Search found 71 matches
- Thu Mar 15, 2018 10:42 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 16.69: Pre exponential factor in catalysis
- Replies: 1
- Views: 312
Re: 16.69: Pre exponential factor in catalysis
I think it's just that when doing the problems, we can assume A is the same and that they cancel out, but in real life A is actually different.
- Thu Mar 15, 2018 10:35 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.17 [ENDORSED]
- Replies: 3
- Views: 484
Re: 14.17 [ENDORSED]
David Zhou 1L wrote:For this though, couldn't the Fe be oxidized into Fe2+? Why does it have to be Fe2+ oxidized into Fe3+?
I believe it's because we have FeCl2 as a reactant, which would imply the iron on the reactants side must be Fe2+.
- Thu Mar 15, 2018 10:33 pm
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: 14.55
- Replies: 3
- Views: 862
Re: 14.55
The +1.23 volts is for the oxidation reaction. The standard reduction potential is actually -1.23 V so the solution manual is correct. And you want to get the most negative value of E because in an electrolytic cell, the reaction is non spontaneous. Thanks, I see what you mean. But if the +1.23 vol...
- Thu Mar 15, 2018 9:06 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.91
- Replies: 1
- Views: 318
Re: 14.91
I was confused at this too, but I think it's asking about more about the salt bridge. The external circuit is where the flow of electrons happens, but we need electrolytes to complete the circuit. Since e- go from anode to cathode, electrolytes will then go from cathode to anode. https://www.chem.wi...
- Thu Mar 15, 2018 9:03 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57
- Replies: 1
- Views: 378
Re: 8.57
Did you mean standard enthalpy? We aren't able to directly do Hfproducts-Hfreactants because the question gives values for heat of combustion (not formation).
- Thu Mar 15, 2018 12:38 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.13 c
- Replies: 1
- Views: 248
Re: 14.13 c
The anode should be written on the left and the cathode should be on the right. In this reaction, Cl2 is being reduced, so the half reaction with Cl2 and Cl- is on the right.
- Thu Mar 15, 2018 2:35 am
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: 14.55
- Replies: 3
- Views: 862
14.55
The solution manual states that the cathode should have the least negative E and the anode should have the most negative E. However, it then says that the cathode, which is Ni2+, has a E of -0.23 V while the anode has a E of +1.23 V. Isn't this contradictory? Also, if the point is to make a cell wit...
- Wed Mar 14, 2018 10:13 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.17 [ENDORSED]
- Replies: 3
- Views: 484
14.17 [ENDORSED]
"Write balanced half-reactions for the redox reaction of an acidified solution of potassium permanganate and iron(II) chloride." How do we know to use Fe3+ + e- --> Fe2+ as the half rxn? Why not Fe2+ + 2e- --> Fe or something else?
- Wed Mar 14, 2018 12:34 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73 b and 8.75 c
- Replies: 1
- Views: 282
Re: 8.73 b and 8.75 c
In 8.75c, 4 C-H bonds are still broken. It's just written as 1 in the answer key because it cancels with the 3 C-H bonds formed in CH3Cl in the products.
- Tue Mar 13, 2018 10:44 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: elementary reaction
- Replies: 1
- Views: 266
Re: elementary reaction
We can assume the rate law using the coefficients in an elementary reaction only when proposing a reaction mechanism.
- Tue Mar 13, 2018 10:41 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalyst
- Replies: 2
- Views: 448
Re: Catalyst
Catalysts increase the rate of reaction by providing a new pathway for the reaction with a lower activation energy.
- Wed Mar 07, 2018 7:54 pm
- Forum: First Order Reactions
- Topic: 15.37C
- Replies: 5
- Views: 1494
Re: 15.37C
Nope you're good, if you divide both the initial and final mass by liters to find the concentrations of each, the liters will cancel out when you plug into [A]t=[A]0e-ktanyway.
- Wed Mar 07, 2018 2:55 pm
- Forum: First Order Reactions
- Topic: 15.27 C & D
- Replies: 1
- Views: 339
Re: 15.27 C & D
Since 15% isn't a multiple of 1/2, we have to do more calculations. We first solve for k using the 1st-order equation t 1/2 =0.693/k. If we end up with 15% of the initial concentration, we can say that [A] t =.15, and [A] 0 =1. By rearranging the equation ln[A] t = -kt + ln[A] 0 , we can solve for t...
