How did the hipster chemist burn his hand?
He picked up his beaker before it was cool.
Search found 62 matches
- Sun Mar 17, 2019 12:00 am
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2764602
- Sat Mar 16, 2019 11:56 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Adding an Inert Gas
- Replies: 4
- Views: 561
Re: Adding an Inert Gas
Adding an inert gas will not shift the equilibrium because what we are really seeing occur in these problems with increasing pressure is (usually) a change in the concentration (defined by C=n/v). The volume does not change when an inert gas is pumped into the system, and therefore the concentration...
- Sat Mar 16, 2019 11:53 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Q
- Replies: 6
- Views: 798
Re: Q
The concentrations do not have to be in standard conditions when calculating Q.
- Fri Mar 08, 2019 9:58 pm
- Forum: General Rate Laws
- Topic: Coefficients and Products
- Replies: 1
- Views: 221
Re: Coefficients and Products
Since we write rate laws based on the concentration of reactants, I do not think that products would have much of an effect. Coefficients in the reactants are important because they are directly related to the molecularity which is needed to write the elementary and overall rate laws of reactions (e...
- Fri Mar 08, 2019 9:42 pm
- Forum: First Order Reactions
- Topic: Homework 7B.3 part C
- Replies: 1
- Views: 243
Re: Homework 7B.3 part C
For this question, you have to think of how the changes in concentration for B affect the concentration of A at the same time period, such as an ICE table. Although we're not dealing with a reaction at equilibrium, the concepts still apply. 2A ----> B + C I 0.153M 0 C -2x +0.034 E 0.153-2x 0.034 The...
- Fri Mar 08, 2019 9:19 pm
- Forum: General Rate Laws
- Topic: Third Order Reactions
- Replies: 1
- Views: 279
Re: Third Order Reactions
If we were to keep it in the linearized form that makes it easy to plot reaction data and see what kind of order a reaction is, the rate law for a third order reaction would be 1/[A]^2 = 2kt + 1/[Ai]^2, with Ai representing the initial concentration of some reactant A.
- Sun Mar 03, 2019 12:41 am
- Forum: General Rate Laws
- Topic: Overall sum
- Replies: 7
- Views: 734
Re: Overall sum
Yes. The overall order reaction is the sum of the partial orders from each substance (n+m if there are two). The amount of exponents could be greater depending on how many reactants you have.
- Sun Mar 03, 2019 12:28 am
- Forum: Balancing Redox Reactions
- Topic: 7th edition 6L.5
- Replies: 1
- Views: 295
Re: 7th edition 6L.5
The anode and the cathode separate HCl into H+ and Cl- because HCl is a strong acid and will completely dissociate into its ions (H+ and Cl-) in aqueous solution. Therefore, we specify their location in the anode and the cathode because HCl will not effectively be present in water (it will only be p...
- Sun Mar 03, 2019 12:13 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Reaction order
- Replies: 3
- Views: 408
Re: Reaction order
The reaction order represents how changing the concentrations of the reactants changes the overall reaction rate. Numerically, it is the sum of the orders of reaction with respect to the various reactants (ex: for a system with two reactants A and B where rate = k[A]^x[B]^y, the rate order is x+y).
- Fri Feb 22, 2019 10:27 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Hw Ch.9 #75 (6th Edition) Residual Entropy
- Replies: 2
- Views: 281
Re: Hw Ch.9 #75 (6th Edition) Residual Entropy
By definition, a trans isomer must have atoms that are on opposite sides of each other (ie the bond angle between them is 180 degrees in a MX2Y4 molecule). With an octahedral molecule, there are only three unique ways to orient the atoms so that they remain opposite to each other (left/right, up/dow...
- Thu Feb 21, 2019 9:39 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6L.11 7th Edition
- Replies: 1
- Views: 242
6L.11 7th Edition
The question reads "a) Write balanced half-reactions for the redo reaction of an acidified solution of potassium permanganate and iron(II) chloride b) Write the balanced equation for the cell reaction and devise a galvanic cell to study the reaction." Now I understand how to get the cell d...
