are using 5mL for V1 because that is the volume of the original solution that is used to dilute it to 500 mL.
- Thu Jan 16, 2020 1:15 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6C.13
- Replies: 2
- Views: 104
Re: 6C.13
You could use the pKa to find the pKb by subtracting pKa value from 14. The lower the pKb value, the stronger the base.
- Thu Jan 16, 2020 1:13 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: percentage reacted
- Replies: 3
- Views: 104
Re: percentage reacted
When you are writing out your ICE table, you want to find the concentration of reactants and products at equilibrium. For your initial concentration, you would calculate 0.4 mol/3 L H2 to find the initial concentration of H2 which is 0.133 mol/L. For the change in concentration, you would multiply 0...
- Thu Jan 16, 2020 12:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5.35 part b
- Replies: 4
- Views: 219
Re: 5.35 part b
Yes I think you should always convert from Pa to atm or bar before you continue with calculations.
- Sun Jan 12, 2020 12:50 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Partial pressure
- Replies: 4
- Views: 178
Re: Partial pressure
The partial pressure refers to the pressure of a gas that could be calculated if it were alone in a container of the same volume and at the same temperature as the original mixture.
- Fri Jan 10, 2020 11:25 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE Tables
- Replies: 5
- Views: 158
Re: ICE Tables
When writing out your equilibrium constant, you wouldn't consider liquids and solids present in the reaction since they are pure substances and don't change over the course of the reaction. So when you write out your ICE box, you would only include gases and aqueous products and reactions.
- Thu Jan 09, 2020 9:56 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Changing Kc [ENDORSED]
- Replies: 4
- Views: 133
Re: Changing Kc [ENDORSED]
A change in pressure is brought about by a change in volume. A change in pressure can cause the reaction to shift to the left or to the right based on the number of moles of products and reactants. But this shift occurs to restore the value of the equilibrium constant.
- Wed Jan 08, 2020 9:09 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5I.25 [ENDORSED]
- Replies: 1
- Views: 136
Re: 5I.25 [ENDORSED]
1. First, calculate the concentration of all gasses by dividing the amount in mols (this is given) by the volume (5L) 2. Set up your ICE box (as shown below) y writing down the initial concentration for all gases 3. Write out the equilibrium constant expression 4. calculate the value of x which repr...
- Wed Jan 08, 2020 8:49 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 5.33
- Replies: 3
- Views: 178
Re: 5.33
Hi! Energy is actually required to break bonds which means that a certain amount of energy must be provided to the reaction in order for the dissociation of X2 to occur. As you can see in this image (https://i.stack.imgur.com/jm5uO.png) bound atoms have low potential energy, and energy is required f...
- Sat Dec 07, 2019 10:06 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: Polarity/Dipoles
- Replies: 1
- Views: 149
Re: Polarity/Dipoles
What does it mean when dipoles "cancel out?" Dipoles could cancel out if the are facing the same direction and if the atoms surrounding the central atom are all the same. This would mean that the electronegativity difference between the central atom and each of the bound atoms is the same...
- Sat Dec 07, 2019 10:01 pm
- Forum: Sigma & Pi Bonds
- Topic: Question on Test 2
- Replies: 11
- Views: 877
Re: Question on Test 2
If the question asks "how many atoms on the structure would form a hydrogen bond?" you would just count the number of atoms that would have hydrogen bonding sites. But if it asks for the number of hydrogen bonding sites, you would count each available lone pair where bonding could occur.
- Sat Dec 07, 2019 9:52 pm
- Forum: Trends in The Periodic Table
- Topic: Trend for Polarizability
- Replies: 6
- Views: 588
Re: Trend for Polarizability
Cations that are small and have a high charge have large polarizing power. Anions that are large and have small charge have high polarizability.
- Sat Dec 07, 2019 9:40 pm
- Forum: Properties of Light
- Topic: relationship between frequency and intensity?
- Replies: 3
- Views: 217
Re: relationship between frequency and intensity?
The intensity, or the number of electrons emitted from the metal, can be acknowledged when the frequency of the incident light is at or above the threshold frequency. If the threshold frequency is not reached, no electrons will be ejected no matter how high the intensity of the light is.
- Sat Dec 07, 2019 9:37 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: bond strength
- Replies: 2
- Views: 269
Re: bond strength
You can compare the bond length of binary acids whose central atom varies down a group to determine which acid is stronger or weaker.