- Tue Mar 06, 2018 5:01 pm
- Forum: Second Order Reactions
- Topic: 15.19 Part C
- Replies: 3
- Views: 554
Re: 15.19 Part C
So we have rate=k[A][B] 2 [C] 2 . The initial rate is given in units (mmolA)*L -1 *s -1 . Each of the concentrations are given in units mmol*L -1 . When we multiply all the concentrations of reactants together, we end up with units mmol*L -5 . By dividing (mmolA)*L -1 *s -1 by mmol 5 *L -5 to find u...
- Tue Mar 06, 2018 4:52 pm
- Forum: General Rate Laws
- Topic: 15.3
- Replies: 4
- Views: 593
Re: 15.3
I think the reaction rate is always written as positive. In the solutions manual, it makes up for the decreasing concentration by adding a negative sign in the equation (-∆conc/∆time). So, you would just say that the reactant concentration is decreasing at (input rate here).
- Tue Mar 06, 2018 3:34 pm
- Forum: General Rate Laws
- Topic: Coefficients affecting order & molecularity
- Replies: 2
- Views: 384
Coefficients affecting order & molecularity
Are the coefficients of the reactants correlated to the order/rate law of a reaction? Or is the rate law only dependent on the concentrations and rate constant given? I guess in other words, can we determine the order/rate law solely by looking at an equation and the coefficients? For example, in 15...
- Thu Mar 01, 2018 10:20 am
- Forum: General Rate Laws
- Topic: 15.9
- Replies: 5
- Views: 647
Re: 15.9
The reaction rate = ∆concentration of reactants or products/∆time, which would give you units of molarity/time, which is usually mol*L -1 seconds -1 . If you look at the the general equation for the differential rate law, it is rate = k[R] -1 . For example, for a first order reaction, the rate = k[A...
- Wed Feb 28, 2018 10:17 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.9
- Replies: 3
- Views: 537
Re: 15.9
So to calculate the reaction rate, we usually do reaction rate = ∆concentration of reactants or products/∆time. This would give you Molarity/Time, which is usually mol*L -1 seconds -1 . If you look at the the general equation for the differential rate law, it is rate = k[R] n . Part a of this questi...
- Wed Feb 28, 2018 9:34 pm
- Forum: First Order Reactions
- Topic: 15.15
- Replies: 4
- Views: 585
Re: 15.15
Doesn't ln[A] vs. time have to be linear to assume it's first order? Or in this case we're just saying that in this question, we can see that multiplying by a factor of 1.2 will increase the rate by a factor of 1.2, which aligns with the first order differential rate law rate = k[A].
- Thu Feb 22, 2018 11:34 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.57
- Replies: 2
- Views: 437
Re: 14.57
The values are given to you for pH=7 at the beginning of the problem set!
- Thu Feb 22, 2018 11:30 pm
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: 14.55 homework
- Replies: 1
- Views: 535
14.55 homework
In this question, how do we know one of the half reactions should include water?
- Thu Feb 22, 2018 2:16 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.33b homework
- Replies: 3
- Views: 486
14.33b homework
Where do we get the equation 3Tl+ --> 2Tl + Tl 3+ from? Wouldn't Tl+ e- --> Tl also be Tl+ "disproportionating" in solution?
- Wed Feb 21, 2018 9:54 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15a cell drawing
- Replies: 1
- Views: 265
14.15a cell drawing
For the anode, why is it written in the order Ag(s)|AgBr(s)|Br-(aq)? Is this because Br- is an ion/aqueous so it's written closest to the salt bridge? If solids are written on the outside, why is Ag(s) written before AgBr? Is this because it's written from reactants to products? But if it's written ...
- Wed Feb 21, 2018 8:59 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.13B
- Replies: 3
- Views: 389
Re: 14.13B
How do we know I2 isn't conducting?
