Search found 61 matches

Tue Mar 12, 2019 12:19 am
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Problem 7A17
Replies: 2
Views: 231

Re: Problem 7A17

When you solve for k, remember that the concentrations have to be in mol/L; in the problem, the concentrations are in mmols, so you would have to convert the values to moles. I'm not sure what you mean by multiplying by .457, but looking at the problem, experiment 4 has a initial rate of 457 mmol/L*...
Tue Mar 12, 2019 12:10 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Graphing the Reactions Profiles
Replies: 2
Views: 57

Re: Graphing the Reactions Profiles

Dr. Lavelle said that anything covered in lecture is fair game, so I would assume that we would need to understand the general idea of reaction profiles (where the activation energy, transition state, change in energy, reactants/products are)
Tue Mar 12, 2019 12:08 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: stability of reactants and products
Replies: 3
Views: 83

Re: stability of reactants and products

Whether the forward or reverse reaction is favored gives us a hint to the stability of the products: if the reaction is spontaneous, then the reaction will tend toward the product, so the product is more stable. On the other hand, if the reaction is not spontaneous, then the reaction will tend towar...
Tue Mar 05, 2019 9:20 am
Forum: General Rate Laws
Topic: 6th edition 15.9
Replies: 4
Views: 132

Re: 6th edition 15.9

I'm not sure which question you're talking about (I have 7th edition), but generally, the units for rate constant k depend on the order of the reaction. The units change so that the rate in the rate law will always equal M/s. For example, for first order reactions where rate = k [A], [A] is in molar...
Tue Mar 05, 2019 9:14 am
Forum: General Rate Laws
Topic: 15.9 6th edition
Replies: 3
Views: 104

Re: 15.9 6th edition

Zeroth order reactions mean that changing the concentration of any reactant will not affect the rate at which the reaction proceeds. In terms of the question, the rate law is rate = k, so k has units of M/s
Tue Mar 05, 2019 9:12 am
Forum: First Order Reactions
Topic: Units
Replies: 3
Views: 131

Re: Units

Looking at the rate law, the units for the rate should end up being M/s, so depending on the order of the reaction, the units of k should be different. If you write out the units of each concentration and the rate, you should be able to figure out the units for k. For example, for a first order reac...
Tue Feb 26, 2019 10:46 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Solid Electrodes
Replies: 1
Views: 55

Re: Solid Electrodes

Platinum is used as an inert electrode if there is no solid species in the half reaction. For example, if the reduction reaction goes from Fe3+ to Fe2+, both of these are aqueous ions, so we would have to use a platinum electrode in the cell. Remember that the two half reactions are in different cel...
Tue Feb 26, 2019 10:43 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagrams and aq
Replies: 1
Views: 52

Re: Cell Diagrams and aq

It depends on your half reaction. There is usually an aqueous next to the salt bridge because there is often times aqueous ions in the half reaction. However, if there is no aqueous ion in the half reaction, there will be no aqueous species next to the salt bridge. Just keep in mind that the solid e...
Tue Feb 26, 2019 10:38 pm
Forum: Balancing Redox Reactions
Topic: Half reactions
Replies: 1
Views: 63

Re: Half reactions

Without multiplying. When it asks for the balanced half reaction, you just have to provide the balanced half reaction with the lowest coefficients. Only when you write the balanced full reaction do you need to make sure the electrons on both sides are equal (so they cancel out).
Mon Feb 18, 2019 3:49 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 6th edition 9.65
Replies: 2
Views: 77

Re: 6th edition 9.65

What the solution is doing is looking at the decomposition reaction of each molecule. In essentiality, they are looking at deltaG = deltaH - TdeltaS; if the delta S is positive, then as temperature rises, then the reaction becomes more favorable. If the delta S is negative, then as temperature rises...
Mon Feb 18, 2019 3:42 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Pressure Change
Replies: 4
Views: 114

