Search found 21 matches

by Destiny Dare 1D
Sat Mar 12, 2016 9:45 am
Forum: *Complex Reaction Coordinate Diagrams
Topic: Intermediates
Replies: 1
Views: 964

Re: Intermediates

The graph of page 88 of the course reader demonstrates an electrophilic addition. In an electrophile addition, an electrophile is added to another molecule twice (hence the two transition states). The intermediates in this case are the new electrophile and the new nucleophile that result from the fi...
by Destiny Dare 1D
Sun Mar 06, 2016 4:28 pm
Forum: *Ethers
Topic: Question 2.46
Replies: 1
Views: 470

Re: Question 2.46

Ether is not listed because the oxygen bonded to two carbons at the top of the structure is part of an ester functional group. Here is the general form of an ester functional group:

Image
by Destiny Dare 1D
Sat Feb 27, 2016 4:48 pm
Forum: *Alkanes
Topic: IUPAC and Common Names
Replies: 1
Views: 226

Re: IUPAC and Common Names

You can combine both IUPAC and common names if the problem does not specifically ask you for the IUPAC name.
by Destiny Dare 1D
Sat Feb 27, 2016 4:44 pm
Forum: *Electrophiles
Topic: nucleophiles vs electrophiles
Replies: 2
Views: 476

Re: nucleophiles vs electrophiles

No, but you should be able to recognize whether something acts as a nucleophile or an electrophile
by Destiny Dare 1D
Sun Feb 21, 2016 8:12 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: pg. 78 of course reader
Replies: 1
Views: 316

Re: pg. 78 of course reader

We recognize NO as a catalyst because it is one of the initial reactants that react to make the first product (it disappears) and then it shows up later as a product in the second step of the reaction (it reappears). The two intermediates, NO2 and O, do the opposite. They appear first as products, a...
by Destiny Dare 1D
Sat Feb 13, 2016 5:15 pm
Forum: General Rate Laws
Topic: Quiz prep winter 2013 #1
Replies: 1
Views: 356

Re: Quiz prep winter 2013 #1

Reaction rate of a reactant is found by the equation -\frac{1}{a} \frac{d[R]}{dt} where a is the stoichiometric coefficient of the reactant. When solving, you multiply by -\frac{1}{a} in order to account for the amount of moles that are reacting. For the reaction 2 MBr_{2}+O_{2}\rightarrow 2MO+2Br_{...
by Destiny Dare 1D
Sat Feb 06, 2016 5:24 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Writing Cell Diagrams
Replies: 1
Views: 321

Writing Cell Diagrams

How do you know whether to include Platinum when writing out a cell diagram? Also, why is there Platinum on both sides in some cell diagrams while there is only Platinum on one side in others?
by Destiny Dare 1D
Sun Jan 31, 2016 2:57 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: General Cell Reaction Question
Replies: 1
Views: 321

Re: General Cell Reaction Question

Platinum is not taken into account when calculating the cell potentials because it is just used in redox reactions as an electrode, a conductor through which electricity enters or leaves, that transfers electrons. A line is used in cell diagrams when there is a change in phase. Since there is not ch...
by Destiny Dare 1D
Sat Jan 23, 2016 6:11 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Quiz 1 Prep 2015 #11
Replies: 1
Views: 450

Re: Quiz 1 Prep 2015 #11

Question 11 tells us that for the reaction, PbO(s) + C(s) --> Pb(s) + CO(g), \Delta H =-106.9 kJ. By looking at the reaction given, we know that the given \Delta H occurs for 1 mol of PbO, so we can rewrite it as \frac{-106.9kJ}{1 mol PbO} . To figure out the heat released from converting 49.7g PbO ...
by Destiny Dare 1D
Sun Jan 17, 2016 2:54 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Degeneracy
Replies: 3
Views: 665

Re: Degeneracy

A higher degeneracy means high entropy. A high entropy means greater disorder which increases the chance that a spontaneous change(a reaction) will occur. Therefore, higher degeneracy is only better if you want a reaction to occur.
by Destiny Dare 1D
Sat Jan 09, 2016 12:20 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Change in internal energy: Really Big or Really Small
Replies: 1
Views: 271

