Search found 20 matches
- Sun Mar 13, 2016 12:15 pm
- Forum: *Alcohols
- Topic: 2015 Final 9A second molecule
- Replies: 1
- Views: 552
Re: 2015 Final 9A second molecule
Yes, the functional group takes priority over the cycloheptane in this problem.
- Sun Mar 06, 2016 10:01 pm
- Forum: *Nucleophiles
- Topic: Double Bonds
- Replies: 3
- Views: 783
Re: Double Bonds
A double bond is considered a nucleophile because it is a region that is electron rich.
Re: Suffixes
-ane indicates it is an alkane with a single bond between carbons
-ene indicates it is an alkene with a double bond between carbons
-yne indicates it is an alkyne with a triple bond between carbons
-ol indicates it is an alcohol with a functional group -OH attached
-ene indicates it is an alkene with a double bond between carbons
-yne indicates it is an alkyne with a triple bond between carbons
-ol indicates it is an alcohol with a functional group -OH attached
- Sat Feb 20, 2016 10:34 pm
- Forum: *Organic Reaction Mechanisms in General
- Topic: illustrating electron movement with arrows
- Replies: 1
- Views: 417
Re: illustrating electron movement with arrows
Yes, the base of an arrow should start at the bond rather than at an element such as carbon. It should end pointing to the element that is receiving the electron.
- Sun Feb 14, 2016 8:01 pm
- Forum: General Rate Laws
- Topic: Meaning of orders of rate laws
- Replies: 1
- Views: 500
Re: Meaning of orders of rate laws
The order specifies what relationship there is between the rate of a reaction and the concentration of a substance. For a first order reaction, the rate is proportional to the first power of the concentration. For second order, the rate is proportional to the second power of the concentration. For z...
- Sun Feb 07, 2016 1:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: HW Question 14.13 (c)
- Replies: 1
- Views: 364
HW Question 14.13 (c)
14.13 Write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following equations: Cl2(g)+H2(g)->HCl(aq) I know that cell diagrams are normally written in the order oxidation reduction, so in the answer Pt(s)/H2(g)/H+(aq)//Cl-(aq)/Cl2(g)/Pt(s) if H...
- Fri Jan 29, 2016 2:37 pm
- Forum: Balancing Redox Reactions
- Topic: Identifying Which Half-Reaction to Reverse
- Replies: 3
- Views: 744
Re: Identifying Which Half-Reaction to Reverse
In these problems you want to get a positive voltage. From lecture today, Lavelle mentioned that if there is a negative voltage, then the reaction is not favorable.
- Sat Jan 23, 2016 5:28 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Homework problem 8.59
- Replies: 5
- Views: 2210
Re: Homework problem 8.59
It doesn't add the standard enthalpy of N2 because it is equal to zero. In section 8.17 of the textbook, it says that for a element in its most stable form, the standard enthalpy of formation is zero.
- Sun Jan 17, 2016 5:46 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3000496
Re: Chemistry Jokes
Which one are you?
- Sun Jan 10, 2016 3:13 pm
- Forum: Phase Changes & Related Calculations
- Topic: Melting/Evaporation vs Vaporization
- Replies: 4
- Views: 1151
Re: Melting/Evaporation vs Vaporization
I think that in order for sublimation to occur, the solid must change to a liquid then a gas. In his lecture notes, Lavelle described the enthalpy of sublimation as Hsub=Hfus+Hvap. This indicates that in the process of sublimation, fusion (solid to liquid) and vaporization (liquid to solid) both occ...
- Sun Dec 06, 2015 1:18 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: shape when given a formula for a complex ion
- Replies: 1
- Views: 521
Re: shape when given a formula for a complex ion
Ethylenediamine is bidentate, so it can form two bonds. Because there are two of these and two bromine attached to Cobalt, there are six bonds total. This gives a VSEPR formula of AX6, which is octahedral.
