[QUOTE=joe588;39538782]right cheers, makes more sense now. thanks man.[/QUOTE]
no probs
Would anyone mind if I requested some test correction help in here? I don't want to flood the thread with homework but I believe you guys might enjoy some of the questions.
[QUOTE=RIPBILLYMAYS;39539195]Would anyone mind if I requested some test correction help in here? I don't want to flood the thread with homework but I believe you guys might enjoy some of the questions.[/QUOTE]
Go ahead.
[QUOTE=Angus725;39538055]Google acetic acid powder.
Heh, acid-base chem. There's a bit more to it than H and OH.[/QUOTE]
Acetic acid powder is no more than a powder ingredient infused with a little acetic acid.
Acetic acid in its purest form is a liquid, and is called glacial acetic acid when there's no water in it.
The unit was on bonding as an fyi. I'll post my wrong answers as well.
These questions refer to the following descriptions of bonding in different types of solids.
A. Lattice of positive and negative ions held together by electrostatic forces
B. Closely packed lattice with delocalized electrons throughout
C. Strong single covalent bonds with weak intermolecular forces of attraction
D. Strong multiple covalent bonds (including pi-bonds) with weak intermolecular forces
E. Macromolecules held together with strong polar bonds.
1. Gold Au (answer is not C)
2. Carbon Dioxide CO2 (answer is not B)
3. Methane CH4 (answer is not E)
What is the hybridization exhibited in tellurium tetrachloride. TeCl4?
A. sp
B. sp2
[del]C. sp3[/del]
D. dsp3
E. d2sp3
Which type of hybridization is not associated with a square planar array of hybrid orbitals?
A. sp
B. sp2
[del]C. sp3[/del]
D. dsp3
E. d2sp3
Which of the following statements concerning nitrous acid, HNO2, is [I][B]false?[/B][/I]
I. It possesses 18 valence electrons
II. The arrangement of bonds is:
[img]http://i.imgur.com/bAWG3ng.png[/img]
III. Delocalized pi overlap is indicated
A. I
B. II
C. III
[del]D. I and II[/del]
E. I and III
F. II and III
G. I, II, and III
H. All statements are [B][U][I]true[/I][/U][/B]
Which of the following statements regarding the melting of solid iodine, I2, is [I][B]false[/B][/I]?
1.I2 is higher melting than NaI because the I--I covalent bonds of I2 are stronger than the Na--I ionic bonds
2. I2 is a molecular solid while Br2 is liquid because London force are stronger in I2.
3. I2 is higher melting than Br2 because the I-I covalent bond is stronger than the Br-Br covalent bond
[del]A. 1[/del]
B. 2
C. 3
D. 1 and 2
E. 1 and 3
Hydrogen bonding is the most significant intermolecular for in all but which of the following:
A: H2O
B: HCl
[del]C. CH3CH2OH[/del]
D. NH3
E. HF
(From the previous question) Which of the compound choices would you expect to be the lowest boiling and why?
Although propane and acetaldehyde have the same molar mass, their boiling points vary by about 62C. Based upon their structures, which of the following statements is [I][B][U]false?[/U][/B][/I]
[img]http://i.imgur.com/uJ3UGAn.png[/img]
1. Acetaldehyde is higher boiling than propane because it exhibits significant hydrogen bonding among molecules.
2. The boiling point of propane is significantly less than that of acetaldehyde because the dipole moment of propane is signficantly less than that of acetaldehyde.
3. Propane is higher boiling because the surface area of the propane molecule is greater than the surface area of acetaldehyde molecules.
A. 1
[del]B. 2[/del]
C. 3
D. 1 and 2
E. 1 and 3
Which of the following best describes non-polar, molecular solids?
I. In the molten state they can conduct electricity
II. They have high melting points
III. Most would be soluble in liquid carbon tetrachloride
A. I
B. II
C. III
D. I and II
[del]E. I and III[/del]
F. II and III
G. I, II, and III
H. All statements are [B][U][I]true[/I][/U][/B]
[QUOTE=JohanGS;39538796]Bottom left and top middle
CH3 and H3C are the same thing and it's the same for the CH3O and OCH3.
