Search found 55 matches
- Fri Mar 16, 2018 12:13 am
- Forum: Zero Order Reactions
- Topic: 15.99
- Replies: 1
- Views: 418
Re: 15.99
F is linear because it shows the correlation between the initial rate (k[A] in the first order) and A itself. Since k is multiplied by [A] to get the initial rate and k is constant, this would be linear. G is very similar to part b, since they both have the concentration of A against time in the 0th...
- Fri Mar 16, 2018 12:05 am
- Forum: *Enzyme Kinetics
- Topic: 15.85
- Replies: 2
- Views: 858
Re: 15.85
We might be expected to know it because I think it was mentioned briefly during lecture. I think you just draw the formulas of the products with a line connecting them.
Ex. [CH3CHO] (Products are CH3 and CHO)
the structure is CH3 --- CHO
Ex. [CH3CHO] (Products are CH3 and CHO)
the structure is CH3 --- CHO
- Fri Mar 16, 2018 12:00 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: When to use standard potentials and when to balance
- Replies: 2
- Views: 391
Re: When to use standard potentials and when to balance
I think that if the products have oxygen or hydrogens that are unaccounted for in the reactants then you would have to balance by adding H+ and OH-. Especially when it says something along the lines of: 1. Balance each of the following skeletal equations by using oxidation and reduction half-reactio...
- Fri Mar 09, 2018 1:43 am
- Forum: Zero Order Reactions
- Topic: Definition of Reaction Rate
- Replies: 4
- Views: 753
Re: Definition of Reaction Rate
The formula for the reaction rate is the change in concentration over the change in time, so the units are M/s.
- Fri Mar 09, 2018 1:41 am
- Forum: First Order Reactions
- Topic: pseudoreactions [ENDORSED]
- Replies: 2
- Views: 376
Re: pseudoreactions [ENDORSED]
I think that it is mainly conceptual, since it was mentioned in class how we can use large concentrations of most reactants and a small concentration of one reactant to determine the effect of that one reactant on the rate law. That being said, you may be given k' and the large concentration of one ...
- Fri Mar 09, 2018 1:25 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.65 (b)
- Replies: 1
- Views: 262
Re: 15.65 (b)
https://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Gibbs_free_energy_graph I think this image sums up the reason why it is endothermic. Since the reverse reaction has a lower activation energy barrier than the forward reaction, the products have a higher Gibbs free energy than the reactan...
- Thu Mar 01, 2018 11:16 pm
- Forum: Zero Order Reactions
- Topic: Graph of Zero Order vs First and Second Order [ENDORSED]
- Replies: 3
- Views: 513
Re: Graph of Zero Order vs First and Second Order [ENDORSED]
Table 15.2 sums these up really well if you want to remember rate laws, integrated rate laws, plots, and half-life equations for every order of a reaction.
- Thu Mar 01, 2018 11:13 pm
- Forum: Zero Order Reactions
- Topic: Graph of Zero Order vs First and Second Order [ENDORSED]
- Replies: 3
- Views: 513
Re: Graph of Zero Order vs First and Second Order [ENDORSED]
We could be asked that and you would have to look at both the axes and the slope. A positive slope of k would mean a second order reaction (when the axes are 1/[A] vs time), a slope of negative k with axes of [A] vs time would be zero order, and a slope of k with axes ln[A] vs time would be a first ...
- Thu Mar 01, 2018 11:08 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Fractional Reaction Orders [ENDORSED]
- Replies: 2
- Views: 414
Re: Fractional Reaction Orders [ENDORSED]
This example is included in one of the sections we have to study for the next test, but we did not really cover it in class so I don't think that it will make up a significant portion of an exam. A fractional reaction order is when the order of a reaction is a fraction rather than an integer (we hav...
- Thu Feb 22, 2018 5:51 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Units of Reaction Rate
- Replies: 7
- Views: 924
Re: Units of Reaction Rate
Yes, because the rate is just a constant multiplied by the change in concentration over the change in time.
Concentration = mol/L
Rate = mol/(L* seconds)
Concentration = mol/L
Rate = mol/(L* seconds)
- Thu Feb 22, 2018 5:41 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.11d
- Replies: 1
- Views: 302
Re: 14.11d
H+ is added on the O2 side, so that the left side of the equation has 2H2O ------> O2 + 4H+. The reduction reaction is similar except O2 is added to 2H2O instead, to equal 4OH-. The oxygens should cancel with each other and leave you with the final equation.
