Search found 54 matches
- Fri Mar 16, 2018 3:54 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Finding E
- Replies: 3
- Views: 588
Re: Finding E
You can use either method of solving for E, but just make sure you stay consistent and don't combine the two. If you use cathode-anode, then don't flip the sign of the E for the anode.
- Fri Mar 16, 2018 3:48 pm
- Forum: Calculating Work of Expansion
- Topic: Conversion of R
- Replies: 7
- Views: 1048
Re: Conversion of R
It still depends on what you're solving for. If you're not given pressure and solving for pressure, then you would use 8.206 x 10-2 L·atm·K-1·mol-1, because that would cancel out so that you get atmospheres for the units of pressure. If you're not given pressure and trying to solve for work, then yo...
- Fri Mar 16, 2018 3:42 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Predicting if a metal will be dissolved in a solution
- Replies: 2
- Views: 650
Re: Predicting if a metal will be dissolved in a solution
A metal will dissolve in the solution if the standard cell potential of the redox reaction is positive, meaning the standard Gibbs free energy is negative because that would mean the reaction is spontaneous.
Re: Carbonyls
A carbonyl is a functional group with C double bonded to O. Aldehydes and ketones are both carbonyls. Aldehyde has the carbonyl group bonded with either one or two hydrogens. Ketone has the carbonyl group bonded with two carbons.
- Sat Mar 10, 2018 8:26 pm
- Forum: Balancing Redox Reactions
- Topic: 14.5 Part a
- Replies: 4
- Views: 605
Re: 14.5 Part a
Redox reactions in basic solutions don't need to have OH- in the final balanced equation if they cancel out. OH- would only show up in the final equation when the two sides of the equation don't require the same amount of OH- when balancing.
- Sat Mar 10, 2018 8:22 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Example 15.8 [ENDORSED]
- Replies: 2
- Views: 472
Re: Example 15.8 [ENDORSED]
That information should be given on a test because we are not allowed to use graphing calculators.
- Sat Mar 10, 2018 8:19 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Equations: ln Q vs. log Q [ENDORSED]
- Replies: 5
- Views: 1742
Re: Equations: ln Q vs. log Q [ENDORSED]
They are not interchangeable. If you want to use log instead of ln then remember that ln(X) = 2.303log(X). If an equation has ln, then you should use ln. If an equation has log, then you should use log. This way it's easier to avoid error.
- Sun Mar 04, 2018 3:36 pm
- Forum: First Order Reactions
- Topic: Deriving these Equations
- Replies: 7
- Views: 778
Re: Deriving these Equations
Unless it's given [A]0 then [A] is just the concentration at the given time, not the initial concentration.
- Sun Mar 04, 2018 3:27 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Rate Law
- Replies: 2
- Views: 378
Re: Rate Law
When determining the rate laws of a reaction with multiple reactants, we use the reactant of the smallest concentration because that reactant will determine the rate of the overall reactions. Since the other reactants are in excess, they don't really affect the rate of the reaction. This way, we are...
- Sun Mar 04, 2018 3:22 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Independent of rate
- Replies: 3
- Views: 548
Re: Independent of rate
A reaction rate is independent when an increase of concentration doesn't affect of the rate of the reaction. This type of reaction would be a zero order. When given a graph, it is when the graph of concentration vs. time is linear.
- Sun Mar 04, 2018 3:19 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 0 order
- Replies: 6
- Views: 2570
Re: 0 order
A 0 order reaction would have a linear graph of concentration vs time. The slope of the line equals -k, and the y-intercept equals the initial concentration.
- Sun Mar 04, 2018 3:09 pm
- Forum: First Order Reactions
- Topic: orders of a reaction
- Replies: 3
- Views: 465
Re: orders of a reaction
You would have to look at the experimental data to find the order or reactions because the chemical equation doesn't tell you anything about the order. You would have to look at a table of data and solve for the order or look at different graphs and determine what the order is.
- Fri Feb 23, 2018 2:53 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Standard State with Nernst Equation
- Replies: 4
- Views: 601
Re: Standard State with Nernst Equation
It should be safe to assume that that it's in standard state, if no additional information about concentration or partial pressure is given.
- Fri Feb 23, 2018 2:37 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.111
- Replies: 3
- Views: 1687
Re: 11.111
When you take the natural log (ln) of both sides, you get rid of e because ln is equal to loge. Once you do this, you just solve for deltaG.