- Thu Feb 21, 2019 9:19 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Work max and delta G
- Replies: 2
- Views: 266
Re: Work max and delta G
Delta G is defined as the amount of energy available to do useful work (ie non expansion work) for a spontaneous process under a constant temperature and pressure. Therefore, by definition delta G equals the maximum possible work that the system can do. A proof for this exists that involves some dif...
- Sat Feb 16, 2019 9:02 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2764602
Re: Post All Chemistry Jokes Here
Why does a pound of hamburger have less energy than a pound of steak?
Because it is in the ground state.
Because it is in the ground state.
- Sat Feb 16, 2019 8:50 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Constant Pressure Calorimeter
- Replies: 5
- Views: 922
Re: Constant Pressure Calorimeter
I think it would be considered an isolated system because the reaction must not lose heat to the environment in order for us to make calculations. If it was able to exchange energy as heat (like one and closed systems can), measuring the heat of the reaction would be impossible.
- Sat Feb 16, 2019 7:10 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: When does delta U equal zero?
- Replies: 17
- Views: 8153
Re: When does delta U equal zero?
I would just like to add that while delta U equals zero in an isothermal process (or in any isolated system), the two have no change in U for different reasons. In an isothermal process, the energy exchanged as heat (q) equals the amount of work done by the system (q = -w) so any heat energy from th...
- Sat Feb 09, 2019 8:38 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Entropy and Equilibrium
- Replies: 2
- Views: 365
Re: Entropy and Equilibrium
For an isolated system, when a system is at equilibrium, it has searched out all of the possible states/positions of atoms. In other words, an isolated system at equilibrium would have maximum degeneracy, which would result in the maximum entropy.
- Sat Feb 09, 2019 8:22 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Homework Question 6th ED
- Replies: 1
- Views: 196
Re: Homework Question 6th ED
You first have to start by finding the enthalpy of the reaction, which you just calculate normally with the values in the appendix and the given one for TNT. You should get -13168kJ. Per mole of TNT (there are 4 in the original reaction), 1/4 of this amount of energy, or 3292 KJ, will be released pe...
- Sat Feb 09, 2019 8:07 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57 6th Edition
- Replies: 1
- Views: 223
Re: 8.57 6th Edition
The c subscript is supposed to represent combustion, so the problem is giving you each molecule's enthalpy of combustion. To do Hess' Law with this, you would have to write out the combustion reaction for each molecule (molecule + O2 --> CO2 + H2O for all molecules except H2) and then do it like you...
- Sat Feb 09, 2019 7:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: homework 6th edition 12.31
- Replies: 1
- Views: 261
Re: homework 6th edition 12.31
When the solution is diluted from the 5.00 mL that were transferred to 500.0 mL, we are given what the pH of that solution is. because we know that pH + pOH = 14 and that [OH-] = 10^-pOH, we would just find the pOH of the solution (14-13.25 = 0.75) and then put that into [OH-] = 10^-pOH to find the ...
- Fri Feb 01, 2019 9:20 pm
- Forum: Ideal Gases
- Topic: pV=nRT
- Replies: 12
- Views: 1989
Re: pV=nRT
In addition to the many uses labeled above, you can also use the law to calculate work in a constant-pressure system (-P*deltaV) when you are not given either the change in volume or the pressure because assuming that you have been given the change in the moles of gas in the reaction, you can equate...
- Fri Feb 01, 2019 8:48 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 4A11
- Replies: 2
- Views: 292
Re: 4A11
The first step would be to realize that the amount of energy supplied to the calorimeter is the same amount of energy as the amount of energy lost by the surroundings (q(cal) = -q(surroundings)). The surroundings lost energy, so the calorimeter gained -(-22.5kJ) = 22.5kJ. We can then divide this num...
- Fri Feb 01, 2019 8:32 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Thermo Test/Midterm
- Replies: 9
- Views: 716
Re: Thermo Test/Midterm
I don't think we will be, considering we learned how to do it in 14A. My TA mentioned that we would be given the bond dissociation energies for all of the bonds needed to make the calculation, but we would have to know based on Lewis structures whether we had to use the energy for a triple, double, ...
