Search found 21 matches
- Sat Mar 18, 2017 1:13 pm
- Forum: *Cyclohexanes (Chair, Boat, Geometric Isomers)
- Topic: Winter 2011 Final Q8C
- Replies: 2
- Views: 1572
Winter 2011 Final Q8C
Question 8C asks you to draw the most stable conformation possible for cis-1-ethyl-2-methylcyclohexane and the answer shows a chair structure with an axial 2-methyl group and an equatorial 1-ethyl group, if I drew a chair structure with an equatorial 2-methyl group and an axial 1-ethyl group would t...
- Wed Mar 15, 2017 9:55 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Final 2014 Q3
- Replies: 3
- Views: 816
Re: Final 2014 Q3
Saw that this question was answered before: viewtopic.php?t=5476, hope this helps!
- Sat Mar 11, 2017 4:19 pm
- Forum: *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
- Topic: gauche and anti
- Replies: 1
- Views: 552
Re: gauche and anti
I think you can consider a staggered conformation another low energy conformation if you compare it to an eclipsed conformation.
- Sat Mar 04, 2017 1:00 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3621787
Re: Post All Chemistry Jokes Here
If all the good chemistry jokes argon then we should probably zinc of new ones
- Sun Feb 26, 2017 12:20 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalysts and equilibrium constant
- Replies: 4
- Views: 946
Re: Catalysts and equilibrium constant
A catalyst just provides a different pathway for the reaction to occur at a lower activation energy, this will not affect the value of the equilibrium constant.
- Thu Feb 16, 2017 11:21 pm
- Forum: General Rate Laws
- Topic: Reaction Rate Constant Units
- Replies: 2
- Views: 789
Re: Reaction Rate Constant Units
Yeah I'm pretty sure you're right!
- Sat Feb 11, 2017 12:33 pm
- Forum: Balancing Redox Reactions
- Topic: Writing Cell Diagrams
- Replies: 5
- Views: 1253
Re: Writing Cell Diagrams
I could be wrong but I don't think the placement of the aqueous ions matter in terms of which one comes before or after matters. As long as you have the rest of the notation down: (solid being oxidized | aqueous ions in anode beaker || aqueous ions in cathode beaker | solid being reduced), the aqueo...
- Sun Feb 05, 2017 8:23 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Chapter 14 Question #85
- Replies: 1
- Views: 541
Re: Chapter 14 Question #85
I saw that this was a question already answered: viewtopic.php?t=12118, hope this helps!
- Mon Jan 30, 2017 4:25 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Spontaneity
- Replies: 4
- Views: 1064
Re: Spontaneity
If you have a system where delta G = 0, the system is at equilibrium. In order for the reaction to be a spontaneous process delta G of the system must be negative/delta S universe is positive.
- Sat Jan 21, 2017 4:56 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies: 1
- Views: 520
Re: Hess's Law
Not sure if I'm understanding the question correctly, but I think you might be mixing up another concept with Hess' Law. Maybe in the question you were referring to the H2 was cancelled out during the process of manipulating the multi-step reactions to match the net reaction? Hope this helps
- Sun Jan 15, 2017 12:25 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Specific heat capacity [ENDORSED]
- Replies: 2
- Views: 826
Re: Specific heat capacity [ENDORSED]
I believe the answer is negative because the formation of water by its elements is exothermic, and yes because delta H is negative, it would imply that the reaction is exothermic. Hope this helps!
- Thu Dec 01, 2016 9:15 pm
- Forum: Lewis Acids & Bases
- Topic: 12.13
- Replies: 1
- Views: 1329
Re: 12.13
We know that Ag+ is a Lewis acid because it can act as an electron pair acceptor. For example, in a reaction where ammonia reacts with a silver ion to form a silver diamine ion, ammonia would act as an electron pair donor to the silver ion, making the silver ion an electron pair acceptor (aka. a Lew...
- Sat Nov 26, 2016 3:22 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Homework Help 12.21
- Replies: 2
- Views: 751
Re: Homework Help 12.21
In the equation Kw = [H30+] [OH-], the value 1.0 x 10^-14 is the value for Kw (equilibrium constant for water), it is equal to the product of the molar concentrations of hydronium and hydroxide ions and applies to parts a through c for question 12.21.
- Wed Nov 16, 2016 10:29 am
- Forum: Ideal Gases
- Topic: Ice Box Change in Concentrations
- Replies: 2
- Views: 764
Re: Ice Box Change in Concentrations
I think I understand what you're asking but I think the change in molar concentration just depends on the number of moles each molecule has. For example, with the reaction 2N2 + O2 <-> 2N2O, the change in concentration for N2 would be 2x because it has two moles, x for O2 since it only has one mole,...
Re: 17.35
I believe a chelating complex is a structure that has molecules that can form several bonds to a single metal ion. In question 17.35, only (b) would be considered a chelating complex because (a) and (c) have NH3 in positions where they wouldn't be able to bind to the metal ion more than once. Hope t...
- Fri Nov 04, 2016 5:29 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Polar/Non Polar 4.25d
- Replies: 2
- Views: 1260
Re: Polar/Non Polar 4.25d
I think where you might have made an error was how you drew the lewis structure of SF4. Since SF4 has an expanded octet, it would have one lone pair around the central sulfur and it would have a see-saw molecular shape, not a tetrahedral shape (which would indicate a nonpolar molecule). Since SF4 ha...
- Fri Oct 28, 2016 1:51 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Do we need to know exact bond angles?
- Replies: 3
- Views: 902
Re: Do we need to know exact bond angles?
I think what Prof. Lavelle said was that we didn't need to know exact bond angles but we should be able to predict if they're greater or less than the common bond angles (90°,120°,180°). Hope that helps!
- Sat Oct 22, 2016 12:36 am
- Forum: Ionic & Covalent Bonds
- Topic: HW 3.25 part a
- Replies: 1
- Views: 833
Re: HW 3.25 part a
I think in simpler terms they're just asking you to write the chemical formula of magnesium arsenide. To do so, you need to know the charges of the two ions that make up the compound; in this case, you need to figure out the charge of a magnesium (Mn) ion and an arsenic (As) ion. Looking at the peri...
- Wed Oct 12, 2016 6:32 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Question 2.19 part b and c [ENDORSED]
- Replies: 6
- Views: 1411
Re: Question 2.19 part b and c [ENDORSED]
b) The relationship between the angular momentum quantum number (L) and the magnetic quantum number (m sub L) is that m sub L can take on values of L, L-1,..., -L. Since the question is asking for m sub L values for an electron in a 6d subshell, that would mean that L=2. Because we know that L=2, we...
- Thu Oct 06, 2016 5:00 pm
- Forum: Photoelectric Effect
- Topic: Given the kinetic energy, how do you find the energy? [ENDORSED]
- Replies: 4
- Views: 1692
Re: Given the kinetic energy, how do you find the energy? [ENDORSED]
I think the energy of the incident light is referring to the threshold energy added with the kinetic energy. In the course reader, the formula is E(photon) = threshold energy + Ek; E(photon) refers to the energy of the incident light and Ek refers to the kinetic energy. Hope that helps!
- Thu Sep 29, 2016 8:09 pm
- Forum: Properties of Light
- Topic: Chapter 1 homework question
- Replies: 1
- Views: 516
Re: Chapter 1 homework question
I think the rounding might've thrown off your answer because when I used the value 2.998 x 10^8 m/s for the value for the speed of light, I got v = 1.1992 x 10^17 s. And if you use 6.626 x 10^-34 J as the value for planck's constant and multiply that by the frequency value you should end up with E =...