- Fri Feb 23, 2018 2:31 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Spontaneous Directions [ENDORSED]
- Replies: 7
- Views: 1046
Re: Spontaneous Directions [ENDORSED]
You want the standard potential of the reaction to be positive because that would make deltaG be negative (deltaG=-nFE) and k be greater than 1.
- Sat Feb 17, 2018 1:20 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Delta G=-nFE
- Replies: 5
- Views: 2684
Re: Delta G=-nFE
The n in the equation is the moles of electrons transferred. To find the electrons transferred, you'll have to write the two half-reactions and balance them.
- Sat Feb 17, 2018 1:14 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidizing vs. Reducing Agent
- Replies: 7
- Views: 855
Re: Oxidizing vs. Reducing Agent
The oxidizing agent is the substance that's getting reduced. The reducing agent is the substance that's getting oxidized.
- Sat Feb 17, 2018 1:09 pm
- Forum: Balancing Redox Reactions
- Topic: 14.1
- Replies: 3
- Views: 413
Re: 14.1
The charges should equal each other on each side. If they don't equal each other, then the redox reaction is not balanced.
- Sat Feb 10, 2018 4:57 pm
- 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
A salt bridge connects two separate beakers while a porous disc is a separation between two solutions in the same beaker. They both serve the same purpose of allowing ions to flow between the two solutions.
- Thu Feb 08, 2018 9:48 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Spontaneity
- Replies: 9
- Views: 1291
Re: Spontaneity
A negative delta G is spontaneous because there is a lost in free energy, and going from high free energy to low free energy is favorable. A positive delta G is nonspontaneous because there is a gain in free energy. For that to happen, energy would have to be added, making the reaction nonspontaneous.
- Thu Feb 08, 2018 9:20 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: interpretation of galvanic cell diagram
- Replies: 2
- Views: 363
Re: interpretation of galvanic cell diagram
||represents a salt bridge. | separates the elements of different phases. A comma separates differences in oxidation states.
- Sun Feb 04, 2018 3:12 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Heating Capacity
- Replies: 3
- Views: 411
Re: Heating Capacity
The heat capacities, 3/2*R=Cv and 5/2*R=Cp, relate to when there is an ideal gas at constant volume/pressure. For this problem, the water is in its liquid state, so the heat capacity of an ideal gas would not apply.
- Sun Feb 04, 2018 3:09 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Delta G not
- Replies: 5
- Views: 794
Re: Delta G not
∆G° is the standard free energy of the reaction. This means that the reactants and products are at their standard state: 1 atm for gases and 1 M of aqueous solutions, typically at 25°C. ∆G is just the free energy of the reaction that doesn't have to be under the standard state conditions.
- Sun Feb 04, 2018 3:00 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.25
- Replies: 5
- Views: 586
Re: 9.25
The 6 is raised to avogadro's number because the question is asking for the molar entropy, so that's the entropy per mole of SO2F2.
- Mon Jan 29, 2018 8:45 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.13
- Replies: 3
- Views: 487
Re: 9.13
The question says to assume ideal gas behavior, but no moles of gas are given, so it's safe assume that there's only one mole of N2 gas to make the calculations easier.
- Mon Jan 29, 2018 8:43 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Entropy vs. Enthalpy [ENDORSED]
- Replies: 2
- Views: 516
Re: Entropy vs. Enthalpy [ENDORSED]
Change in enthalpy, delta H, is equal to heat loss/gained under constant pressure. If there's constant pressure than in the equation delta S=q/T, q can be replaced by delta H. Under these condition, change in entropy would be directly proportional with change in entropy.
- Mon Jan 29, 2018 8:38 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.13 [ENDORSED]
- Replies: 1
- Views: 276
Re: 9.13 [ENDORSED]
If you look on Lavelle's website, there is a link to solution manual errors. For question 9.13, R should be multiplied by 5/2, so the equation would be S=nCln(T2/T1). The work in the solution manual is just wrong for that part.
- Thu Jan 18, 2018 9:15 pm
- Forum: Calculating Work of Expansion
- Topic: Question 8.27 Conversion
- Replies: 2
- Views: 313
Re: Question 8.27 Conversion
The 101.325 J.L-1.atm-1 is just the conversion from L.atm to Joules. On the test, this number will be given to you on the constant and equations sheet.
- Thu Jan 18, 2018 9:22 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Self-Test 8.5B
- Replies: 3
- Views: 394
Re: Self-Test 8.5B
If the system is heated up, then the work is positive because work has to be done on the system for it to take in heat. If the system loses heat, then the work is negative because the system did work to lose heat.