[IMG]http://f2.braxupload.se/g8nybh.6ff812a5df.png[/IMG]
[editline]10th February 2013[/editline]
ninja :l[/QUOTE]
yeah i somehow had it in my head that it was simply a mirror image
[editline]10th February 2013[/editline]
ok another question, a very basic one too embarrassingly.
why is gold not reactive? having one outer valence electron you'd expect it to be reactive no? is it to do with the energy in the bond being very low?
So any of you trying to get into med school? That's my goal for the next few years.
[QUOTE=joe588;39539540]yeah i somehow had it in my head that it was simply a mirror image
ok another question, a very basic one too embarrassingly.
why is gold not reactive? having one outer valence electron you'd expect it to be reactive no? is it to do with the energy in the bond being very low?[/QUOTE]
I think it's due to the fact that it's valence electron is in the 6s orbital, which is below the energy level of the paired 5d electrons. This means that while it would normally react due to it having one valence electron, it can't as that electron is effectively blocked by the 5d electrons.
This means it only reacts under specific circumstances. Such as it being dissolved by aqua regia.
Did some of the ones I knew
[quote]What is the hybridization exhibited in tellurium tetrachloride. TeCl4?
A. sp
B. sp2
C. sp3
D. dsp3
E. d2sp3[/quote]
TeCL4 is a trigonal bipyramidal molecule with 4 bonding domains and 1 nonbonding pair, therefore the answer is dsp3
[quote]Hydrogen bonding is the most significant intermolecular for in all but which of the following:
A: H2O
B: HCl
C. CH3CH2OH
D. NH3
E. HF[/quote]
H-bonding is when H is bonded to N, O, or F. Therefore, the answer is HCl
[QUOTE=RIPBILLYMAYS;39539525]The unit was on bonding as an fyi. I'll post my wrong answers as well.
These questions refer to the following descriptions of bonding in different types of solids.
A. Lattice of positive and negative ions held together by electrostatic forces
B. Closely packed lattice with delocalized electrons throughout
C. Strong single covalent bonds with weak intermolecular forces of attraction
D. Strong multiple covalent bonds (including pi-bonds) with weak intermolecular forces
E. Macromolecules held together with strong polar bonds.
1. Gold Au (answer is not C) [B]B[/B]
2. Carbon Dioxide CO2 (answer is not B) [B]D[/B]
3. Methane CH4 (answer is not E) [B]C[/B]
What is the hybridization exhibited in tellurium tetrachloride. TeCl4?
A. sp
B. sp2
[del]C. sp3[/del]
D. dsp3 [B]"See-saw" geometry, Tellurium has an additional lone pair.[/B]
E. d2sp3
Which type of hybridization is not associated with a square planar array of hybrid orbitals?
A. sp
B. sp2
[del]C. sp3[/del]
D. dsp3
E. d2sp3 [B]Assuming axial lone pairs. Or platinum[/B]
Which of the following statements concerning nitrous acid, HNO2, is [I][B]false?[/B][/I]
I. It possesses 18 valence electrons
II. The arrangement of bonds is:
[img]http://i.imgur.com/bAWG3ng.png[/img]
III. Delocalized pi overlap is indicated
A. I
B. II [B]This one, HNO2 is represented as HO-N=O[/B]
C. III
[del]D. I and II[/del]
E. I and III
F. II and III
G. I, II, and III
H. All statements are [B][U][I]true[/I][/U][/B]
Which of the following statements regarding the melting of solid iodine, I2, is [I][B]false[/B][/I]?