- Fri Jan 25, 2019 8:04 pm
- Forum: Phase Changes & Related Calculations
- Topic: 6th edition 8.13
- Replies: 1
- Views: 222
Re: 6th edition 8.13
The internal energy of a system (the cylinder in this case) is defined as deltaU = q + w, where q is the energy transferred as heat and w is the work done. The total energy change is given to us, and we know it must be negative because combustion reactions are exothermic. (deltaU = -2573). We are al...
- Fri Jan 25, 2019 7:53 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6D.15 part b 7th
- Replies: 1
- Views: 255
Re: 6D.15 part b 7th
In water, AlCl3 ionizes to form Al3+ and Cl- ions. The Cl- ions do not affect the pH, but in aqueous solution, water molecules surround the highly positively charged Al3+ and form coordinate covalent bonds with the Al atom. This is written as Al(H2O)6 3+, which is the behavior of the Al3+ ion in aqu...
- Tue Jan 22, 2019 10:28 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6A.19, 7th Edition
- Replies: 1
- Views: 171
6A.19, 7th Edition
"Calculate the molar concentrations of OH- in solutions with the following molar concentrations of H3O+: a) 0.020 mol*L^-1 b) 1.0 x 10^-5 c) 3.1 mol * L^-1." I get a) and b), but why is the answer in the back of the book for c) 3.2 x 10^-12 mol * L^-1? When I solved it by doing Kw/[H3O+], ...
- Thu Jan 17, 2019 7:06 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6D.3 7th ed, calculating Ka and pKa
- Replies: 1
- Views: 164
Re: 6D.3 7th ed, calculating Ka and pKa
It says that [H3O]=[ClO2] because for every ClO2- ion that is made, a H3O+ ion must be made as well (They have a molar ratio of 1:1). We subtract .06 from 0.10 based on the ICE chart for this model. It looks like this (ignore water because it's the solvent of the reaction): HClO2(aq) + H2O(l) ------...
- Thu Jan 17, 2019 6:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Question
- Replies: 2
- Views: 139
Re: Textbook Question
For a), you would start out with balancing the equation of Cl2 dissociating into Cl atoms (Cl2 --> 2CL). We'll use this later. Then, we are given the amount of mols and the volume, so we can calculate the initial concentration of Cl2, which is (2.0x10^-3 mol) / (2.0L) = 0.0010 mol*L^-1. From here, w...
- Thu Jan 17, 2019 6:20 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Conjugate base
- Replies: 7
- Views: 655
Re: Conjugate base
Conjugate bases are what remain when an acid donates a proton. In a reaction, you can spot them because they are often negatively charged because they have donated a proton during the reaction. For example, formic acid, HCOOH, has a conjugate base of HCO2- (the H+ is removed from the carboxyl group ...
- Sat Jan 12, 2019 2:45 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: partial pressures vs concentrations
- Replies: 4
- Views: 276
Re: partial pressures vs concentrations
If the problems ask for Kc or Kp, that gives a pretty clear indication of what is desired. If they just ask for K, look at the units that they give. There is a good chance that you can figure out what they are asking for based on if the amounts given to you are in mol/L, atm, bar, etc. Also, the che...
- Sat Jan 12, 2019 2:37 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Homework 11.79 6th edition
- Replies: 2
- Views: 245
Re: Homework 11.79 6th edition
In regards to the first part, the use of bar as your units is not really significant; the bar is just a certain kind of unit of pressure (note that bar is a non-SI unit) and it won't affect how we solve the problem. To start the problem, calculate K based on the equilibrium pressures that are given ...
- Fri Jan 11, 2019 11:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Minimum and Maximum
- Replies: 2
- Views: 228
Re: Minimum and Maximum
Theoretically, there is no maximum or equilibrium constant that exists. However, there is a point where K is so large (the concentration of products is so much larger than the concentration of products) that the reaction is essentially running to completion. There is also a point where K becomes so ...
