Search found 40 matches

by Divya Prajapati 1E
Thu Mar 10, 2016 8:57 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Catalyst in a Reaction Mechanism
Replies: 1
Views: 489

Re: Catalyst in a Reaction Mechanism

A catalyst can be identified as something that is put into the reaction and then regenerated by the end of the reaction. You can think of it as the opposite of an intermediate - which is generated in one step and then used up in the next step. For example in the reaction below, Cl is the catalyst an...
by Divya Prajapati 1E
Sun Mar 06, 2016 3:54 pm
Forum: *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
Topic: Steric v Torsional
Replies: 1
Views: 290

Re: Steric v Torsional

Hi! Torsional strain is the force that prevents rotation in a bond. For example, ethane has torsional strain about its C-C bond when its H atoms are directly across from each other (eclipsed conformation). This leads to instability which forces one side of the molecule to shift. In this case, one se...
by Divya Prajapati 1E
Fri Feb 19, 2016 4:55 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: reaction order
Replies: 3
Views: 374

Re: reaction order

All rate constants do not have the same units. If we are using M and s, each order's units are:

Zero order: M/s
First order: 1/s
Second order: 1/M*s

Therefore, if you are given the k value for a reaction, you can look at its units to determine its reaction order!
by Divya Prajapati 1E
Fri Feb 12, 2016 10:40 am
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Gibbs Free Energy for Ideal Gases
Replies: 1
Views: 306

Re: Gibbs Free Energy for Ideal Gases

Ideal gases are hypothetical small (low molecular mass), nonpolar gases. The real gases which are closest to ideal include H2, N2, F2, O2, and He. ALL of these gases have a Gibbs free energy of 0 because these are their most stable, elemental states. Conversely, gases like CH4, CO, or SO2 have non-z...
by Divya Prajapati 1E
Sun Feb 07, 2016 10:46 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Potential Calculations?
Replies: 1
Views: 223

Re: Potential Calculations?

Hi! It all depends on what the E values are talking about. There are two possible E values for any given 1/2 reaction. One E value is the reduction potential, the potential created when a substance gains electrons. The other E value is the oxidation potential, the potential created when a substance ...
by Divya Prajapati 1E
Sun Feb 07, 2016 10:38 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Nerst Equation
Replies: 1
Views: 318

Re: Nerst Equation

Hi! The Nernst equation has several applications, many of which you noted: - Finding Ecell when we don't have 1.0 M solutions or 1.0 bar partial pressures - Finding Eo when we have Ecell and the non-standard concentrations/pressures - Finding Q when we know Ecell and Eo - Finding Q when we know Ecel...
by Divya Prajapati 1E
Sun Feb 07, 2016 10:33 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 14.37a Nernst Equation
Replies: 1
Views: 328

Re: 14.37a Nernst Equation

If you know that the Nernst equation needs to be used, start with this equation to solve the problem: E cell = Eo cell - (RT/nF)lnQ Because this is a concentration cell (anode/cathode involve the same products and reactants), we know that Eo cell = 0. This is because if the concentration of all prod...
by Divya Prajapati 1E
Sat Jan 30, 2016 10:20 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Calculating the standard potential of a half reaction
Replies: 1
Views: 217

Re: Calculating the standard potential of a half reaction

You can still use the Hess' law method IF the reaction is occurring under standard conditions. This means half reactions must be occurring in 1M solutions. If this is not the case, we can also use this equation (pg. 53 of course reader): E^o = -deltaG^o/nF (standard cell potential is equal to standa...
by Divya Prajapati 1E
Fri Jan 22, 2016 3:58 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: gibbs free energy
Replies: 2
Views: 259

Re: gibbs free energy

For a reaction to be spontaneous, the entropy of the entire universe must increase (this is based on the 2nd Law of Thermodynamics). In the last section, we determined this by summing entropy of the system and surroundings to find total entropy. When total entropy was negative, meaning a decrease in...
by Divya Prajapati 1E
Fri Jan 22, 2016 3:49 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: molar heat capacity
Replies: 1
Views: 180

Re: molar heat capacity

You can distinguish this using molecular geometry (after you figure out the Lewis structures). For example, oxygen and carbon dioxide are linear molecules, because all atoms lie on the same axis. In contrast, water or methane are non-linear because they either have bent geometry and 3-D tetrahedral ...
by Divya Prajapati 1E
Fri Jan 22, 2016 3:43 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Entropy change 9.35
Replies: 1
Views: 258