- Thu Jan 18, 2018 9:10 am
- Forum: Calculating Work of Expansion
- Topic: Homework 8.11
- Replies: 8
- Views: 865
Re: Homework 8.11
Part b does more work because a reversible process is the maximum output for the process because it's done slowly, so less energy is lost. In a irreversible process, more heat is lost because it's done quickly.
- Fri Jan 12, 2018 8:11 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heating Curve
- Replies: 3
- Views: 379
Re: Heating Curve
There is a plateau even though heat is being added because all the heat being added will go to breaking the bonds first, before the temperature will start changing. While the bonds are breaking, the temperature isn't changing because the energy added will only go to breaking the bonds.
- Fri Jan 12, 2018 8:05 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat capacity
- Replies: 4
- Views: 361
Re: Heat capacity
Heat capacity is an extensive property because it does depend on the amount of substance. However, specific heat capacity is per gram of substance and molar heat capacity is per mole of substance, so they are intensive properties.
- Thu Jan 11, 2018 10:18 am
- Forum: Phase Changes & Related Calculations
- Topic: General Question
- Replies: 3
- Views: 346
Re: General Question
Heat, q, refers the amount of work the system can do. Enthalpy, H, refers to the amount of heat absorbed or released by the system. Heat can equal enthalpy if under constant pressure.
- Wed Dec 06, 2017 6:11 pm
- Forum: Conjugate Acids & Bases
- Topic: Identifying Acid / Base
- Replies: 1
- Views: 234
Re: Identifying Acid / Base
Since the pH is related to the amount of concentration of hydronium ions, the pH of an acid in a solution could be greater than 7. If the concentration of acid is very little than the solution would have a higher pH because there wouldn't be much acid in the solution.
- Tue Dec 05, 2017 9:49 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: How to tell if bidentate, tri- etc
- Replies: 5
- Views: 3635
Re: How to tell if bidentate, tri- etc
To see if a ligand if a bidentate, tridentate, or hexadentate, you look to see how many lone pairs there are one different atoms. The best way to see this is by drawing a lewis structure. For example, a bidentate would have two lone pairs, each on different atoms.
- Thu Nov 30, 2017 8:58 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.13
- Replies: 3
- Views: 304
Re: 11.13
It wouldn't be wrong to use concentration, but normally you would use partial pressure if all the reactants and products are gas. If you were given concentrations, then you could just use the concentrations.
- Thu Nov 30, 2017 8:51 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: P4
- Replies: 2
- Views: 480
Re: P4
P4 is non polar. Since all of the atoms in the bond are the same, there is no polarity in the bonds between the atoms, so the entire molecule is non polar.
- Sun Nov 19, 2017 2:04 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Polydentate
- Replies: 3
- Views: 249
Re: Polydentate
A polydentate ligand has more than one donor atom, so it can occupy more than one binding site at a time.
- Sun Nov 19, 2017 1:57 pm
- Forum: Naming
- Topic: (H2O) or (OH2)? [ENDORSED]
- Replies: 3
- Views: 1802
Re: (H2O) or (OH2)? [ENDORSED]
The reason OH2 is used is to how that the oxygen is being bonded to the central atom not the hydrogen. It would be written OH2 when attached to the right of the central atom and written H2O when attached to the left of the central atom.
- Sun Nov 12, 2017 2:48 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR Notation
- Replies: 3
- Views: 603
Re: VSEPR Notation
If two molecules have the same VSEPR formula, then they have the same shape. From the formula, we can only estimate what the bond angle is, though they may vary a bit for different molecules. We can't find out the true bond angle with the VSEPR formula, that has to be found experimentally.
- Sun Nov 12, 2017 11:14 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular shapes
- Replies: 2
- Views: 181
Re: Molecular shapes
Polar molecules have a molecular shape where the dipole moments don't cancel each other. Nonpolar molecules have molecular shapes there the dipole moments do cancel each other out. It also depends on what the atoms are because dipole moments are different, they might not always cancel out.
- Sun Nov 05, 2017 3:41 pm
- Forum: Lewis Structures
- Topic: Biradical vs Lone Pair
- Replies: 3
- Views: 499
Re: Biradical vs Lone Pair
A biradical is a molecule with two unpaired electrons and a lone pair is when two valence electrons that aren't part of bonding for an atom. For a biradical, the two unpaired electrons on usually on different atoms.