- Thu Feb 22, 2018 5:33 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.107
- Replies: 1
- Views: 336
Re: 14.107
I think for that for this problem we assume that it is 1 mol since it is not otherwise specified.
- Fri Feb 16, 2018 12:20 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Question 14.19
- Replies: 4
- Views: 555
Re: Question 14.19
You can find the E˚anode value for Cu in Table 14.1.
- Fri Feb 16, 2018 12:18 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Question 14.19
- Replies: 4
- Views: 555
Re: Question 14.19
You would use the equation E˚cell = E˚cathode - E˚anode. The E˚cell is the value that you were given for the cell potential. The E˚anode is the potential of Cu is this case, because it is being oxidized. You would substitute these values and then solve for E˚cathode, which is the E˚of the metal.
- Fri Feb 16, 2018 12:03 am
- Forum: Balancing Redox Reactions
- Topic: 14.5 d
- Replies: 2
- Views: 353
Re: 14.5 d
I think that which side you normally add the water molecule on is determined by whether it is an oxidizing or reducing agent. Since P4 is both, the water molecule and OH- are added to the same side.
- Thu Feb 15, 2018 11:59 pm
- Forum: Balancing Redox Reactions
- Topic: 14.5 c
- Replies: 2
- Views: 390
Re: 14.5 c
Cr3+ loses 3 electrons in the reaction and MnO gains 2 electrons. So Cr3+ is the reducing agent and MnO is the oxidizing agent.
- Fri Feb 09, 2018 12:17 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.35 Explained
- Replies: 2
- Views: 350
Re: 9.35 Explained
Ideal gas = highest entropy
Vibrationally active (generating heat and disorder by moving) = second highest entropy
Not vibrationally active = least entropy
Vibrationally active (generating heat and disorder by moving) = second highest entropy
Not vibrationally active = least entropy
- Fri Feb 09, 2018 12:15 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Exercise 9.19
- Replies: 2
- Views: 439
Re: Exercise 9.19
The water is at 85 degrees in liquid form. To get the entropy of vaporization at 85 degrees, it must be turned into a gas. To do that, you have to heat it up to 100 degrees, change the state to a gas, and then cool the gas to 85 degrees. Heating it up to 100 degrees is done by using the entropy depe...
- Fri Feb 09, 2018 12:08 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Porous Disc vs Salt Bridge
- Replies: 2
- Views: 679
Re: Porous Disc vs Salt Bridge
I think a salt bridge connects two beakers for ion transfer while the porous disk allows ions to flow between two solutions in the same beaker.
- Fri Feb 09, 2018 12:05 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Derivation of isothermal entropy equations
- Replies: 1
- Views: 227
Re: Derivation of isothermal entropy equations
I think that the question means isothermal reversible expansion (nRT * ln(V2/V1) ) rather than irreversible expansion (-PdeltaV), because the temperature is constant and the amount of work is maximized.
- Thu Feb 08, 2018 11:58 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: conceptual entropy questions
- Replies: 5
- Views: 820
Re: conceptual entropy questions
Bromine has more electrons than Fluorine, so HBr has more possible states and disorder than HF. Also, Iodine is bigger than Bromine and has more electrons, so for the same reason it has more possible states and entropy.
- Thu Feb 08, 2018 11:54 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.43 Using Cp vs Cv?
- Replies: 2
- Views: 421
Re: 9.43 Using Cp vs Cv?
I think that it's because you use (5/2R) for C when you are talking about an ideal gas. In this question, they are talking about two amounts of water, both of which are liquid and are not ideal gases.
- Thu Feb 01, 2018 11:46 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Adiabatic Expansion of an Ideal Gas
- Replies: 1
- Views: 321
Re: Adiabatic Expansion of an Ideal Gas
Other words/phrases could be "assume ideal gas behavior", "the volume changes to/the final volume is", and "what is the change in entropy after the change in volume". It could also say adiabatic, but most problems of this type would refer to a change in volume and have ...
