Search found 10 matches
- Sat Mar 18, 2017 3:28 pm
- Forum: *Aldehydes
- Topic: Aldehydes vs. Ketones
- Replies: 2
- Views: 1696
Re: Aldehydes vs. Ketones
An aldehyde can only be at the end of a carbon chain while a ketone can never be at the end.
- Sun Mar 12, 2017 2:19 pm
- Forum: *Alkanes
- Topic: Common Names on Page 94 of course reader
- Replies: 2
- Views: 682
Re: Common Names on Page 94 of course reader
We are expected to know common names. Also, the common name prefixes such as "iso" and "neo" are included when arranging the substituent alphabetically because it relates to the structural arrangement of the substituent.
- Fri Mar 03, 2017 11:40 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalyst
- Replies: 16
- Views: 2438
Re: Catalyst
Intermediates cannot be in the rate law, so we use the pre-equilibrium approach to get rid of the intermediate.
- Sat Feb 25, 2017 3:41 pm
- Forum: *Organic Reaction Mechanisms in General
- Topic: Electrophiles
- Replies: 3
- Views: 817
Re: Electrophiles
Fluorine and Chlorine are electrophiles because they need one more electron to satisfy the octet rule. Electrophiles are attracted to electrons.
- Sat Feb 25, 2017 3:30 pm
- Forum: *Electrophiles
- Topic: electrophile vs neutrophile
- Replies: 2
- Views: 741
Re: electrophile vs neutrophile
Electrophiles are positively charged and are attracted to electrons. Nucleophiles are negatively charged and donate electrons.
- Sun Feb 12, 2017 9:32 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams
- Replies: 2
- Views: 639
Re: Cell Diagrams
Yes, the anode is on the left while the cathode is on the right. They are separated by the salt bridge, represented by the double bars (II). I believe the order doesn't matter; you just separate the species in terms of phase and separate them with commas if they are the same phase.
- Sun Feb 12, 2017 9:09 pm
- Forum: Balancing Redox Reactions
- Topic: K and K(sub)a
- Replies: 2
- Views: 700
Re: K and K(sub)a
From the solution, it appears that K is for 2 moles 2 HF (aq) ->2 H+(aq) + 2F-(aq) while k(sub)a is for 1 mole HF (aq) ->2 H+(aq) + 2F-(aq) K(sub)a = ([H+][F-])/[HF]. K = ([H+]^2*[F-]^2])/[HF]^2 That is why in the end you take the square root of K to get K(sub)a since K has the products and reactant...
- Sun Feb 05, 2017 3:08 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3655958
Re: Post All Chemistry Jokes Here
Did you hear about the man who got cooled to absolute zero?
He's 0K now.
He's 0K now.
- Sun Jan 29, 2017 2:34 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3655958
Re: Post All Chemistry Jokes Here
Pick up line
- Sun Jan 22, 2017 4:17 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HW 8.51
- Replies: 1
- Views: 472
Re: HW 8.51
Since the question asks to to calculate the enthalpy released per liter for the reaction (exothermic reaction), the negative sign is not necessary.