- Sun Dec 01, 2019 11:15 am
- Forum: Lewis Acids & Bases
- Topic: Conjugate acid and conjugate base
- Replies: 2
- Views: 138
Re: Conjugate acid and conjugate base
The acid on the reactants side of the equation will have a proton that it will lose on the products side. This is how you can identify the Bronsted acid on the reactants side, and the opposite is true for the Bronsted base. Once you have found the Bronsted acid and base, you should remember that the...
- Thu Nov 28, 2019 11:52 pm
- Forum: Calculating the pH of Salt Solutions
- Topic: Error in 6B.5 e)?
- Replies: 2
- Views: 506
Re: Error in 6B.5 e)?
Make sure you have the correct decimal places for your molarity. 13.6 mg --> 0.0136 g
moles NaOH = (0.0136 g)/(40 g/mol) = 3.4 * 10^-4 mol
Molarity = (3.4 * 10^-4 mol)/(0.350 L) = 9.7143 *10^-4 mol/L
pOH = -log [9.7143 *10^-4 mol/L] = 3.01
moles NaOH = (0.0136 g)/(40 g/mol) = 3.4 * 10^-4 mol
Molarity = (3.4 * 10^-4 mol)/(0.350 L) = 9.7143 *10^-4 mol/L
pOH = -log [9.7143 *10^-4 mol/L] = 3.01
- Thu Nov 28, 2019 11:34 pm
- Forum: Bronsted Acids & Bases
- Topic: Weak Acid Strength
- Replies: 3
- Views: 208
Re: Weak Acid Strength
The greater the electronegativity of A in A-H for binary acids, across a period, the stronger the acid. The stronger the bond between A and H in A-H down a group, the weaker the acid. For oxoacids, the greater the number of oxygen atoms bound to the central atom, the stronger the acid. Oxoacids can ...
- Tue Nov 26, 2019 10:21 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Atoms in the same plane
- Replies: 4
- Views: 8996
Re: Atoms in the same plane
Also, if the structure contains trigonal planar or linear arrangement, the atoms involved will be in the same plane.
- Tue Nov 26, 2019 10:15 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Shapes
- Replies: 4
- Views: 301
Re: Shapes
If the central transition metal is bound to four ions or molecules, it is considered either square planar or tetrahedral. Generally, you don't assume that transition metals have lone pairs, which is why the shapes of coordination compounds aren't as extensive.
- Fri Nov 22, 2019 3:47 pm
- Forum: Ionic & Covalent Bonds
- Topic: Direction in anions increase in polarizability
- Replies: 3
- Views: 243
Re: Direction in anions increase in polarizability
Anions that have the greatest polarizability are those that are large and less electronegative. Cations that have greater polarization power are those that are smaller and have a high charge.
- Fri Nov 22, 2019 3:36 pm
- Forum: Dipole Moments
- Topic: How to tell
- Replies: 2
- Views: 251
Re: How to tell
Hydrogen bonds occur between a Hydrogen atom that is covalently bonded to a more electronegative atom (F, N, O) and another electronegative atom that has lone pairs. An ion-dipole bond is an electrostatic interaction between an ion and a molecule that has a dipole.
- Fri Nov 22, 2019 3:18 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: polarizability
- Replies: 9
- Views: 707
Re: polarizability
Larger molecules will have stronger London Dispersion Forces, which means that their boiling point will be high.
- Fri Nov 22, 2019 3:16 pm
- Forum: Dipole Moments
- Topic: Polar or Nonpolar
- Replies: 5
- Views: 446
Re: Polar or Nonpolar
Generally, you can look at the terminal toms to determine whether a structure will be polar or polar. With linear, trigonal planar, and tetrahedral structures, if the terminal atoms are the same the molecules will be nonpolar. But if the terminal atoms are different, the molecules will be polar. Reg...
- Fri Nov 22, 2019 3:12 pm
- Forum: Hybridization
- Topic: 2F.15
- Replies: 2
- Views: 131
Re: 2F.15
S character is directly proportional to the bond angle, so as the s character of a hybrid orbital increases the bond angle increases as well. In VSEPR, electrons are arranged to experience minimum repulsion. A high s character would increase the electron density across the central atom, so the bond ...