- Thu Feb 01, 2018 11:39 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9. 43
- Replies: 5
- Views: 613
Re: 9. 43
I don't think that you will have to calculate the value, it will usually be given to you (either by a table in the textbook or in the problem itself).
- Thu Feb 01, 2018 11:35 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Free Energy and Pressure
- Replies: 4
- Views: 482
Re: Free Energy and Pressure
I think it might be because free energy is dependent on entropy and enthalpy. Pressure has an inverse relationship with entropy, so if pressure goes up, entropy goes down. This would make the delta G more positive. If pressure decreases, entropy increases, and the delta G becomes more negative.
- Thu Jan 25, 2018 9:42 pm
- Forum: Calculating Work of Expansion
- Topic: Work Equation
- Replies: 3
- Views: 451
Re: Work Equation
If the pressure is varying and not constant, we use the equation w=-nRT(Vfinal/Vinitial). This is used for reversible isothermal expansion of an ideal gas.
- Thu Jan 25, 2018 9:36 pm
- Forum: Calculating Work of Expansion
- Topic: Work Equation
- Replies: 3
- Views: 451
Re: Work Equation
w=integralPdv is the same as w=-PdeltaV when there is a constant pressure P. In order words, the pressure stays the same and is not variable/changing. Since P is constant, it can be taken out of the integral so w=P X integral dv. The integral of dv equals Vinitial - Vfinal , so w=-P X (Vfinal - Vini...
- Thu Jan 25, 2018 9:24 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Chem 14A final
- Replies: 2
- Views: 401
Re: Chem 14A final
I thought that Professor Lavelle said Week 3 of this quarter, so maybe he'll say something about that tomorrow. Or it could be discussion next week.
- Wed Jan 17, 2018 11:03 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73a
- Replies: 3
- Views: 314
Re: 8.73a
It is -6*518 because you are forming six c-c bonds and the enthalpy of each one (with resonance) is 518. 3*837 refers to the enthalpy of c-c triple bonds broken. The enthalpy of bonds formed is subtracted from that of bonds broken.
- Wed Jan 17, 2018 10:53 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73a
- Replies: 2
- Views: 181
Re: 8.73a
Actually, you could put the h-c bonds in the broken bonds category and calculate the enthalpies, but you would have to put them in the bonds formed category as well, since the structure denoted on the right side of the equation is benzene (ring structure with carbon-carbon bonds and each carbon bond...
- Wed Jan 17, 2018 10:47 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73a
- Replies: 2
- Views: 181
Re: 8.73a
They are important, but if you broke those bonds, you'd just have to form them again. Each carbon is still has to be attached to a hydrogen, so it's easier to just keep those c-h bonds intact.
- Sat Jan 13, 2018 3:30 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess Law
- Replies: 4
- Views: 377
Re: Hess Law
Hess's Law states that enthalpy is a state function, so the total change in enthalpy after a specific reaction will be the same regardless of the number of steps taken to achieve that reaction or how much work was involved. If you add together the change in enthalpy at each step, you will get the to...
- Sat Jan 13, 2018 3:19 am
- Forum: Phase Changes & Related Calculations
- Topic: Using Bond Enthalpies for Resonance Structures
- Replies: 2
- Views: 193
Re: Using Bond Enthalpies for Resonance Structures
Update: The bond enthalpies are given in the table, but only for individual bonds, so if you wanted to form 6 bonds in a resonance structure (ex. benzene in 8.73), you would look at the table to find the resonance enthalpy value of 518, which you would then multiply by 6 to get the total enthalpy of...
- Sat Jan 13, 2018 3:13 am
- Forum: Phase Changes & Related Calculations
- Topic: Using Bond Enthalpies for Resonance Structures
- Replies: 2
- Views: 193
Re: Using Bond Enthalpies for Resonance Structures
You can find bond enthalpies for resonance structures in Table 8.7 in the textbook.
- Thu Dec 07, 2017 11:56 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Electron Withdrawing Power
- Replies: 3
- Views: 1000
Re: Electron Withdrawing Power
Electron withdrawing power is how good an atom is at pulling an electron away from another atom. It is like electronegativity in the sense that an atom with higher electronegativity and usually a smaller radius will have more electron-withdrawing power. Example: Fluorine has greater electron withdra...