Re: Pressure Change

If you look at Boyle's Law, P1V1 = P2V2, then dividing V1 and P2 from both sides, we get V2/V1 = P1/P2. Therefore, if we substitute that in the equation, then deltaS=nRln(V2/V1) will become deltaS = nRln(P1/P2). In essentiality, it is just using Boyle's Law in conjunction with the change in entropy ...
Mon Feb 18, 2019 3:39 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Compound stability with respect to decomposition
Replies: 3
Views: 70

Re: Compound stability with respect to decomposition

Looking at a decomposition reaction, let's say A --> B + C, if the reaction has a negative delta G, then the decomposition reaction is spontaneous, or favorable. This means that A has a tendency to decompose into B and C; in other words, we can say that A is unstable.
Wed Feb 13, 2019 7:54 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: ΔU= 3/2nRT
Replies: 5
Views: 222

Re: ΔU= 3/2nRT

For the most part, that equation is used to show that ΔU = 0 during isothermal reactions. Since ΔT = 0 and ΔU = 3/2nrΔT, ΔU = 0.
Wed Feb 13, 2019 7:43 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Isothermal work
Replies: 2
Views: 338

Re: Isothermal work

To add on, in an isothermal reaction, the graph shows a curved line from point A to point B while the line for an irreversible reaction is like an L, going down at constant pressure and across at constant volume (See Lavelle's notes). Comparing the areas under these curves, the area under the curved...
Wed Feb 13, 2019 7:36 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Internal energy (U) of an isolated system
Replies: 17
Views: 789

Re: Internal energy (U) of an isolated system

The First Law of Thermodynamics states that the change in internal energy of an isolated system is 0. So to answer your question, if a system had, let's say +5kJ, after any amount of time, it will still have +5kJ.
Thu Feb 07, 2019 9:15 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Work on an adiabatic process?
Replies: 2
Views: 91

Re: Work on an adiabatic process?

Adiabatic just means that there is no heat transfer, so q = 0. Therefore, there can be work done. In fact, since deltaU = q + W, and q = 0, deltaU = W in adiabatic systems.
Thu Feb 07, 2019 9:11 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Degeneracy
Replies: 2
Views: 89

Re: Degeneracy

In a given system, degeneracy W would equal the number of possible microstates to the power of the number of particles in the system. For example, if you look at the example given in class, there were 4 CO molecules at T = 0K; each CO can have atoms in two possible positions. In this problem, the 0K...
Tue Feb 05, 2019 6:25 pm
Forum: Administrative Questions and Class Announcements
Topic: Practice Midterm
Replies: 3
Views: 175

Practice Midterm

Last quarter in 14A, TA/UA Lyndon always had a midterm review/practice midterm with a funky name. Does anyone know if there is one this midterm; if so, what is it called. Thanks!
Thu Jan 31, 2019 5:48 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: delta u eqn
Replies: 3
Views: 119

Re: delta u eqn

Like Jonathan said, the change in internal energy is equal to the heat transferred to the system added to the work done on the system. In the case of your question, q will be zero when there is no heat transfer happening so delta U would be equal to work. On the flip side, if there was no work being...
Thu Jan 31, 2019 5:43 pm
Forum: Phase Changes & Related Calculations
Topic: HW problem 8.5
Replies: 3
Views: 99

Re: HW problem 8.5

To expand, change in internal energy is the work done on the system plus the change in enthalpy (or heat at constant pressure). In other words, delta(U) = delta(H) + W. In the problem, the 524 kJ of heat is your delta(H) and the 340 kJ of work done by the piston is your W. Therefore, your change in ...
Thu Jan 31, 2019 5:20 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Work Calculation
Replies: 3
Views: 97

Re: Work Calculation

There must be a negative sign since our system is the gas inside the piston. When we compress the gas, we do work on the gas and increase its internal energy. However, the final volume minus the initial volume would give us a negative number (because the gas is compressed). Without a negative sign, ...
Wed Jan 23, 2019 1:41 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Calculating pH of a weak acid and its salt
Replies: 4
Views: 292