- Wed Feb 14, 2018 10:11 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Zero Degeneracy
- Replies: 2
- Views: 487
Re: Zero Degeneracy
Degeneracy can't be 0, but entropy can be 0 if it's a perfect crystal (aka when W=1 because there is only one state for it to exist in). When W=1, ln(1)=0, making S equal to 0.
- Wed Feb 14, 2018 9:48 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.101 sign?
- Replies: 1
- Views: 283
Re: 9.101 sign?
I believe the signs don't change due to the way the question is worded. We don't assign the interior or the exterior to be the "system," so we don't say the ∆S of the "surroundings" is -∆S of the "system." The question simply asks which is greater: the change in entropy...
- Wed Feb 14, 2018 9:12 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.47 Isothermal Irreversible Free Expansion
- Replies: 3
- Views: 575
Re: 9.47 Isothermal Irreversible Free Expansion
Since entropy is a state function, the pathway taken from final to initial doesn't matter. ∆Ssys for irreversible and reversible should be the same.
- Wed Feb 07, 2018 7:13 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.63
- Replies: 4
- Views: 466
Re: 9.63
In order for decomposition of compounds into their elements to be stable, ∆G of decomposition would have to be (+) aka it won't spontaneously break down. Since decomposition is the opposite of formation, ∆G f would have to be (-). In other words, the formation is spontaneous and the decomposition is...
- Wed Feb 07, 2018 4:39 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.35 solution manual errors [ENDORSED]
- Replies: 1
- Views: 145
9.35 solution manual errors [ENDORSED]
In the solution manual error doc, a different value of Cv is used for each gas. Why is this?
- Tue Feb 06, 2018 9:44 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: How to approach 9.53
- Replies: 1
- Views: 958
Re: How to approach 9.53
Yes, we are supposed to use the equation ∆G = ∆Hvap - T∆Svap. We use the vaporation values because NH3 is going from a liquid to a gas.
- Tue Feb 06, 2018 9:42 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.13
- Replies: 6
- Views: 698
Re: 9.13
The solution manual error says Cv but uses 5/2 R, which is the value for Cp. Maybe the doc has a typo too?
- Tue Feb 06, 2018 12:01 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.25
- Replies: 5
- Views: 580
Re: 9.25
There are 6 different states the molecule can exist in (you can switch the Os and the Fs around the S atom 6 different ways).
- Mon Feb 05, 2018 11:51 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: HW 9.13 Cp vs Cv
- Replies: 2
- Views: 122
Re: HW 9.13 Cp vs Cv
In the solution manual errors PDF, the equation ∆S=nCvln(T2/T1) is used, with Cv being 5/2 R
- Mon Feb 05, 2018 10:47 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: HW 9.13 Cp vs Cv
- Replies: 2
- Views: 122
HW 9.13 Cp vs Cv
In 9.13, why is Cv used instead of Cp when finding the change in entropy resulting from the increase in temperature? Cv is the heat capacity at constant volume, and volume in this question isn't constant
- Fri Feb 02, 2018 1:37 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Micro states
- Replies: 5
- Views: 627
Re: Micro states
Also, the more microstates there are, the higher the entropy.
- Tue Jan 30, 2018 11:15 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Question 9.89
- Replies: 1
- Views: 204
Re: Question 9.89
I believe it stands for standard molar entropy, the entropy of 1 mol of substance at standard state
- Tue Jan 30, 2018 11:01 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.27a
- Replies: 4
- Views: 461
Re: 9.27a
In general, more complex molecules will have higher entropy. Since HBr is larger than HF, it is more complex, has more particles (and thus more possible states it can exist in), and higher entropy.
- Tue Jan 30, 2018 10:56 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 9.33 (a)
- Replies: 3
- Views: 1596
Re: 9.33 (a)
The way I thought about it was that when we first look at the equation, there are 2 mols on the left and 2 mols on the right. However, the reactants side has a mol of gas, which has more entropy than liquids are solids. Aqueous just means it's dissolved in water. A gas would have more entropy than a...
- Wed Jan 24, 2018 1:56 pm
- Forum: Calculating Work of Expansion
- Topic: 8.93
- Replies: 3
- Views: 446
8.93
When writing the chemical equation, how do you know water is in the liquid state?