Re: Entropy change 9.35

Hi! I had the same confusion. However, if you read the problem carefully, it says there is 1 mole of ATOMS (not molecules). So, we actually have 1 mole of monoatomic compound A and 0.5 moles of diatomic compounds B and C.
by Divya Prajapati 1E
Sun Jan 17, 2016 7:39 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Homework Problem 8.53
Replies: 1
Views: 311

Re: Homework Problem 8.53

Based on the given information, we will be using the equation "q rxn = - q cal = - C x delta T". This equation tells us that the heat released in the reaction is the negative of the total heat capacity of the calorimeter times the change in temperature. The problem gives us C = 3.00 kJ/C a...
by Divya Prajapati 1E
Sun Jan 17, 2016 7:32 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Homework Problem 8.25
Replies: 2
Views: 492

Re: Homework Problem 8.25

Hi! Based on your calculations, you found C, the heat capacity of the entire calorimeter (rather than its specific heat capacity) is 0.478 kJ/K. We can use the equation q = - C x delta T to find how much heat the reaction released into the calorimeter. We know delta T is 2.49 K using the given chang...
by Divya Prajapati 1E
Sun Jan 17, 2016 7:24 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Homework Question
Replies: 3
Views: 737

Re: Homework Question

Hi! We need to find the change in internal energy (delta U), which we can calculate using the equation "delta U = delta H + delta nRT". Using the reaction, we can find the change in moles of gas, which in this case is : 5 moles of gaseous products - 3 moles of gaseous reactants = 2 moles g...
by Divya Prajapati 1E
Wed Jan 06, 2016 11:50 pm
Forum: Phase Changes & Related Calculations
Topic: Heating Curves
Replies: 1
Views: 344

Re: Heating Curves

Most heating curves follow the same general shape as that for water. Water's heating curve, however, can be identified by the following features: - Melting point at 0 C and boiling point at 100 C - Takes about 6 kJ for solid-liquid transition and about 41 kJ for liquid-gas transition - Plateau in so...
by Divya Prajapati 1E
Sat Dec 05, 2015 12:37 pm
Forum: *Biological Importance of Buffer Solutions
Topic: Mixed Solutions.
Replies: 2
Views: 700

Re: Mixed Solutions.

We can answer all of these questions using Le Chatelier's principle, which applies to any aqueous solution in a state of equilibrium. Because we are dealing with weak acid and base solutions, we know that this is a situation of aqueous equilibrium, because both reactants and products will be present...
by Divya Prajapati 1E
Fri Nov 27, 2015 10:34 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Video Explaining HW 12.75
Replies: 2
Views: 371

Re: Video Explaining HW 12.75

Here is the mp4 version of the video found on YouTube.

Chem 14A HW Problem 12.75.mp4 [ 82.07 MiB | Viewed 361 times ]

by Divya Prajapati 1E
Wed Nov 25, 2015 6:06 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Video Explaining HW 12.75
Replies: 2
Views: 371

Video Explaining HW 12.75

Hi everyone!

I wanted to explain HW problem 12.75 because it deals with a acid/base constants as well as the conjugate seesaw. I have posted a video to YouTube, accessible through this link: https://youtu.be/7f9_cpjkIBU.

Hopefully you find it useful!
by Divya Prajapati 1E
Tue Nov 24, 2015 9:58 am
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Formic Acid vs. Acetic Acid - HW 12.53
Replies: 1
Views: 1053

Re: Formic Acid vs. Acetic Acid - HW 12.53

I found the following explanation on an earlier post on Chem Community. It seems that acetic acid ADDS electron density, which keeps the H atom (positively charged) attached to the molecule. Formic acid, on the other hand, lacks such electron density near the O-H bond, so the H is more easily releas...
by Divya Prajapati 1E
Sun Nov 22, 2015 9:28 pm
Forum: Amphoteric Compounds
Topic: Amphoteric vs amphiprotic
Replies: 5
Views: 12394

Re: Amphoteric vs amphiprotic

Most of the time, you will be using the term amphoteric, as it does not require a substance to accept and donate protons. An amphiprotic substance, on the other hand, is specifically an amphoteric substance that transfers H+ ions. I found the following explanation on this website: http://www.chemgui...
by Divya Prajapati 1E
Sun Nov 22, 2015 9:19 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Anion exceptions
Replies: 1
Views: 282