Re: Calculating pH of a weak acid and its salt

Yes, the steps are exactly the same as calculating without a salt (though the calculations might be a bit more difficult). In these types of problems, the thing to note is that the initial concentration for the salt (or product of the salt) will not be zero. Therefore, the equilibrium concentration ...
Wed Jan 23, 2019 1:36 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Knowing When to Approximate
Replies: 4
Views: 180

Re: Knowing When to Approximate

The general rule of thumb according to Dr. Lavelle is that approximations are typically valid if the equilibrium constant is less that 10^-3. After approximating, in order to see if it was a valid approximation, you use the 5% rule. To use this rule, you calculate x/initial, or in other words, the a...
Wed Jan 23, 2019 1:32 am
Forum: Ideal Gases
Topic: OH and H3O
Replies: 10
Views: 551

Re: OH and H3O

There are many ways to determine if something is acidic or basic. One way is definitely to write out its reaction with water and see if it creates H3O+ or OH-. Another way can be to look at its Ka or Kb. The larger the Ka, the stronger the acid; the larger the Kb, the stronger the base. On the flip ...
Wed Jan 23, 2019 1:26 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6A.19, 7th Edition
Replies: 1
Views: 65

Re: 6A.19, 7th Edition

You are definitely right. To solve the problem, you do Kw/[OH-] which equals 3.2 x 10^-15 M. I came across this when I was doing this problem too. I assume that it's just an error in the solution manual.
Mon Jan 14, 2019 10:23 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Concentration Affecting K
Replies: 7
Views: 269

Re: Concentration Affecting K

Increasing concentration does not affect K since when you add more reactants or products, the reaction is no longer in equilibrium. Remember that K is the ratio of the concentration of products to the concentration of reactants. After adding either reactants or products, the ratio would be imbalance...
Mon Jan 14, 2019 10:20 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Exothermic Reaction
Replies: 3
Views: 82

Re: Exothermic Reaction

An exothermic reaction tends to make products since the energy of the products are lower than the energy of the reactants (hence the release of energy). Since it's more favorable to be at a lower energy state, the reaction will often be spontaneous and make products. However, in regards to equilibri...
Mon Jan 14, 2019 10:14 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Example from lecture on Wednesday
Replies: 2
Views: 81

Re: Example from lecture on Wednesday

In the example, ATP + H20 <> ADP + Pi and given to us are the concentrations of ATP in healthy and in dead tissue. The question is asking for the concentrations of ADP and Pi in dead tissue. For this example, we are assuming that there is no (or negligible amounts of) ADP and Pi in the healthy tissu...
Mon Jan 07, 2019 3:14 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Help on 11.13
Replies: 2
Views: 82

Re: Help on 11.13

When considering substances in your equilibrium expression, always include aqueous and gas phase molecules while excluding solids and liquids. Aqueous solutions are not pure liquids, rather, they are the substance dissolved in water. Therefore, they have concentrations that change with the reaction ...
Mon Jan 07, 2019 3:09 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Help on 11.9: b and c
Replies: 4
Views: 118

Re: Help on 11.9: b and c

I'm not sure where that equation came from, but I would just find the equilibrium expressions like how Lavelle taught us in class. Given a hypothetical chemical reaction aA + bB--> cC + dD, the equilibrium constant Kc = ([C]^c [D]^d) / ([A]^a [B]^b), or simply just Kc = [products] / [reactants]. So ...
Mon Jan 07, 2019 2:57 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework Problem 11.3 for 6th Edition
Replies: 3
Views: 90

Re: Homework Problem 11.3 for 6th Edition

For gas-phase molecules, you can use either partial pressures or concentrations depending on what they give you. Additionally, you can convert between partial pressure and concentration using the ideal gas law PV=nRT, where n/V is concentration, so P = (conc)RT or (conc) = P/RT. I'm not sure which q...
Mon Dec 03, 2018 11:17 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Electronegativity vs. Size in Acid Strength
Replies: 1
Views: 202

Re: Electronegativity vs. Size in Acid Strength

You have to remember that the hydrogen are not bonded to Br and Ge. Instead, they are bound to the oxygens, so that makes the bond length in each relatively the same. The reason why a higher electronegativity constitutes a stronger acid is because after the hydrogen is lost, a higher electronegativi...
Mon Dec 03, 2018 11:03 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: pH and Molarity
Replies: 2
Views: 121