- Sun Nov 29, 2015 5:09 pm
- Forum: Bronsted Acids & Bases
- Topic: Bronsted and Arrhenius Acids/Bases
- Replies: 1
- Views: 948
Re: Bronsted and Arrhenius Acids/Bases
The main difference between Bronsted and Arrhenius acids and bases is that Arrhenius' definition is specific to aqueous solvents (aka water). He defined an acid as a compound that contains hydrogen and reacts with water to form hydrogen ions and a base as a compound that produces hydroxide ions in w...
- Sun Nov 22, 2015 3:33 pm
- Forum: Amphoteric Compounds
- Topic: Determining if oxides are acidic, basic, or amphoteric
- Replies: 2
- Views: 6770
Re: Determining if oxides are acidic, basic, or amphoteric
A general guideline to determine if an oxide is acidic, basic, or amphoteric is to use the periodic table. Typically, metals such as Ba form basic oxides (BaO), while nonmetals such as S form acidic oxides (SO3). The elements in and around the diagonal line of metalloids do tend to form amphoteric o...
- Sun Nov 15, 2015 4:14 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: "Steady State"
- Replies: 1
- Views: 618
Re: "Steady State"
From what I have read it looks like the main difference is this: In equilibrium, the rates of the forward and reverse reactions are equal and therefore there is no net change. In a steady state reaction, which can be represented as A->B->C, the reactants (A) and products (C ) are changing, while the...
- Sun Nov 08, 2015 2:31 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Homogeneous vs. Heterogeneous Equilibria
- Replies: 1
- Views: 821
Re: Homogeneous vs. Heterogeneous Equilibria
To the best of my understanding, a phase is the form of matter (solid, liquid, gas), so "in the same phase" means that the reactants and products are all gases, all liquids, or all solids. For example, a homogeneous equilibria could have all reactants and products in the gas phase, whereas...
- Sun Nov 01, 2015 4:06 pm
- Forum: Hybridization
- Topic: What hybrid orbitals are involved
- Replies: 1
- Views: 496
Re: What hybrid orbitals are involved
This question is basically asking for the hybridization of carbon, but it is a little more specific because it provides the molecule, Benzene, and the specific bonds to look at, C-C and C-H. You can solve this by drawing the Benzene molecule, and figuring out which hybridization carbon must undergo ...
- Sun Oct 25, 2015 4:59 pm
- Forum: Electronegativity
- Topic: Electronegativity Values
- Replies: 1
- Views: 564
Re: Electronegativity Values
I don't believe we have to memorize the specific values for each element. If anything, I would try to remember the trends in the periodic table regarding electronegativity. For example, electronegativity seems to decrease moving down a group and increase moving right across a period. This is excludi...
- Sun Oct 18, 2015 2:26 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron configuration
- Replies: 2
- Views: 1300
Re: Electron configuration
From my understanding, the reason it is [Kr] 4d10 5s1 and not [Kr] 4d9 5s2 is because 9 electrons in the d orbital is just one electron away from having a full orbital. Because it is so close, there is a tendency for one of the 5s electrons to fill the 4d orbital, bringing it to 10 electrons. Lookin...
- Sun Oct 11, 2015 3:06 pm
- Forum: DeBroglie Equation
- Topic: Determining whether or not there are wavelike properties
- Replies: 2
- Views: 595
Re: Determining whether or not there are wavelike properties
From my understanding, the smallest amount the solution can be is _x10^-15 in order to be detected. A number smaller than this, such as x10^-38 in the example on page 12 is too small to be detected, but larger numbers such as x10^-10 in the second example are large enough to be detected. Generally, ...
- Sun Oct 04, 2015 5:29 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: h/2pi, How does this play a part in Heisenber's Equation?
- Replies: 1
- Views: 962
Re: h/2pi, How does this play a part in Heisenber's Equation
To the best of my understanding, hbar is equal to h/2pi. Therefore, hbar/2 can be rewritten as (h/2pi)x(1/2), which when multiplied out is equal to h/4pi. When Heisenberg's equation mentions h/4pi, it is simply another way to write out hbar/2 in an easier to understand format.