Re: Anion exceptions

Keep in mind that acids like like HSO4- and H2PO4- have very small K values. As such, almost all H atoms do NOT dissociate from the anion. They still, however, have the potential to dissociate because they have H atoms available to lose. Structurally, these H atoms are usually located at the end of ...
by Divya Prajapati 1E
Thu Nov 12, 2015 4:09 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Determining monodentate versus polydentate by shape
Replies: 1
Views: 308

Re: Determining monodentate versus polydentate by shape

Shape determines how easily a molecule is able to surround a transitional metal atom and bind to multiple sites. Take, for example, C2O4 which we know to be a bidentate. It's Lewis structure consists of two C atoms bonded to each other. Each C atom has one single-bonded O atom and one double bonded ...
by Divya Prajapati 1E
Thu Nov 12, 2015 3:51 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Changing Concentrations of Solids
Replies: 1
Views: 207

Re: Changing Concentrations of Solids

Changing the concentration of a solid leads to a change in the concentration of gaseous reactants and products. Take the following example: P4S10 (s) + 16H2O (l) <--> 4H3PO4 (aq) + 10H2S (aq) When we increase the concentration of P4S10, the rightward reaction will be favored. This will produce more ...
by Divya Prajapati 1E
Thu Nov 12, 2015 3:46 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Reaction Quotient and Heterogeneous Equilibria (HW 11.13)
Replies: 1
Views: 166

Re: Reaction Quotient and Heterogeneous Equilibria (HW 11.13

I believe in part (a) BCl3 is the only gaseous species. All other reactants and products are in the solid or liquid phase. So, Q = 1/[BCl3], as this is the only reactant/product we take into consideration.
by Divya Prajapati 1E
Thu Nov 12, 2015 3:41 pm
Forum: Ideal Gases
Topic: Ratios and Change
Replies: 1
Views: 232

Re: Ratios and Change

At a given temperature, the ratio of products to reactants at equilibrium is a constant (it will not change). This quantity only changes if the temperature of the system is raised or lowered. So, if the concentration of products and reactants is not changing, then the system is probably already at e...
by Divya Prajapati 1E
Sun Nov 01, 2015 3:52 pm
Forum: Bond Lengths & Energies
Topic: Question 3.117 a
Replies: 1
Views: 386

Re: Question 3.117 a

You can determine bond character by using resonance structures. In this particular case, each isomer of S2F2 has two resonance structures. A characteristic of molecules with resonance structures is that electrons constantly change in electron density, sometimes having single and sometimes having dou...
by Divya Prajapati 1E
Sun Nov 01, 2015 3:46 pm
Forum: Bond Lengths & Energies
Topic: Exciting Bonding
Replies: 1
Views: 272

Re: Exciting Bonding

While this may not exactly answer your question, molecular orbital theory does offer an explanation for how excited electrons function in a molecule. If you look at page 20 in the course reader, excited electrons transition from the highest unoccupied molecular orbital (HOMO) to the lowest unoccupie...
by Divya Prajapati 1E
Sun Nov 01, 2015 3:41 pm
Forum: Ionic & Covalent Bonds
Topic: molecular orbital theory
Replies: 1
Views: 293

Re: molecular orbital theory

While this may not answer your question, I think it is important to know that "in molecular orbital theory all valence electrons are delocalized over the whole molecule, not confined to individual bonds" (pg. 131 in textbook). In contrast, ionic bonds have extremely unequal sharing of elec...
by Divya Prajapati 1E
Sun Nov 01, 2015 3:30 pm
Forum: Hybridization
Topic: Hybrid orbitals when given a long formula
Replies: 3
Views: 519

Re: Hybrid orbitals when given a long formula

When you are given a "long formula" you first need to write out the Lewis structure. The reason the formula is so long is to guide you in creating this structure. As you read the formula left to right, place atoms in order of how they are given to you, usually with carbon atoms attached to...
by Divya Prajapati 1E
Sun Nov 01, 2015 3:22 pm
Forum: Hybridization
Topic: Unhybridized orbitals
Replies: 6
Views: 10328

Re: Unhybridized orbitals

The main function of unhybridized orbitals is usually to form double bonds. Consider each of the following cases: - sp3 hybridization has 4 sp3 hybridized orbitals and 0 unhybridized orbitals. These allow molecules to form 4 sigma bonds. - sp2 hybridization has 3 sp2 hybridized orbitals and 1 unhybr...
by Divya Prajapati 1E
Sat Oct 24, 2015 7:17 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Homework Question 4.1
Replies: 1
Views: 196