- Sun Dec 09, 2018 4:48 pm
- Forum: Bronsted Acids & Bases
- Topic: Difference
- Replies: 1
- Views: 507
Re: Difference
With enough knowledge about dipole-dipole interactions, you could calculate the dipole moment on the separate parts of the oxoacid and see how that affects the wave functions of the other atoms. However, that's getting into pretty advanced physical chemistry, which is pretty interesting (albeit outs...
- Sun Dec 09, 2018 4:38 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: acid rain formula
- Replies: 4
- Views: 472
Re: acid rain formula
Based on what I have in my notes, the equation is
SO2(g) + H20(l) ---> H2SO3(aq). There should be an arrow going in the opposite direction (Sulphurous acid is a weak acid so the reaction can go the other way), but I don't know how to type that out.
SO2(g) + H20(l) ---> H2SO3(aq). There should be an arrow going in the opposite direction (Sulphurous acid is a weak acid so the reaction can go the other way), but I don't know how to type that out.
- Sun Dec 09, 2018 4:35 pm
- Forum: Amphoteric Compounds
- Topic: Examples
- Replies: 5
- Views: 754
Re: Examples
Another example of an amphoteric compound (besides water) is Al2O3, which is also an amphoteric oxide.
- Sun Dec 02, 2018 12:29 pm
- Forum: Lewis Acids & Bases
- Topic: Weak Acids and Bases 12.53
- Replies: 1
- Views: 273
Re: Weak Acids and Bases 12.53
For a), trichloroacetic acid has a -CCl3 group that is bonded to a carboxyl group. That -CCl3 group can attract electrons more effectively than the CH3 group in acetic acid (Chlorine is far more electronegative than carbon and creates a polar bond, whereas the C-H bond in CH3 is non-polar), making i...
- Sat Dec 01, 2018 11:56 pm
- Forum: Naming
- Topic: 7th Edit 9C1
- Replies: 2
- Views: 265
Re: 7th Edit 9C1
With coordination compounds, NH3 is actually a neutral ligand with a charge of 0 (with the lone pair on NH3, its formal charge amounts to zero, and it will often donate that lone pair, making it a neutral ligand). So in this ligand, the sulfate would have a 2- charge, and in order to balance the cha...
Re: Aqua
I'm pretty sure that the textbook writes aqua as OH2 as a convention to indicate that the lone pairs on the oxygen atom are where the aqua ligand bonds to the central metal atom. Therefore, writing it either way would be okay and would just be down to personal preference (regardless of how you write...
- Sun Nov 25, 2018 1:02 pm
- Forum: Bond Lengths & Energies
- Topic: Bonds in Water
- Replies: 5
- Views: 563
Re: Bonds in Water
The difference between the two is that the covalent bonds of the water molecule occur within the molecule whereas the hydrogen bonds are intermolecular forces. Specifically, between the two hydrogen molecules and the O atom in a water molecule, there is an unequal sharing of the electrons due to the...
- Sun Nov 25, 2018 12:52 pm
- Forum: Sigma & Pi Bonds
- Topic: HW 4.43
- Replies: 1
- Views: 104
Re: HW 4.43
For this question, I thought of the "s-character" as the fraction of the s-orbitals in a hybridized atom (ex. of the three orbitals in an sp2 hybrid, 1 is an s-orbital, making the hybrid have 1/3, or about 33.3% "s-character"). So as shown in the problem, the larger the s-charact...
- Sun Nov 25, 2018 12:44 pm
- Forum: Sigma & Pi Bonds
- Topic: Sigma and pi bonds and hybridization
- Replies: 2
- Views: 509
Re: Sigma and pi bonds and hybridization
Sigma and pi bonds are ways to describe bonds that form in molecules, and knowing where a sigma or a pi bond will be can help us predict orbital hybridization. A single bond is a sigma bond, a double bond is a pi-bond and a sigma bond, and a triple bond is a sigma bond plus two pi bonds. This hybrid...
- Fri Nov 16, 2018 12:17 am
- Forum: Lewis Structures
- Topic: Axial Bond Angles
- Replies: 3
- Views: 640
Re: Axial Bond Angles
Axial bond angles are a unique type of bond angle usually seen in more complicated lewis structures. They are the bond angles involving the atoms on the axis of the molecule at a 90 degree angle to the equatorial plane. An example of an axial bond angle is the one seen in PCl5, which has a trigonal ...