1.I2 is higher melting than NaI because the I--I covalent bonds of I2 are stronger than the Na--I ionic bonds
2. I2 is a molecular solid while Br2 is liquid because London force are stronger in I2.
3. I2 is higher melting than Br2 because the I-I covalent bond is stronger than the Br-Br covalent bond
[del]A. 1[/del]
B. 2 [B]Gonna have to say this one[/B]
C. 3
D. 1 and 2
E. 1 and 3
Hydrogen bonding is the most significant intermolecular for in all but which of the following:
A: H2O
B: HCl [B]In water, H+ and Cl- dissociate completely.[/B]
[del]C. CH3CH2OH[/del]
D. NH3
E. HF [I]HF is a weak acid, unlike its halogen brothers.[/I]
(From the previous question) Which of the compound choices would you expect to be the lowest boiling and why?
Although propane and acetaldehyde have the same molar mass, their boiling points vary by about 62C. Based upon their structures, which of the following statements is [I][B][U]false?[/U][/B][/I]
[img]http://i.imgur.com/uJ3UGAn.png[/img]
1. [del]Acetaldehyde[/del] Ethanal is higher boiling than propane because it exhibits significant hydrogen bonding among molecules.
2. The boiling point of propane is significantly less than that of [del]Acetaldehyde[/del] Ethanal because the dipole moment of propane is signficantly less than that of [del]Acetaldehyde[/del] Ethanal.
3. Propane is higher boiling because the surface area of the propane molecule is greater than the surface area of [del]Acetaldehyde[/del] Ethanal molecules.
A. 1
[del]B. 2[/del]
C. 3
D. 1 and 2 [B]3 is false, propane is a gas at room temp, ethanal is a solid.[/B]
E. 1 and 3
Which of the following best describes non-polar, molecular solids?
I. In the molten state they can conduct electricity [B]They do not need to be molten to conduct, see: graphene[/B]
II. They have high melting points
III. Most would be soluble in liquid carbon tetrachloride [del][B]Methane is a gas at room temperature[/B][/de]
A. I
B. II
C. III [B]CCl4 would definitely solvate many solid-phase molecular solids. It essentially "gets in the way" between two dipole moments[/B]
D. I and II
[del]E. I and III[/del]
F. II and III
G. I, II, and III
H. All statements are [B][U][I]true[/I][/U][/B][/QUOTE]
[QUOTE=RIPBILLYMAYS;39539525]The unit was on bonding as an fyi. I'll post my wrong answers as well.
These questions refer to the following descriptions of bonding in different types of solids.
A. Lattice of positive and negative ions held together by electrostatic forces
B. Closely packed lattice with delocalized electrons throughout
C. Strong single covalent bonds with weak intermolecular forces of attraction
D. Strong multiple covalent bonds (including pi-bonds) with weak intermolecular forces
E. Macromolecules held together with strong polar bonds.
1. Gold Au (answer is not C) [highlight]B[/highlight] - As a metal, gold has the "sea of electrons" structure
2. Carbon Dioxide CO2 (answer is not B) [highlight]D[/highlight] CO2 has double bonds which means it has both sigma and pi bonds. It is nonpolar so it has weak IMFs.
3. Methane CH4 (answer is not E) [highlight]C[/highlight] CH4 is all single bonds and is nonpolar so it has weak IMFs
What is the hybridization exhibited in tellurium tetrachloride. TeCl4?
A. sp
B. sp2
[del]C. sp3[/del]
D. dsp3
E. d2sp3
[highlight]D[/highlight] TeCl4 has a single lone pair on the central atom as well as the 4 bonds, so it's dsp3.
Which of the following statements regarding the melting of solid iodine, I2, is [I][B]false[/B][/I]?
1.I2 is higher melting than NaI because the I--I covalent bonds of I2 are stronger than the Na--I ionic bonds
2. I2 is a molecular solid while Br2 is liquid because London force are stronger in I2.
3. I2 is higher melting than Br2 because the I-I covalent bond is stronger than the Br-Br covalent bond
[del]A. 1[/del]
B. 2
C. 3
D. 1 and 2
E. 1 and 3
[highlight]E[/highlight] - When talking about phases of matter, only IMFs matter, not the strength of the bonds. (In any case ionic bonds are almost always way stronger than covalent so I is not true). And I2 has stronger LDFs because it has a large electron cloud. So 2 is true, and 1 and 3 are false. Woops.