- Sun Dec 03, 2017 6:10 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 11.73 Concept Question
- Replies: 3
- Views: 327
Re: 11.73 Concept Question
I think that a decrease in volume would inadvertently increase the concentration, since there are the same number of moles in a smaller volume of solvent. Thus, if the pressure on the left side increased, the reaction would move towards the right because reactions move in terms of higher to lower co...
- Sun Dec 03, 2017 6:02 pm
- Forum: Bronsted Acids & Bases
- Topic: How do I identify if a molecule is a Bronsted Acid or Base?
- Replies: 5
- Views: 1702
Re: How do I identify if a molecule is a Bronsted Acid or Base?
Another way to think of it is that the N atom in NH3 has a lone pair of electrons so it is an electron donor, which makes it a Lewis base. This also means that it is a proton acceptor and a Bronsted base. H2CO3 is the formula for carbonic acid, so that would be a Bronsted acid.
- Sun Nov 26, 2017 3:12 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Homework Question: 17.35 & 17.36
- Replies: 1
- Views: 203
Re: Homework Question: 17.35 & 17.36
I think b would form a chelating complex. I'm not sure about c because the bonds between a potential metal ion and the n atoms would have to be really long to form a complex, which I don't think will be favorable.
- Sun Nov 26, 2017 3:05 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Non-equilibirum situations
- Replies: 2
- Views: 468
Re: Non-equilibirum situations
Q uses the same equation as k, so you would calculate it in the same way as the equilibrium constant. If Q turns out to be less than K, there is a higher concentration of reactants and the reaction will go forward from reactants to products. If Q is greater than K, there is a higher concentration of...
- Sat Nov 18, 2017 10:24 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: HW Problem 4.15c
- Replies: 2
- Views: 404
Re: HW Problem 4.15c
This is also better in terms of formal charge, because each atom has a formal charge of zero when oxygen has a double bond/two single bonds, carbon has four bonds, and fluorine has a single bond.
- Sat Nov 18, 2017 10:09 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Chelating Complexes
- Replies: 5
- Views: 657
Re: Chelating Complexes
I think chelating complexes can be formed if ligands have lone pairs of electrons which they can use to form additional bonds with a metal ion. In 17.35, nitrogen has a lone pair left after bonding to the complex and the two hydrogen atoms (3 bonds, one lone pair). Therefore, this lone pair could be...
- Sat Nov 18, 2017 9:49 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Expanded Octet Rule and Coordination Compounds
- Replies: 1
- Views: 531
Re: Expanded Octet Rule and Coordination Compounds
I think that because Nickel has two electrons from the 4s subshell and 8 electrons from 3d, it can form 10 bonds in a coordination compound. It is similar to how Sulphur can form 6 bonds (2 electrons in 3s + 4 electrons in 3p) due to the expanded octet rule and its empty d-orbitals.
- Sat Nov 11, 2017 11:45 pm
- Forum: Empirical & Molecular Formulas
- Topic: Practice Midterm Question
- Replies: 4
- Views: 650
Re: Practice Midterm Question
Hi,
Is the answer by any chance C8H14O7? If it is, then you just need to find the number of moles of Carbon (grams/molar mass), divide it by the least number of moles, and multiply each answer by seven to get a whole number.
Is the answer by any chance C8H14O7? If it is, then you just need to find the number of moles of Carbon (grams/molar mass), divide it by the least number of moles, and multiply each answer by seven to get a whole number.
- Sat Nov 11, 2017 11:35 pm
- Forum: Ionic & Covalent Bonds
- Topic: Covalent Bonding: Polar and Non-polar
- Replies: 6
- Views: 1074
Re: Covalent Bonding: Polar and Non-polar
A non-polar covalent bond occurs when the atoms are sharing the electrons more equally, which happens if both atoms have similar electronegativity (ex. carbon and hydrogen). A polar covalent bond is when there is a significant difference in electronegativity between the two atoms (one of them is bet...
- Sun Nov 05, 2017 12:19 am
- Forum: Electronegativity
- Topic: Electronegativity related to solubility
- Replies: 2
- Views: 1011
Re: Electronegativity related to solubility
If there is a large difference in electronegativity (greater than 2) between two atoms, then they usually form an ionic bond. Atoms which form ionic bonds tend to have a more solid, lattice-like structure, and can dissolve more easily in water. Thus, higher difference in electronegativity = higher s...