- Fri Nov 15, 2019 3:03 pm
- Forum: Resonance Structures
- Topic: Contribution of each structure?
- Replies: 4
- Views: 404
Re: Contribution of each structure?
I think that if you draw a Lewis Structure for the molecule, you will see that the atom that forms a double or triple bond with the central atom is contributing more electrons than the atom that is forming a single bond with the central atom.
- Fri Nov 15, 2019 3:01 pm
- Forum: Electronegativity
- Topic: Problem 3F.15
- Replies: 2
- Views: 176
Re: Problem 3F.15
Polarity plays a part in determining which structure has a higher boiling point. AsF3 has the higher boiling point because it is a polar molecule and thus has dipole-dipole interactions, while AsF5 is a nonpolar molecule with London dispersion forces. London dispersion forces are weaker than dipole-...
- Fri Nov 15, 2019 9:57 am
- Forum: Resonance Structures
- Topic: Orbitals
- Replies: 4
- Views: 343
Re: Orbitals
In terms of resonance Lewis Structure, electrons can be delocalized if they are free to move throughout the plane. This means that a double bond between Oxygen and Nitrogen, for example in the structure NO3, can be moved between any of the Oxygens while the remaining two form a single bond with Nitr...
- Fri Nov 15, 2019 9:51 am
- Forum: Lewis Structures
- Topic: Oxygen
- Replies: 9
- Views: 708
Re: Oxygen
Oxygen can form double, single, or triple bonds depending on the atom it is bonding with. Oxygen can form a triple bond with Carbon since the total number of valence electrons is 10 which means that there are 5 pairs of electrons. The only way that you can form five pairs of electrons for a CO Lewis...
- Fri Nov 15, 2019 9:46 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bent or angular?
- Replies: 2
- Views: 171
Re: Bent or angular?
I think you can also call it V shaped, although that sounds a little informal.
- Sun Nov 10, 2019 10:29 am
- Forum: Formal Charge and Oxidation Numbers
- Topic: Chlorine in center: electronegativity vs. formal charge
- Replies: 4
- Views: 270
Re: Chlorine in center: electronegativity vs. formal charge
I think it has to do with the fact that Chlorine's atomic radius is larger than that of Oxygen, so Oxygen's outermost electrons are held more tightly to the nucleus than Chlorine's outermost electrons. Also, the change in electronegativity is greater as you move up or down a group, than if you move ...
- Sun Nov 10, 2019 10:25 am
- Forum: Resonance Structures
- Topic: Unequal Contribution
- Replies: 3
- Views: 139
Re: Unequal Contribution
If you are referring to ionic bonds then you just draw both atoms separately with their complete octet and place them next to each other, without bonds. The anion will pull electrons away from the cation. If you are talking about resonance structures, the atom that is either double bonded or triple ...
- Sun Nov 10, 2019 10:09 am
- Forum: Ionic & Covalent Bonds
- Topic: Textbook question 2.25
- Replies: 2
- Views: 272
Re: Textbook question 2.25
I think for the first question, the answer is that the CN bond in H3CNH2 is longer than the CN bond in HCN because Carbon and Nitrogen share a triple bond in HCN while they only share a single bond in H3CNH2. Single bonds are longer than triple bonds because less electrons are being shared. For part...
- Thu Nov 07, 2019 11:38 am
- Forum: Formal Charge and Oxidation Numbers
- Topic: How to know where a double bond should go?
- Replies: 10
- Views: 1072
Re: How to know where a double bond should go?
If the central atom is in row one and row two, with the exception of the incomplete octet elements, you should calculate the number of valence electrons of the structure and then divide by two to find the number of electron pairs. When you add all of the atoms around the central atom and account for...
- Thu Nov 07, 2019 11:22 am
- Forum: Formal Charge and Oxidation Numbers
- Topic: Chlorine in center: electronegativity vs. formal charge
- Replies: 4
- Views: 270
Re: Chlorine in center: electronegativity vs. formal charge
Also since Oxygen is more electronegative that Chlorine, it can carry the -1 charge.
- Thu Nov 07, 2019 11:18 am
- Forum: Properties of Light
- Topic: Rydberg Equation
- Replies: 5
- Views: 337
Re: Rydberg Equation
The energy gaps between the energy levels of the Balmer series are much smaller than the gaps for the Lyman series, so that's why the frequency produced is also smaller.