Re: pH and Molarity

For part a, you'd want to find out the pH of the intended solution, which is the 200mL of 0.025M of HCl. Using -log [H+] to find pH, you'll get 1.6 (like you mentioned). For part b, the actual solution would be 250mL instead of 200mL, but you don't know the molarity. To find the molarity, you'd need...
Mon Dec 03, 2018 10:50 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Proton transfer reactions?
Replies: 2
Views: 67

Re: Proton transfer reactions?

To expand off of the previous answer, in a net ionic equation, any species that doesn't take part in the reaction (and therefore appears on both sides) are not shown. This just highlights the main species participating in the reaction. If it asks for the "chemical equation" like it does in...
Tue Nov 27, 2018 11:12 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Dipole moment
Replies: 1
Views: 73

Re: Dipole moment

If you look at the lewis structure of CN2H2, the carbon is triple bonded to one of the nitrogens on one side and single bonded to the other nitrogen on the other side. The single bonded nitrogen is bonded to the two hydrogens. There is a net dipole since the single bonded nitrogen attracts the elect...
Tue Nov 27, 2018 11:04 pm
Forum: Hybridization
Topic: Lone Pairs
Replies: 1
Views: 117

Re: Lone Pairs

The type of hybridization of the orbitals is sp2. 2sp2 just specifies which orbitals within a specific energy level is participating in the hybridization. Thus, when it asks for the hybridization, just writing sp2 is fine. (I'm sure you won't be docked for writing 2sp2 either)
Tue Nov 27, 2018 10:57 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: 4.17
Replies: 3
Views: 154

Re: 4.17

There isn't double bonds on both terminal oxygens because if you count the number of electrons, 6x3 = 18 electrons. This means that the structure that you proposed is impossible; if there were double bonds on both terminal oxygens, there would only be a total of 16 electrons. Thus, there must be a d...
Wed Nov 21, 2018 1:44 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Drawing out Molecular Shape
Replies: 3
Views: 118

Re: Drawing out Molecular Shape

No, we will not be asked to draw VSEPR models, though it does help to visualize the molecule. When asked to draw the lewis structure, the standard dots and lines are fine.
Wed Nov 21, 2018 1:41 pm
Forum: Hybridization
Topic: Using hybrid orbitals
Replies: 2
Views: 74

Re: Using hybrid orbitals

It depends on the question since without the quantum number, it signifies the type of hybrid orbital the atom makes. With the quantum number, it specifies the hybridization of two specific subshells. However, to be safe, I would include the quantum number.
Wed Nov 21, 2018 12:45 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 2E.1
Replies: 2
Views: 38

Re: 2E.1

Option B can have lone pairs since if there are five electron regions with three lone pairs, the lone pairs will reside in the equatorial plane which would create a linear structure. Additionally, if there were 6 electron regions with four lone pairs, the resulting structure will also be linear. Sin...
Wed Nov 14, 2018 12:34 am
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Cations and polarizing power
Replies: 2
Views: 117

Re: Cations and polarizing power

Another way to look at it is that polarizing power is the ability of the cation to attract electrons (which cause distortions in the electrons of anions). Looking at it this way, smaller, highly-charge cations have the ability to attract electrons more, which makes them have a higher polarizing powe...
Wed Nov 14, 2018 12:29 am
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Determining the polarizability
Replies: 4
Views: 181

Re: Determining the polarizability

Charge is less relevant when looking at anions, rather you should look at the number of electrons in relation to the number of protons in the nucleus, in addition to the size of the anion. For example, in the problem 2D.11 in 7th edition that was brought up, the polarizability of N3- is more than O2...
Wed Nov 14, 2018 12:25 am
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Polarization of cation's
Replies: 2
Views: 86