Re: Change in internal energy: Really Big or Really Small

When solving for change in internal energy, a negative sign indicates that energy is lost by the system. A system with a large negative change in internal energy loses a greater amount of energy then a system with a small negative change in internal energy.
by Destiny Dare 1D
Sun Dec 06, 2015 9:24 pm
Forum: *Making Buffers & Calculating Buffer pH (Henderson-Hasselbalch Equation)
Topic: Halfway Point
Replies: 1
Views: 748

Re: Halfway Point

You can use the knowledge, that the volume at the halfway point is half the volume at the stoichiometric point, to find the pH at the halfway point. First, use the given to pH to solve for [H3O+] at the stoichiometric point. pH= 9.25 -log[H3O+]= 9.25 [H3O+]= 10^(-9.25)= 5.62x10^(-10)M Next, Multiply...
by Destiny Dare 1D
Sun Nov 29, 2015 5:49 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Strengths of Acids
Replies: 3
Views: 532

Re: Strengths of Acids

S-orbitals are closer to the nucleus than p-orbitals. So the more s-character a hybrid orbital has, the closer the electrons are to the atom. Such atoms are less willing to share their electron density (have greater electronegativity), and thus have greater acidic character.
by Destiny Dare 1D
Sun Nov 22, 2015 7:25 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Strengths of Acids
Replies: 3
Views: 532

Re: Strengths of Acids

The greater the s-character of a hybrid orbital, the greater the acidity of the acid. So an acid with hybridization will be stronger than an acid with hybridization, and an acid with hybridization will be stronger than an acid with hybridization.
by Destiny Dare 1D
Sun Nov 15, 2015 5:31 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: hw problem 11.39
Replies: 3
Views: 529

Re: hw problem 11.39

What you want to do first is write K as an expression for each given equation. For 2 BrCl (g) + H_{2} (g) \rightleftharpoons Br_{2} (g) + 2 HCl (g) (Equation 1): K_{1} = \frac{[Br_{2}][HCl]^{2}}{[BrCl]^{2}[H_{2}]} For H_{2} (g) + Cl_{2} (g) \rightlefth...
by Destiny Dare 1D
Sun Nov 08, 2015 7:27 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Chelates
Replies: 2
Views: 381

Re: Chelates

A chelate is a complex containing one or more ligands that form a ring of atoms that includes the central metal atom. For more information, you can read pg. 744 in the textbook.
by Destiny Dare 1D
Thu Oct 29, 2015 5:27 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: HW 4.93(d)
Replies: 1
Views: 283

Re: HW 4.93(d)

Based on the lecture given by Lavelle, we know that sigma bonds form in an end-to-end arrangement. The statement given by the solution manual for part D is describing an imaginary line, called the internuclear axis, that connects the nucleus of one atom to the nucleus of another atom. In a sigma bon...
by Destiny Dare 1D
Sun Oct 25, 2015 1:22 am
Forum: Formal Charge and Oxidation Numbers
Topic: Understanding Formal Charge
Replies: 1
Views: 348

Re: Understanding Formal Charge

For both lone pairs and bonded electrons you count the electrons individually. For example, in a water molecule we know that oxygen has 2 lone pairs and 2 single bonds (1 single bond with each hydrogen). Using the formula given by Dr. Lavelle, FC=V-(L+\frac{S}{2}) , we find that the formal c...
by Destiny Dare 1D
Sun Oct 18, 2015 7:03 pm
Forum: Dipole Moments
Topic: Charge in relation to bond length
Replies: 1
Views: 714

Re: Charge in relation to bond length

Bond length and charge difference are inversely related. A larger difference in charge would make for a stronger bond, meaning a shorter bond length.
by Destiny Dare 1D
Sun Oct 11, 2015 1:30 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Electron Configurations of Ions
Replies: 1
Views: 402

Re: Electron Configurations of Ions

Before adjusting the amount of electrons, a molecule has a neutral charge meaning that there is the same amount of protons as there is electrons. By removing an electron from the molecule, you are actually making the molecule "less negative". Thus, the resulting cation will have a positive...
by Destiny Dare 1D
Sun Oct 11, 2015 12:45 pm
Forum: Photoelectric Effect
Topic: Clarification of the kinetic energy of an ejected electron
Replies: 1
Views: 396

Re: Clarification of the kinetic energy of an ejected electr

m=mass and v=velocity
The mass of an electron is a constant that can be looked up, and the velocity of the electron should be given in the problem.

Go to advanced search