Hydrogen bonding is the most significant intermolecular for in all but which of the following:
A: H2O
B: HCl
[del]C. CH3CH2OH[/del]
D. NH3
E. HF
[highlight]B[/highlight] - Not completely sure why, but H bonding usually occurs with N, F, and O. I think it may have something to do with Chlorine's larger amount of electrons, making the molecule very polar.
(From the previous question) Which of the compound choices would you expect to be the lowest boiling and why?
Although propane and acetaldehyde have the same molar mass, their boiling points vary by about 62C. Based upon their structures, which of the following statements is [I][B][U]false?[/U][/B][/I]
[img]http://i.imgur.com/uJ3UGAn.png[/img]
1. Acetaldehyde is higher boiling than propane because it exhibits significant hydrogen bonding among molecules.
2. The boiling point of propane is significantly less than that of acetaldehyde because the dipole moment of propane is signficantly less than that of acetaldehyde.
3. Propane is higher boiling because the surface area of the propane molecule is greater than the surface area of acetaldehyde molecules.
A. 1
[del]B. 2[/del]
C. 3
D. 1 and 2
E. 1 and 3
[highlight]D[/highlight] That O will be take part in hydrogen bonding, so 1 is true. For 2, acetaldehyde has a much more significant charge difference from end to end than propane, which is rather symmetrical.
Which of the following best describes non-polar, molecular solids?
I. In the molten state they can conduct electricity
II. They have high melting points
III. Most would be soluble in liquid carbon tetrachloride
A. I
B. II
C. III
D. I and II
[del]E. I and III[/del]
F. II and III
G. I, II, and III
H. All statements are [B][U][I]true[/I][/U][/B]
[highlight]C[/highlight] - Usually only ionic solids conduct electricity when molten. And they would usually be soluble in CCl4 because it's also a nonpolar molecular compound.
[/QUOTE]
Did the ones that I know pretty well. It's been about a year since I've done this stuff, so my apologies if anything is wrong.
Oh balls, misread Carbon Tetrachloride as Carbon Tetrahydride.
[QUOTE=RIPBILLYMAYS;39539525]
A. Lattice of positive and negative ions held together by electrostatic forces [B](= a salt, none of the answers)[/B]
B. Closely packed lattice with delocalized electrons throughout [B](= a metal, so gold is the answer)[/B]
C. Strong single covalent bonds with weak intermolecular forces of attraction [B](= methane, since hydrogen atoms only have one valence electron and therefore only have single covalent bonds)[/B]
D. Strong multiple covalent bonds (including pi-bonds) with weak intermolecular forces [B](= carbon dioxide, since the structure is (C=O=C, no angle!) and because of the angle it's not polar => Weak intermolecular forces[/B]
E. Macromolecules held together with strong polar bonds. [B](= macromolecules are way bigger than the examples)[/B]
1. Gold Au (answer is not C) [B]B[/B]
2. Carbon Dioxide CO2 (answer is not B) [B]D[/B]
3. Methane CH4 (answer is not E) [B]C[/B]
What is the hybridization exhibited in tellurium tetrachloride. TeCl4? [B]No idea what this is, we don't do it here (in Sweden) at least[/B]
Which of the following statements concerning nitrous acid, HNO2, is [I][B]false?[/B][/I]
I. It possesses 18 valence electrons [B](= H contains 1, N contains 5, the two O's contain 12 which is a total of 18, not sure whether you can conclude that the molecule has 18 valence electrons though, anyone else?)[/B]
II. The arrangement of bonds is:
[img]http://i.imgur.com/bAWG3ng.png[/img] [B](= the molecule contains a hydroxide group so this isn't correct)[/B]
III. Delocalized pi overlap is indicated [B](= don't know what this means)[/B]
[B]One of these[/B]
A. I
C. III
E. I and III
Which of the following statements regarding the melting of solid iodine, I2, is [I][B]false[/B][/I]?