Re: Polarization of cation's

A cation's polarizing power depends on its charge as well as the size of the ion. This is because the higher the positive charge, the stronger it will attract negative electrons. Additionally, the same logic applies to size, where the smaller the ion, the less shielding will be caused by electron sh...
Thu Nov 08, 2018 1:20 pm
Forum: Electronegativity
Topic: Trend Exceptions
Replies: 5
Views: 228

Re: Trend Exceptions

Expanding off of the previous answer, an exception for electron affinity is that the electron affinity of carbon is higher than that of nitrogen. This is because the addition of an electron to carbon makes the shell half-filled, which is much more favorable compared to nitrogen having an electron ad...
Thu Nov 08, 2018 1:13 pm
Forum: Octet Exceptions
Topic: Octet
Replies: 6
Views: 233

Re: Octet

In addition to expanded valence shells and radicals, as mentioned above, there are molecules that contain elements in group 13 of the periodic table, namely boron and aluminum, that are exceptions to the octet rule. These atoms only need 6 valence electrons in their shell (BF3, AlCl3). However, they...
Thu Nov 08, 2018 1:06 pm
Forum: Sigma & Pi Bonds
Topic: four bonds
Replies: 3
Views: 292

Re: four bonds

A single bond is a sigma bond, and the bonds that come after (the second and third ones in double and triple bonds) are pi bonds. Only the first bond will be a sigma bond while the other ones are pi bonds. Depending on the bonds and atoms in the molecule, there will be different amounts of sigma and...
Wed Oct 31, 2018 1:45 am
Forum: Formal Charge and Oxidation Numbers
Topic: FC in relation to Bonds
Replies: 2
Views: 90

Re: FC in relation to Bonds

The formal charge is calculated by Formal Charge = [# of valence electrons] – [non-bonding electrons + number of bonding electrons/2]. With more bonds, you'd have more bonding electrons. Just plug it in and calculate the formal charge.
Wed Oct 31, 2018 1:40 am
Forum: Resonance Structures
Topic: Bond Angels
Replies: 1
Views: 69

Re: Bond Angels

Most likely, we will have to memorize some of them and from the ones we memorize, we can deduce the bond angle in other cases. For example, for an atom with 3 atoms bonded to it and zero lone pairs, the bond angle is 120 degrees. When there is 2 atoms bonded to it and one lone pair, since the lone p...
Wed Oct 31, 2018 1:28 am
Forum: Lewis Structures
Topic: NH4+
Replies: 1
Views: 112

Re: NH4+

The fourth hydrogen is attached with dative covalent bond, otherwise called a co-ordinate bond. In this bond, both the electrons come from a single atom. How the bond between the fourth hydrogen and the nitrogen is formed is that the hydrogen would be ionized to H+ (hydrogen nucleus) and it would ac...
Wed Oct 24, 2018 3:13 pm
Forum: Photoelectric Effect
Topic: eV
Replies: 2
Views: 146

Re: eV

Yes, the conversion will be given on the first sheet of the test. 1eV = 1.602x10^-19 J
You should know how to convert between the two, but you do not need to memorize the conversion.
Wed Oct 24, 2018 3:10 pm
Forum: *Particle in a Box
Topic: Energy of shells vs. subshells
Replies: 2
Views: 404

Re: Energy of shells vs. subshells

In a given energy level, the energy of the subshells follow the order of s<p<d<f, given that the energy level has those subshells. However, if the orbitals are degenerate, then the orbitals in all the subshells in a given energy level would be the same energy. In other words, in hydrogen, energy lev...
Wed Oct 24, 2018 3:03 pm
Forum: Properties of Electrons
Topic: Degeneracy
Replies: 11
Views: 498

Re: Degeneracy

Degeneracy of orbitals means that there are no electrons in the orbitals at ground state and as a result, have the same energy as other orbitals in the energy level. For example, hydrogen has degenerate orbitals for everything higher energy than the 1s, meaning that its orbitals in energy level 2 al...
Wed Oct 17, 2018 1:02 pm
Forum: *Shrodinger Equation
Topic: Schrodinger's Equation Confused
Replies: 1
Views: 100