- Tue Jan 23, 2018 12:11 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Work, reversible and irreversible
- Replies: 2
- Views: 255
Re: Work, reversible and irreversible
Also, in reversible expansion, the external pressure always matches the internal pressure. If there is any increase in pressure, external and internal energy will match each other. So, reversible reactions are always doing the maximum possible amount of work. In a reversible reaction, the process is...
- Tue Jan 23, 2018 12:04 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: HW #41 vs #21
- Replies: 3
- Views: 456
Re: HW #41 vs #21
In lecture, we discussed how in a perfect system, q system + q surroundings = 0. If this is the case, q system = - q surroundings . In this sense, you could say it's true that the negative sign can go on either one, because q system = - q surr is the same as q surroundings = - q system . If you assi...
- Mon Jan 22, 2018 10:28 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.91
- Replies: 1
- Views: 214
8.91
In the textbook, the question states that it takes 10.5 hours to reach 5C, but the solution manual uses 10 h. Is this a typo or am I missing something? I used these equations (4.184)(150 g)(5C - 0C)= 3135 J for 150 g and then I did ( \frac{3135 J}{150 g}*\frac{18.02 g}{1 mol}*\frac{1 kJ}{1000 J}*\fr...
- Wed Jan 17, 2018 9:48 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.19(b)
- Replies: 1
- Views: 182
Re: 8.19(b)
1.30*105 J is the amount of energy it takes to raise the temperature of water which is calculated by (400.0 g)(4.18)(100.0-22.0). 1.45*105 is the total amount of energy it takes to raise the temperature of both the water and the kettle itself.
- Wed Jan 17, 2018 9:43 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.17
- Replies: 3
- Views: 1126
Re: 8.17
When going from a solid to liquid, aka melting, heat is required without a change in temperature because it is a phase change. So, q is positive. For work, say ice is the solid. Ice, the system, gains energy, while the surroundings lose energy. This implies work is negative (because the system loses...
- Wed Jan 17, 2018 8:56 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: system and surroundings [ENDORSED]
- Replies: 3
- Views: 402
Re: system and surroundings [ENDORSED]
The kettle should be part of the system, which is why the kettle is included when calculating the heat that must be supplied. I guess if you think about it, the kettle must be heated first before the water can be heated.
- Wed Jan 17, 2018 8:49 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Transfer at Constant Volume/Pressure
- Replies: 4
- Views: 509
Re: Heat Transfer at Constant Volume/Pressure
Here is a mathematical approach in addition to the responses above... At constant volume, the heat transfer is interpreted as ∆U. This is true because ∆U = q + w, and w = -P ex ∆V, where P is external pressure and V is volume. If the volume is constant, ∆V=0, which implies w=0, and thus ∆U=q. At con...
- Thu Jan 11, 2018 9:44 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Relating energy and heat [ENDORSED]
- Replies: 2
- Views: 226
Re: Relating energy and heat [ENDORSED]
The terms energy and heat are also related in the equation ∆U = q + w, with ∆U being the change in internal energy, q being heat, and w being work. In other words, heat and work are both means of transferring energy.
- Thu Jan 11, 2018 9:38 pm
- Forum: Calculating Work of Expansion
- Topic: Reversible vs. Irreversible Reaction
- Replies: 2
- Views: 382
Re: Reversible vs. Irreversible Reaction
Also, reversible reactions occur slowly, while irreversible reactions occur quickly. So, if the process is reversible (done slowly), minimal energy is lost to the surrounding as heat, allowing for maximum expansion work.
- Wed Jan 10, 2018 12:07 pm
- Forum: Calculating Work of Expansion
- Topic: Work for reversible vs irreversible process
- Replies: 3
- Views: 407
Re: Work for reversible vs irreversible process
Reversible processes are very slow. Because it is done slowly, less energy is lost to the surrounding as heat. In the gas and piston example, an extremely small increase in external pressure would cause the gas to be compressed and the piston to move in. If the process were to be done quickly (aka i...