- Sun Nov 05, 2017 12:10 am
- Forum: Lewis Structures
- Topic: Nitrogen Forms Four Bonds or Three?
- Replies: 1
- Views: 169
Re: Nitrogen Forms Four Bonds or Three?
I think that nitrogen prefers three bonds, as it has five valence electrons. However, it may not necessarily be limited to only three, and can form four covalent bonds if necessary to complete the octet.
- Sun Oct 29, 2017 5:13 pm
- Forum: Lewis Structures
- Topic: Sulfate Ion
- Replies: 4
- Views: 633
Re: Sulfate Ion
This is an anion and thus the compound has gained two more electrons. Suphur has 6 valence electrons, plus is sharing 6 electrons with oxygen. Therefore, the total is 12. Sulphur has two single bonds and two double bonds distributed between the 4 oxygen atoms. since each single bond shares 2 electro...
- Sun Oct 29, 2017 4:31 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Post-Module Q.18 [ENDORSED]
- Replies: 3
- Views: 465
Re: Post-Module Q.18 [ENDORSED]
You could use the uncertainty equation (delta p * delta x is greater than or equal to h/4pi) to find delta p by plugging in the error in the diameter for delta x. Using the result for delta p, divide that by the mass of the electron to find the uncertainty in the speed.
- Sat Oct 21, 2017 7:01 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Negative Energy Near the Nucleus [ENDORSED]
- Replies: 2
- Views: 489
Re: Negative Energy Near the Nucleus [ENDORSED]
I think it might be because electrons closer to the nucleus are in lower energy orbitals, so they have less potential energy. These electrons are also more stable than valence electrons (less likely to be shared/lost to another atom) which might also explain why they have lower energy.
- Sat Oct 21, 2017 6:27 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 2. 39 [ENDORSED]
- Replies: 5
- Views: 771
Re: 2. 39 [ENDORSED]
In order for the atom to be shown in its ground state, the number of arrows/the direction they are facing/how they are spaced must all be correct. For part a, the number of arrows (6 - the atomic number of Carbon) is correct, but the two arrows in 2p should be in two separate boxes and should be fac...
- Thu Oct 12, 2017 12:33 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectra Post Assessment Question 41
- Replies: 2
- Views: 564
Re: Atomic Spectra Post Assessment Question 41
Update: I watched the video for this topic, and it appears that using another equation, E=(-h X R)/n^2, is preferred for problems such as those posed in the first two parts of this problem. The equation which I used in the previous reply was actually derived from this equation by dividing both sides...
- Wed Oct 11, 2017 9:26 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectra Post Assessment Question 41
- Replies: 2
- Views: 564
Re: Atomic Spectra Post Assessment Question 41
I think you would use this equation: v=R((1/n1^2)-(1/n2^2)) v=frequency, which you will find after solving the equation R is a constant, and is equal to 3.29 X 10^15. As for n1 and n2, you take the two values of n which you are given in the question and plug them in. After plugging in R and the two ...
- Wed Oct 11, 2017 9:11 pm
- Forum: Photoelectric Effect
- Topic: Mass of an electron
- Replies: 4
- Views: 547
Re: Mass of an electron
Yes, mass is measured in kilograms for the photoelectric effect equation problems. This is indicated in Example 1.5 for analyzing the photoelectric effect in the textbook.
- Tue Oct 03, 2017 5:49 pm
- Forum: Balancing Chemical Reactions
- Topic: Help on HW question H1
- Replies: 5
- Views: 828
Re: Help on HW question H1
I think that it is because adding oxygen by itself would imply that it is an individual product of the reaction, which is different than saying it is part of the product copper oxide. You would be adding oxygen to the right side of the equation separately. Therefore, the equation has a different mea...
- Tue Oct 03, 2017 5:34 pm
- Forum: SI Units, Unit Conversions
- Topic: Formula Units [ENDORSED]
- Replies: 9
- Views: 1375
Re: Formula Units [ENDORSED]
I would follow it with 'formula units', just because an answer should have a measure of units afterward if possible. In other cases, this allows you to ensure that you are giving the right amount ex. 2 mL is different from 2L.