- Sat Nov 04, 2017 4:23 pm
- Forum: Resonance Structures
- Topic: Resonance and energy
- Replies: 2
- Views: 271
Re: Resonance and energy
Resonance structures are delocalized, which means the bond pair is distributed to multiples pairs of atoms and not just one pair of atoms. Because of this, the energy is lower which also makes the molecule more stable.
- Sat Oct 28, 2017 11:25 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: "No two electrons are the same..." [ENDORSED]
- Replies: 5
- Views: 1126
Re: "No two electrons are the same..." [ENDORSED]
This only refers to electrons in individual atoms. For each atom, no two electrons can have all four quantum numbers be the same due to the Pauli exclusion principle. Because of this, each orbital has only two electrons, and they have opposite spin.
- Sat Oct 28, 2017 11:15 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Zn1+
- Replies: 2
- Views: 961
Re: Zn1+
Zn1+ has that configuration because when you're writing the electron configuration for an cation you take electrons from the outermost orbital. In this case, you're taking the electron from the 4s orbital.
- Fri Oct 20, 2017 2:51 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: 4s to 3d
- Replies: 5
- Views: 698
Re: 4s to 3d
In the cases where 4s comes before 3d, it just has to do with the charge of the atom. When n increase, the subshells are closer together and the charge of the atom can cause 4s to be filled first like with potassium and calcium.
- Fri Oct 20, 2017 2:47 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: 2.19: Part d, Subshells
- Replies: 1
- Views: 271
Re: 2.19: Part d, Subshells
For part b and c they ask about the values of m l which is the orbital. Those values are from -l,-l+1,..., l. The orbital is the different possible shapes for each subshell. Subshells refer to l. l=0 is the s-orbital, l=1 is the p orbital, and l=3 is the d orbital. That's why for part d, n=4 so l=3,...
- Sat Oct 14, 2017 10:24 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectroscopy vs Photoelectric Effect
- Replies: 3
- Views: 591
Re: Atomic Spectroscopy vs Photoelectric Effect
In atomic spectroscopy, the electron doesn't completely leave the atom because it is still being pulled towards the positively charged nucleus of the atom. Also with the photoelectric effect, the detector that detects the speed of the ejected electron is slightly positively charged so it'd draw the ...
- Thu Oct 12, 2017 11:20 am
- Forum: Properties of Light
- Topic: HW 1.25b
- Replies: 2
- Views: 357
Re: HW 1.25b
First, you would convert the 5.00 mg to atoms. Once you have that, you just multiply by the Joules per atoms that you got from part a. The atoms of sodium should cancel leaving you with just Joules.
- Thu Oct 12, 2017 11:16 am
- Forum: Properties of Light
- Topic: Photons and Photoelectric effect
- Replies: 3
- Views: 324
Re: Photons and Photoelectric effect
Photons don't have mass so you wouldn't really be able to use the equation p=mv. Sorry, the first time I read that as proton.
- Thu Oct 05, 2017 6:25 pm
- Forum: Limiting Reactant Calculations
- Topic: m11
- Replies: 7
- Views: 1038
Re: m11
When you use 5.77g P4 again in the second question, it's just to convert it to P4O6 because you want to find how much P4O6 came out of the first equation to use in the second equation. You would use the mole ratio so that 1 mole P4 equals 1 mole of P4O6, and then solve for how much oxygen you need, ...
- Wed Oct 04, 2017 11:27 pm
- Forum: Limiting Reactant Calculations
- Topic: M. 19
- Replies: 1
- Views: 321
Re: M. 19
You need to solve for the masses of carbon, hydrogen, and nitrogen there is based on the masses of CO2, H2O, and N2 given, and then add up the masses of C, H, and N. You should see that the sum of the masses for C, H, and N isn't equal to the mass of caffeine given in the question, so that means som...
- Tue Oct 03, 2017 10:13 pm
- Forum: Limiting Reactant Calculations
- Topic: m11
- Replies: 7
- Views: 1038
Re: m11
There are two equations to the problem. For the first equation, if you use all the P4 and calculate the amount of O2 needed, you get 4.47g O2. Since you originally had 5.77g O2, you still have 1.30g O2 left. For the second equation, you use 5.77g P4 again but in the context of the second equation to...
- Tue Oct 03, 2017 11:39 am
- Forum: Molarity, Solutions, Dilutions
- Topic: G25 homework problem
- Replies: 7
- Views: 831
Re: G25 homework problem
I don't still don't really get how to do this problem. Can someone explain how to do it more?