1.I2 is higher melting than NaI because the I--I covalent bonds of I2 are stronger than the Na--I ionic bonds [B](= false, because I--I is nonpolar and very weak compared to Na--I (polar and strong) and a high melting point is a consequence of polarity, not intramolecular bonds)[/B]
2. I2 is a molecular solid while Br2 is liquid because London force are stronger in I2. [B](= not sure, but I think it's true)[/B]
3. I2 is higher melting than Br2 because the I-I covalent bond is stronger than the Br-Br covalent bond [B](= I think this is false because the electronegativity difference is the same in both cases)[/B]
E. 1 and 3 [B]<- this is it[/B]
Hydrogen bonding is the most significant intermolecular for in all but which of the following: [B](= in order for hydrogen bonding to be present, a hydrogen must bond (not sure what word to use here) with either nitrogen, oxygen or fluoride)[/B]
A: H2O [B](hydrogen and oxygen bonding)[/B]
B: HCl [B](nope, this does NOT contain a hydrogen bond)[/B]
[del]C. CH3CH2OH[/del][B] (this molecule contains a hydroxide group, which is hydrogen and oxygen bonding)[/B]
D. NH3 [B](hydrogen and nitrogen)[/B]
E. HF [B](hydrogen and fluoride)[/B][/QUOTE]
I came this far before I saw an answer in the thread but I figured I'd post it anyway because it contain some explanations.
Cakebatyr, what's with your explanation on the last one about nonpolar molecular solids? They are not known for having high melting points or for conducting electricity, and I'm a little confused as to why you mentioned methane in the last choice.
EDIT: Ninja'd me with a correction.
[QUOTE=account;39540181]Cakebatyr, what's with your explanation on the last one about nonpolar molecular solids? They are not known for having high melting points or for conducting electricity, and I'm a little confused as to why you mentioned methane in the last choice.
EDIT: Ninja'd me with a correction.[/QUOTE]
I was mostly picking at carbon allotropes such as graphite. It is known that graphene (a layer of graphite) is an excellent conductor of electricity. Graphite itself has a melting point in the mid 3000s. Diamond is in the low 4000s.
Edit: Now that I think of it, they don't really fit the definition of the question...
[QUOTE=Cakebatyr;39540300]I was mostly picking at carbon allotropes such as graphite. It is known that graphene (a layer of graphite) is an excellent conductor of electricity. Graphite itself has a melting point in the mid 3000s. Diamond is in the low 4000s.
Edit: Now that I think of it, they don't really fit the definition of the question...[/QUOTE]
I was thinking more along the lines of Glucose and big hydrocarbons.
On some of these I have no idea how I missed them. I must have frozen up on test day.
I really appreciate the help you guys are giving me. There's only a few problems that you guys seem to disagree on/haven't answered.
Hydrogen bonding is the most significant intermolecular for in all but which of the following:
A: H2O
B: HCl [B]Consensus[/B]
C. CH3CH2OH
D. NH3
E. HF
This was unanswered:
[B][U](From the previous question) Which of the compound choices would you expect to be the lowest boiling and why?[/U][/B]
Which of the following statements concerning nitrous acid, HNO2, is false? (1 person said B, the other said it could be A/C/E)
I. It possesses 18 valence electrons
II. The arrangement of bonds is:
[img]http://i.imgur.com/bAWG3ng.png[/img]
III. Delocalized pi overlap is indicated-[B]To whoever asked, here's an example: [img_thumb]http://i.imgur.com/eTBWBkG.png[/img_thumb][/B]
A. I
B. II
C. III
D. I and II
E. I and III
F. II and III
G. I, II, and III
H. All statements are true
[quote]
A: H2O
B: HCl
C. CH3CH2OH
D. NH3
E. HF
(From the previous question) Which of the compound choices would you expect to be the lowest boiling and why?[/quote]
HCl is just LDF + dipole-dipole, and the rest are H-bonding. Stronger IMF = higher boiling/melting points, H-bonding > LDF/D-D, therefore HCl should be the lowest boiling point
I just found out [url=http://www.acdlabs.com/resources/freeware/chemsketch/]Chemsketch[/url] has a freeware version!