- Fri Nov 16, 2018 12:05 am
- Forum: Electronegativity
- Topic: noble gases
- Replies: 2
- Views: 310
Re: noble gases
Purely based on periodicity, noble gases would be more electronegative than oxygen; however, all of the noble gases have a "filled" valence shell of eight electrons and they would not want to attract any more (it would not improve their stability or be energetically favorable to have one m...
- Thu Nov 15, 2018 11:57 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Shortcut
- Replies: 4
- Views: 929
Re: Shortcut
The shortcut that you can use is Formal Charge = # of Valence Electrons - (# of Lone Pair Atoms + (1/2)(# of Bonding electrons))
- Thu Nov 15, 2018 11:29 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Discussion posts grading (Anyone have a grade for this yet?)
- Replies: 3
- Views: 475
Re: Discussion posts grading (Anyone have a grade for this yet?)
As long as you're posting 3 times a week, you should be fine. I've heard that they count up the amount of times that you participate at the end of the quarter and then give you the points accordingly.
- Fri Nov 09, 2018 9:06 am
- Forum: Bond Lengths & Energies
- Topic: 2D.3 (7th ed)
- Replies: 2
- Views: 216
Re: 2D.3 (7th ed)
The answer to this question mainly comes down to which pair of these atoms has a larger electronegativity difference between them. Barium is down towards the left of the periodic table (where electronegativity is low) and Beryllium is higher up in the group than Barium. Thus, the difference in elect...
- Fri Nov 09, 2018 12:25 am
- Forum: Electronegativity
- Topic: electronegativity trends
- Replies: 6
- Views: 590
Re: electronegativity trends
As you go down a group, more sub shells are added that are farther and farther away from the nucleus, making the atoms less likely to attract electrons. As you move across a period, the amount of protons in the nuclei of the atoms increase, increasing the nuclear charge and the ability of the atom t...
- Fri Nov 09, 2018 12:20 am
- Forum: Ionic & Covalent Bonds
- Topic: Proportionality of bond strength
- Replies: 3
- Views: 497
Re: Proportionality of bond strength
The relationship between the number of bonds and the bond strength is not linear because as the amount of bonds increase, the amount of electron repulsion between the bonds increases as well, so each pair is not as effective at bonding (and thus has a lower individual bond strength) as a pair of ele...
- Fri Nov 02, 2018 10:55 pm
- Forum: Lewis Structures
- Topic: Radicals
- Replies: 3
- Views: 388
Re: Radicals
Adding on to what people are saying above, it becomes easy to identify radicals when you look at the Lewis structure. Because radicals always have an unpaired electron, the Lewis diagram for a radical will always have an odd number of valence electrons.
- Fri Nov 02, 2018 10:50 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: 2.51 (6th edition)
- Replies: 1
- Views: 676
Re: 2.51 (6th edition)
To solve this problem, think about the Aufbau principle in filling up the orbitals. Based on the periodic table, Silicon has an electron configuration of [Ne] 3s2 3p2. For silicon, the 3s orbital is filled, leaving us with the p orbital to look at. Based on Hund's rule, the last two electrons will o...
- Thu Nov 01, 2018 10:31 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration of Silver
- Replies: 1
- Views: 475
Re: Electron Configuration of Silver
Silver is similar to Copper (it is in the same group as Copper) in that to maximize stability, it will completely fill its d orbital before filling its last s orbital, which is why we get the 4d105s1 configuration. The reason why it does this has to do with the relative energies of the 4s and 3d orb...
- Fri Oct 26, 2018 12:08 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration Exceptions [ENDORSED]
- Replies: 3
- Views: 479
Re: Electron Configuration Exceptions [ENDORSED]
The exceptions are slightly different for every element, but some of them revolve around prioritizing stability in the sub shells by either filling them up halfway or taking an electron from another sub shell to completely fill a sub shell. This is the case with Copper; to get a more symmetric distr...