- Fri Dec 08, 2017 5:48 pm
- Forum: Bronsted Acids & Bases
- Topic: Determining acidity
- Replies: 3
- Views: 538
Re: Determining acidity
The example given in class compared HClO with HBrO and HIO and said that HClO is more acidic. Since Cl is more electronegative, it takes e- density away from O, thus stabilizing it (a stable resulting anion would drive the reaction forward). Taking e- density away from O makes the compound a stronge...
- Tue Dec 05, 2017 12:40 am
- Forum: Bronsted Acids & Bases
- Topic: Ch 12 #27
- Replies: 2
- Views: 307
Re: Ch 12 #27
We multiply 200 ml by .025 M to find the number of moles that were actually made. Then divide by 250 ml to get the molarity of the actual solution.
- Tue Dec 05, 2017 12:13 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Endothermic v. Exothermic Reactions
- Replies: 9
- Views: 2608
Re: Endothermic v. Exothermic Reactions
Reactions that include bond breaking are generally endothermic. EX: Cl2 <--> 2Cl. Combustion reactions are almost always exothermic.
- Mon Dec 04, 2017 3:46 pm
- Forum: Naming
- Topic: Using ion in names
- Replies: 2
- Views: 411
Re: Using ion in names
Yes, I believe if it just asks you to name the compound and there is no charge, you don't need to write "ion." If it's a complex ion (aka has a + or - charge), you'll need to write ion at the end of the name.
- Sat Dec 02, 2017 12:17 am
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: Delocalized Pi Bonds [ENDORSED]
- Replies: 2
- Views: 597
Re: Delocalized Pi Bonds [ENDORSED]
Pi bonds in benzene would be delocalized. When you draw the Lewis structure, it alternates between single and double bonds. In reality, there are no alternating bonds, which is why benzene is usually drawn with a circle inside a hexagon. The p-orbitals overlap side-to-side and electrons are delocali...
- Sat Dec 02, 2017 12:12 am
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: 11.57 [ENDORSED]
- Replies: 1
- Views: 811
Re: 11.57 [ENDORSED]
If you're given moles and liters, you'll have to convert to molarity. K can only be found with concentrations (molarity) or partial pressure (where the units will be something like atm, bar).
- Thu Nov 23, 2017 1:22 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: HW 4.43
- Replies: 1
- Views: 209
Re: HW 4.43
The s-character in sp2 is more than that in sp3. The way I see it is in sp2, the ratio of s:p is 1:2, while the ratio in sp3 is 1:3. If the bond angle in sp2 is larger, that would mean an increased s-character would result in a larger bond angle.
- Thu Nov 23, 2017 11:15 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Shifted VS Lie
- Replies: 2
- Views: 1583
Re: Shifted VS Lie
When the reaction lies to left, there are more reactants than products. If the equilibrium is shifted to the left, there are more products than reactants. Equilibrium will shift to the side with fewer moles, so if there are products, equilibrium will shift to the reactant side (left side).
- Sun Nov 12, 2017 11:43 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Drawing VSEPR Models
- Replies: 2
- Views: 2012
Re: Drawing VSEPR Models
I believe the lines and triangles you are referring too are just bonds between two atoms. The lines and triangles symbolize depth. If you imagine your paper to be the plane you're drawing the VSEPR structure on, the lines fading out would indicate an atom that is going into your paper, and the trian...
- Sun Nov 12, 2017 5:36 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 3
- Views: 473
Re: Bond Angles
When molecules have lone pairs around the central atom, other bond angles are forced closer together (we won't know the exact angle, but you can see that the bond angles would be a little less than those in a structure without lone pairs). This applies to O3, which has 1 lone pair around the central...
- Sun Nov 12, 2017 5:30 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Thiosulfate ion shape
- Replies: 2
- Views: 1576
Re: Thiosulfate ion shape
There actually is no lone pair around the S. I believe the Lewis structure is just a sulfur atom in the middle, surrounded by 3 oxygen atoms and 1 other sulfur atom. Since there are 4 bonding pairs around the central atom, it would be a tetrahedral shape.