Re: Homework Question 4.1

One example of a linear shape having lone pairs is when the geometry is trigonal bipyramidal. Typically, this geometry involves 5 bonding pairs. But, if we have 3 lone pairs and 2 bonding pairs, the shape is linear, because the bonding pairs will be at 180 degrees from each other. Two molecules that...
by Divya Prajapati 1E
Sat Oct 24, 2015 7:13 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Seesaw Bond Angles
Replies: 6
Views: 5107

Re: Seesaw Bond Angles

For the seesaw shape, we have 5 regions of electron density (trigonal bipyramidal), consisting of 4 bonding pairs and 1 lone pair. Normally, when all these regions are bonding, the molecule has 120 degree angles between the three atoms making up the "trigonal" part of the shape and 90 degr...
by Divya Prajapati 1E
Sat Oct 24, 2015 7:05 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: HW 4.25(a)
Replies: 2
Views: 285

Re: HW 4.25(a)

While the Lewis structure of CH2Cl2 may seem symmetric, this is because we are looking at a 2-D diagram that places each atom on the x-y axis. In reality, CH2Cl2 is a 3-D molecule with a tetrahedral shape. This means that no matter where the Cl atoms are placed, there will always be 2 Cl atoms next ...
by Divya Prajapati 1E
Sat Oct 17, 2015 8:07 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Boron trifluoride --> Tetrafluoroborate
Replies: 2
Views: 479

Re: Boron trifluoride --> Tetrafluoroborate

Hi! As you pointed out, BF3 forms a compound with an F- ION. This is a key point, because it means that BF4 is an ionic, rather than a covalent compound. Typically, formal charges are only assigned to covalent compounds, because they assist us in determining whether electrons are being shared equall...
by Divya Prajapati 1E
Sat Oct 17, 2015 8:03 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Octets
Replies: 1
Views: 287

Re: Octets

Hi! The reason some elements can have more than octet is due to the number of orbitals they have available. Starting with Period 3, elements also have "access" to d orbitals which can be filled with additional electrons. As a member of Period 3, sulfur has an electron configuration at the ...
by Divya Prajapati 1E
Thu Oct 08, 2015 7:23 pm
Forum: Trends in The Periodic Table
Topic: Electronegativity and Electronaffinity
Replies: 1
Views: 326

Re: Electronegativity and Electronaffinity

Electronegativity and electron affinity are basically the same concept. Electronegativity is an element's tendency to attract electrons, while electron affinity is the likelihood of an atom to gain an electron. Both properties deal with how much an element "wants" to gain an electron for g...
by Divya Prajapati 1E
Thu Oct 08, 2015 6:43 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Homework 2.29 - Atomic Orbitals and Electrons
Replies: 1
Views: 465

Re: Homework 2.29 - Atomic Orbitals and Electrons

The quantum numbers of an electron basically are its "address" in an atom. The more quantum numbers provided, the more specifically we can locate an electron. This is just like getting the street name of a person's address, instead of just their city or state. It allows a narrower descript...
by Divya Prajapati 1E
Thu Oct 08, 2015 6:35 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Electron Configuration of Vanadium
Replies: 1
Views: 723

Re: Electron Configuration of Vanadium

As electrons have higher energy levels and orbitals, the distinction between orbitals starts to "blur" a little. If you have seen diagrams of energy levels, the gap between n=1 and n=2 is larger than the one between n=2 and n=3. By the time electrons movie into n=3 and n=4, the energy diff...
by Divya Prajapati 1E
Sat Oct 03, 2015 10:29 pm
Forum: Photoelectric Effect
Topic: Instructional Module Question
Replies: 1
Views: 376

Re: Instructional Module Question

Hi! The steps are very similar to the calculations involved in part A. Here is how I would solve it: 1. Find frequency of incoming light based on the given wavelength (v = c/wavelength) 2. Use frequency to find energy of incoming light (E=hv) 3. The energy of the incoming light is going to be the TO...
by Divya Prajapati 1E
Sat Oct 03, 2015 10:18 pm
Forum: Empirical & Molecular Formulas
Topic: Name of Formula to Molecular Formula, and Vice Versa
Replies: 3
Views: 547

Re: Name of Formula to Molecular Formula, and Vice Versa

There are actually a specific set of rules to naming different compounds. You can find a summary of these rules in the Fundamentals section of the textbook. Typically, the names vary based on the type of compound - ionic or molecular, organic or inorganic, acid or not. The first step to figuring out...

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