- Fri Nov 01, 2019 10:52 am
- Forum: Resonance Structures
- Topic: 2C.3
- Replies: 2
- Views: 131
Re: 2C.3
I think the answer key is saying that all three are possible resonance structures since Chlorine is in the third row of the period table and has an expanded octet. Thus, in the first diagram chlorine can have eight electrons surrounding in, in the second diagram Chlorine can have 10 electron, and in...
- Fri Nov 01, 2019 10:44 am
- Forum: Einstein Equation
- Topic: 1B.15 help w/ part c
- Replies: 2
- Views: 202
Re: 1B.15 help w/ part c
For this question, you have to remember the equation: E photon - work function = Kinetic Energy (of electron). In this situation, the E=hv equation applies to the photon and finding the energy of the photon. But to find the energy of the emitted electron, you need to use KE=(1/2)m(v)^2.
- Thu Oct 31, 2019 7:40 pm
- Forum: Resonance Structures
- Topic: Homework 2C 1
- Replies: 3
- Views: 174
Re: Homework 2C 1
Another thing you could do is calculate the total number of valence electrons for each molecule. If the total number is an odd number, the specie is a radical.
- Thu Oct 31, 2019 7:36 pm
- Forum: Lewis Structures
- Topic: Sulfur Dioxide
- Replies: 2
- Views: 123
Re: Sulfur Dioxide
If the question asks you to draw two resonance structures for Sulfur Dioxide, I think you should do both. But if it just asks for the Lewis structure, draw the one with the lowest formal charge.
- Mon Oct 28, 2019 2:16 pm
- Forum: Lewis Structures
- Topic: Lone Pairs
- Replies: 10
- Views: 397
Re: Lone Pairs
Lone pairs are pairs of electrons in the valence shell that are not shared with another atom through bonding. The lone pairs in a lewis structure are the pairs of electrons that are drawn as dots around the atom. For example, in NH3, N has one lone pair of electrons because it shares six with the th...
- Sat Oct 26, 2019 12:11 pm
- Forum: Einstein Equation
- Topic: 1B. 5
- Replies: 4
- Views: 284
Re: 1B. 5
You should convert the value of electron volts to Joules, and then write the equation of E=hv as E=(hc)/(wavelength). Then you can rewrite this equation as wavelength = (hc)/(E) and solve.
- Thu Oct 24, 2019 4:41 pm
- Forum: Lewis Structures
- Topic: lewis drawings
- Replies: 8
- Views: 430
Re: lewis drawings
I think there are some calculations you can do to tell which arrangement is the most stable. So, I think he might have us draw the most stable configuration.
- Wed Oct 23, 2019 9:35 pm
- Forum: Ionic & Covalent Bonds
- Topic: Double Bond vs Single Bond Length
- Replies: 6
- Views: 516
Re: Double Bond vs Single Bond Length
The length of the bonds is in part based on the distance between two atoms. Atoms that form double bonds with one another have a stinger attraction between them which means they are pulled closer together than if they were connected by a single bond. Thus, the distance between them is smaller which ...
- Tue Oct 22, 2019 11:53 pm
- Forum: DeBroglie Equation
- Topic: Protons and Electrons
- Replies: 4
- Views: 236
Re: Protons and Electrons
Yes, I think that if a proton and neutron had the same wavelength, they would have the same velocity. You can use De Broigle's to determine that wavelength = h/p which is the same as: wavelength = h/mv. v=h/(m*wavelength) and the mass of a proton and neutron are the same. If the wavelength is also t...
- Tue Oct 22, 2019 11:49 pm
- Forum: Properties of Electrons
- Topic: 1A.11
- Replies: 2
- Views: 252
Re: 1A.11
The Lyman series is a spectrum series that produces ultraviolet emission lines from any energy level greater than or equal to 2 (n>=2) to energy level of one (n=1). The Balmer series produces emission lines from any energy level greater than or equal to 3 (n>=3) to energy level of two (n = 2). The a...
- Sat Oct 19, 2019 10:31 pm
- Forum: Photoelectric Effect
- Topic: Chem Video Module-test, Joule Conversion
- Replies: 2
- Views: 123
Re: Chem Video Module-test, Joule Conversion
You can convert the kJ to J by multiplying the energy by 10^3 joules. And the formula for kinetic energy is KE = (1/2)(mv^2) which can be used to solve this problem.