Schrodinger's Equation Confused

Hi, I'm a little confused with the concept of Schrodinger's Equation, so I just wanted to make sure I have it down. In my mind: the wave function is a mathematical function to represent orbitals (and therefore electrons) in an atom, where squaring it gives you the probability of finding an electron ...
Wed Oct 17, 2018 11:38 am
Forum: *Shrodinger Equation
Topic: Shrodinger Equation Meaning
Replies: 3
Views: 157

Re: Shrodinger Equation Meaning

Basically, Schrodinger's equation uses wave functions to describe electrons in an atom. Squaring the wave function represents the probability of find an electron at a given point, in other words, showing us the electron density. Looking at the equation H(psi) = E(psi), the energy represents the ener...
Wed Oct 17, 2018 11:27 am
Forum: Photoelectric Effect
Topic: Definition for photoelectric effect
Replies: 4
Views: 173

Re: Definition for photoelectric effect

In essence, the photoelectric effect happens when light is shone on a material, causing electrons to be emitted. In the experiment, scientists were confused when a low frequency light could not eject an electron since according to the wave model of light, electrons should have been emitted no matter...
Thu Oct 11, 2018 5:54 pm
Forum: Properties of Light
Topic: HW Question 1.23 (6th edition)
Replies: 1
Views: 39

Re: HW Question 1.23 (6th edition)

That's exactly how I would approach this problem. Convert keV to joules and then solve for frequency using E = hv and then for wavelength using c = (lambda)v. Alternatively, there's another value for Planck's Constant in eV where h = 4.1357 × 10^-15 eV s. However, it's all the same.
Thu Oct 11, 2018 5:38 pm
Forum: Properties of Light
Topic: Help with Question 1.57
Replies: 2
Views: 53

Re: Help with Question 1.57

In a Balmer series, n1 = 2 while n2 = 3,4,5,... There, you have the first four terms of the series where n2 = 3, 4, 5, 6. The next one in the series would have n2 = 7. So, doing the calculation, you get v = R(1/2^2 - 1/7^2) which ends up being 7.55 x 10^14. Solving for wavelength, lambda = c/v, we g...
Thu Oct 11, 2018 1:22 pm
Forum: Photoelectric Effect
Topic: Photoelectric effect Module 32
Replies: 1
Views: 67

Re: Photoelectric effect Module 32

Yes, the minimum energy is the E in KE = E - work function when KE is equal to zero. This means that you can just take the minimum frequency given to solve for the minimum energy. The phrase "must absorb radiation" refers to the photon absorbed in order to eject an electron. In this case, ...
Mon Oct 01, 2018 1:06 pm
Forum: Balancing Chemical Reactions
Topic: States of Matter
Replies: 11
Views: 614

Re: States of Matter

Aqueous means that something is dissolved in water. So something like NaOH(aq) would mean an NaOH solution with the solvent being water. Including states of matter when writing chemical equations provides a reference for how the reactants react with one another to form the products, kind of like an ...
Mon Oct 01, 2018 12:03 pm
Forum: Limiting Reactant Calculations
Topic: Audio Visual Question 22
Replies: 3
Views: 131

Re: Audio Visual Question 22

For this type of problem, you'll want to first balance the chemical equations, so balancing it, you get... C6H9Cl3 + 3AgNO3 --> 3AgCl + C6H9(NO3)3 Next, convert the grams of reactants into moles. So, 0.750g C6H9Cl3 / 187.50g mol-1 = 0.004 mol C6H9Cl3 1000g AgNO3 / 169.88g mol-1 = 5.886 mol AgNO3 *Bt...
Mon Oct 01, 2018 11:23 am
Forum: Empirical & Molecular Formulas
Topic: Multiple Ratio Possibility?
Replies: 4
Views: 143

Re: Multiple Ratio Possibility?

If you view the video module for Empirical Formulas on Dr. Lavelle's website, you can find his answer to this question around 22:40. Personally, I like to write down numbers up to three decimal points when calculating moles for empirical formulas; this rarely fails me. Generally, you can tell if the...