- Thu Oct 25, 2018 11:51 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Using Heisenberg's Equation
- Replies: 3
- Views: 411
Re: Using Heisenberg's Equation
If you are talking about the h/4pi term, you can write it in two different ways. The textbook uses the term ħ, which is just a shorthand way of saying h/2pi. Thus, for the Heisenberg uncertainty principle, (1/2)(ħ) = h/4pi. There's no difference between the two numerically; it's just written that wa...
- Thu Oct 25, 2018 11:41 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Vocabulary confusion
- Replies: 2
- Views: 553
Re: Vocabulary confusion
The principal quantum number (denoted by n) specifies the shell that an electron is in, and it is just a whole number starting from 1 (1, 2, 3...). The magnetic quantum number (denoted by ml, with the l in the subscript) specifies which orbital of the sub shell the electron is in. Its values are bas...
- Wed Oct 17, 2018 11:20 am
- Forum: Student Social/Study Group
- Topic: Past Final Exams
- Replies: 6
- Views: 710
Re: Past Final Exams
I would like to know as well. Send them to 321dors@gmail.com please.
- Tue Oct 16, 2018 2:50 pm
- Forum: Properties of Light
- Topic: Question 1B.7
- Replies: 2
- Views: 135
Re: Question 1B.7
I'm pretty sure that we would have to work that information out ourselves because the Einstein equation tells us how many Joules are emitted per atom (remember the 1 atom emitting an electron to 1 photon ratio).
- Tue Oct 16, 2018 2:28 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: G Orbital
- Replies: 2
- Views: 771
Re: G Orbital
The g orbital is an orbital after the f orbital that only exists with excited states of the current 118 known elements. It is not used in writing ground state electron configurations because none of the elements that we know of right now need a g orbital to have all of their electrons accounted for ...
- Thu Oct 11, 2018 12:52 pm
- Forum: Photoelectric Effect
- Topic: Threshold energy
- Replies: 5
- Views: 4394
Re: Threshold energy
The threshold energy is defined as the amount of energy needed to remove an electron from a metal. This energy is called the work function (Φ), and their values are equivalent.
- Thu Oct 11, 2018 12:42 pm
- Forum: Properties of Light
- Topic: Wavelength
- Replies: 10
- Views: 407
Re: Wavelength
The process by which you convert the wavelength does not really matter, but I personally find it easier to keep the wavelength units in meters (and use powers of 10) until the end; it makes it easy to track the units and make conversions involving m/s (usually the speed of light). Converting it to n...
- Mon Oct 08, 2018 11:26 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Question 1.27
- Replies: 2
- Views: 214
Re: Question 1.27
To start this question, find the amount of Joules that the light will give off in 2 seconds. At 32J*s^-1 for 2 seconds, the light will generate 64 Joules. Next, use a combination of Einstein's equation (E = hv) and the definition of the speed of light (c = λ*v, v = c/λ) to formulate the equation E =...
- Wed Oct 03, 2018 12:21 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Question G21 [ENDORSED]
- Replies: 2
- Views: 408
Re: Question G21 [ENDORSED]
After converting the grams to moles, you should have about 2.29 x 10^-2 mols of K from all three of the compounds (this can be found from the molar ratio within each compound). Then, use the definition of molarity (mol/L) to get (2.29 x 10^-2 mols K)/(0.500 L) = 4.58 x 10^-2 M. For the second part o...
- Tue Oct 02, 2018 12:06 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Pre-Module Assessment Q23 and Q25
- Replies: 2
- Views: 678
Re: Pre-Module Assessment Q23 and Q25
23. First for this problem, you will have to find the molar mass of KMnO4. Using the given molar masses, it comes out to be 158.04 g x mol^-1. By multiplying this by the amount of grams given (5.00), you will get the moles of KMnO4 (5.00 g / 158.04 g x mol(KMnO4)^-1 = 0.0316 moles.) To find the mola...
- Mon Oct 01, 2018 11:45 am
- Forum: SI Units, Unit Conversions
- Topic: Unit "t"
- Replies: 2
- Views: 177
Re: Unit "t"
I think the unit "t" here refers to a metric ton, which is equivalent to 1.00 x 10^6 grams. Hope this helps :)