- Sun Nov 12, 2017 4:11 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Textbook Problem Chp 4 #9
- Replies: 3
- Views: 428
Re: Textbook Problem Chp 4 #9
In a trigonal planar shape, there are only 3 bonding pairs surrounding the central atom. As you drew in your Lewis structure, ICl3 has 3 bonding pairs and 2 lone pairs, which would make it T-shaped.
- Tue Oct 31, 2017 11:49 pm
- Forum: Trends in The Periodic Table
- Topic: 2.81 – Oxygen Anomaly
- Replies: 2
- Views: 497
Re: 2.81 – Oxygen Anomaly
Nitrogen also has a half-filled sub shell, which means it has lower energy and is more stable. Since it would be more difficult to take away an electron from a more stable subshell (aka higher ionization energy), Oxygen has the lower ionization energy.
- Tue Oct 31, 2017 11:40 pm
- Forum: Trends in The Periodic Table
- Topic: Writing Electron Config for Chromium after being ionized once?
- Replies: 2
- Views: 572
Re: Writing Electron Config for Chromium after being ionized once?
It should be (1). Higher energy electrons are removed first. In this case, the 4s electrons have a higher energy than 3d.
- Wed Oct 25, 2017 2:49 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: HW Question 2.37
- Replies: 2
- Views: 299
Re: HW Question 2.37
I believe in this case "penetrate" just means how close the electron can get to the atom. So since it "penetrates" more, it is more attracted to the nucleus and therefore more effective at shielding other electrons than orbitals that are further out from the nucleus.
- Wed Oct 25, 2017 1:51 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Helium block
- Replies: 2
- Views: 724
Re: Helium block
^ The electron configuration of Helium is 1s^2, so yes, it is part of the s-block.
- Tue Oct 17, 2017 8:15 pm
- Forum: DeBroglie Equation
- Topic: 1.39 Help
- Replies: 5
- Views: 640
Re: 1.39 Help
I think the only thing you're missing is that 41 should be in m * s^-1, not just meters.
- Tue Oct 17, 2017 7:23 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: ch1 question 43
- Replies: 8
- Views: 1102
Re: ch1 question 43
I also had a question on this^
The solution manual uses a different character for Planck's constant, and it uses the equation ∆p∆x= h/2, whereas in lecture we learned the equation ∆p∆x=h/4pi.
The solution manual uses a different character for Planck's constant, and it uses the equation ∆p∆x= h/2, whereas in lecture we learned the equation ∆p∆x=h/4pi.
- Thu Oct 12, 2017 12:29 am
- Forum: Properties of Light
- Topic: Question 1.15 Rydberg Equation Math Problem
- Replies: 1
- Views: 212
Re: Question 1.15 Rydberg Equation Math Problem
So I used the equation was c/wavelength=R(1/n_1^2 - 1/n_2^2). Essentially, you're using the given wavelength and using c=v*wavelength to find the frequency, and then using v=R((1/n_1^2 - 1/n_2^2) to find n_2. (We also already know n_1 = 1 because a wavelength of 102.6nm is Lyman series
- Wed Oct 11, 2017 2:51 pm
- Forum: Properties of Light
- Topic: Rydberg's Formula
- Replies: 4
- Views: 585
Re: Rydberg's Formula
The formula basically relates the wavelength to the initial and final states of an electron. We can find the inverse of the wavelength by multiplying Rydberg's constant by the change in energy of the electron.
- Wed Oct 04, 2017 3:05 pm
- Forum: Significant Figures
- Topic: Sig Figs in 0.0380 [ENDORSED]
- Replies: 5
- Views: 985
Re: Sig Figs in 0.0380 [ENDORSED]
Leading zeroes are not significant (in 0.032, the zeroes before 3 aren't significant so there would be 2sf). Trailing zeroes ARE significant (in 0.03000, all the zeroes after 3 are significant so there would be 4sf).
- Tue Oct 03, 2017 1:41 pm
- Forum: Empirical & Molecular Formulas
- Topic: F.1 textbook exercise
- Replies: 3
- Views: 422
Re: F.1 textbook exercise
If the compound includes carbon atoms, C will be written first. If it includes hydrogen atoms, H will come second. Any other elements should follow in alphabetical order. Hope that helps!