- Thu Oct 17, 2019 3:35 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Heisenberg's Indeterminacy Equation
- Replies: 3
- Views: 161
Re: Heisenberg's Indeterminacy Equation
Yes it's mostly used to find the indeterminancy in the position or the momentum. But you can use the value you calculated for the indeterminancy of the momentum to solve for the indeterminancy of the velocity of that particle.
- Thu Oct 17, 2019 12:04 pm
- Forum: DeBroglie Equation
- Topic: 1B15
- Replies: 3
- Views: 202
Re: 1B15
Yes, to get the wavelength of the electron you can't use c = 3 * 10^8 because it is the speed of light, thus the speed of a photon. The question has specified the velocity of the electron, so you have to use this velocity in De Broglie's equation.
- Thu Oct 17, 2019 11:59 am
- Forum: Einstein Equation
- Topic: Units for E [ENDORSED]
- Replies: 4
- Views: 650
Re: Units for E [ENDORSED]
Energy in the situations that discuss light is considered as J/photon. If the energy is provided in eV (electron Volts), you will have to convert it to J/photon. The conversion factor is about 1 eV = 1.6 *10^-19 J/photon.
- Thu Oct 17, 2019 11:54 am
- Forum: Properties of Light
- Topic: Rydberg's Equation [ENDORSED]
- Replies: 4
- Views: 167
Re: Rydberg's Equation [ENDORSED]
You use Rydberg's equation when the question talks about an electron transitioning from one energy level to another. Generally, you will use it to sole for the final or the initial energy level, based on the information that is provided. Sometimes, you will also be asked to calculate the frequency o...
- Sat Oct 12, 2019 12:36 pm
- Forum: Photoelectric Effect
- Topic: 1B. 21
- Replies: 2
- Views: 117
Re: 1B. 21
You should use the equation, wavelength = (plank's constant)/(mass x velocity). You can convert the mass of the baseball into grams and then just plug in the rest of the information.
- Sat Oct 12, 2019 12:30 pm
- Forum: Einstein Equation
- Topic: Diffraction Pattern
- Replies: 3
- Views: 269
Re: Diffraction Pattern
Constructive interference refers to waves that are in phase. This means that you can add them to give rise to a larger wave. Destructive interference refers to waves that are out of phase, which means that you can subtract the waves' peaks. If both waves are at their highest peak at a given moment, ...
- Thu Oct 10, 2019 11:48 am
- Forum: Properties of Light
- Topic: Formulas
- Replies: 2
- Views: 118
Re: Formulas
Yes, both equations will essentially give you the same result. Dr. Lavelle suggested using the H-atom equation because it really helps you see whats going on during the transitions, and how the energy is affected. This will allow you to understand the fact that energy is emitted (and thus lost) when...
- Thu Oct 10, 2019 11:42 am
- Forum: Significant Figures
- Topic: Range of Sig Figs
- Replies: 8
- Views: 479
Re: Range of Sig Figs
My TA said to write each calculation answer with at least four decimal places after the decimal point so that when you do further calculations, you have the most precise value from before. But the final answer should be written with the correct number of significant figures.
- Thu Oct 10, 2019 11:40 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Test 1 [ENDORSED]
- Replies: 107
- Views: 16834
Re: Test 1 [ENDORSED]
Make sure you bring a scientific calculator. Graphing calculators aren't allowed.
- Sun Oct 06, 2019 8:08 pm
- Forum: Ionic & Covalent Bonds
- Topic: Ionic vs Covalent
- Replies: 32
- Views: 43027
Re: Ionic vs Covalent
I think in general, ionic bonds are stronger than covalent bonds. But covalent bonds are stronger than ionic bonds in water.
- Thu Oct 03, 2019 11:15 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Dilution Calculations
- Replies: 3
- Views: 114
Re: G.23
The glucose and other sugars aren't relevant because there are no chloride ions present in glucose.
- Thu Oct 03, 2019 11:13 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Empirical and Molecular Formulas
- Replies: 13
- Views: 579
Re: Empirical and Molecular Formulas
Generally if you're given a molecular formula, you will be able to reduce the number of moles of each element by a certain factor. If you're given an empirical formula, you can't divide by a number to reduce the umber of moles of the element in the formula.