Aww yeah!
[img]http://i.imgur.com/a55eneP.png[/img]
[editline]10th February 2013[/editline]
I'm abusing bond angles! (I didn't hit F9 to normalize bonds)
[img]http://i.imgur.com/en8roZQ.png[/img]
[editline]10th February 2013[/editline]
Diethanal, 1,2-cyclopropandione, tetramethanol, propan-1,3-al-2-one, bicyclo[2,2,0]pentane, 1,2-cyclopropandione, acetic acid (methyl acetate), and a carboxylic functional group.
[QUOTE=Yellowamoeba;39539934]I think it's due to the fact that it's valence electron is in the 6s orbital, which is below the energy level of the paired 5d electrons. This means that while it would normally react due to it having one valence electron, it can't as that electron is effectively blocked by the 5d electrons.
This means it only reacts under specific circumstances. Such as it being dissolved by aqua regia.[/QUOTE]
cheers matey! Here's a bit more on it...
[quote]Special relativity is also responsible for gold's resistance to tarnishing and other chemical reactions. Chemistry is mostly concerned with the electrons in the outermost orbitals. With a single 6s electron, you might expect gold to be highly reactive; after all, cæsium has the same 6s1 outer shell, and it is the most alkaline of natural elements: it explodes if dropped in water, and even reacts with ice. Gold's 6s orbital, however, is relativistically contracted toward the nucleus, and its electron has a high probability to be among the electrons of the filled inner shells. This, along with the stronger electrostatic attraction of the 79 protons in the nucleus, reduce the “atomic radius” of gold to 135 picometres compared to 260 picometres for cæsium with its 55 protons and electrons—the gold atom is almost 50% heavier, yet only a little over half the size of cæsium. Only the most reactive substances can tug gold's 6s1 electron out from where it's hiding among the others, and hence not only the colour of gold, but its immunity from tarnishing and corrosion are consequences of special relativity. [/quote]
[editline]11th February 2013[/editline]
Another simple one just checking I'm not being dumb(er).
Asked to calculate molar concentration of h+ ions in sulphirc acid dissolved in 0.5dm-3 water and express this in mol dm-3
This means I have to double my answer so it's the molarity in 1 dm-3 water doesn't it? Or do i leave it and put molarity 0.5dm-3 or something?
I'll shut up now, 12 odd hours learning has disabled my thinking ability.
I want to say the Henderson-Hasselbalch equation would be beneficial here. But I feel that's giving you the wrong answer.
[QUOTE=Cakebatyr;39546881]I want to say the Henderson-Hasselbalch equation would be beneficial here. But I feel that's giving you the wrong answer.[/QUOTE]
Yeah i use that but in a different part, this specifically asks for the h ion concentration (which is ph yeh, but expressed as [H+] = ....) in .5dm3 solute, just wondering if "express this in mol dm-3" means i should double my answer for the concentration in .5dm3 or put 0.5 mol dm-3. or something else?
[QUOTE=joe588;39548054]Yeah i use that but in a different part, this specifically asks for the h ion concentration (which is ph yeh, but expressed as [H+] = ....) in .5dm3 solute, just wondering if "express this in mol dm-3" means i should double my answer for the concentration in .5dm3 or put 0.5 mol dm-3. or something else?[/QUOTE]
Yes you would have to double your answer to get it into the correct units asked for. I think you already understand but: 1mol in 0.5dm3 is the same as 2mol in 1dm3.
Inorganic chemistry appears to be my real weakness in Chemistry. I just can't get my head around some of the theories like crystal field theory and the Jahn-Teller effect, I feel as if would be fine with them if I could visualise them. Also molecular structures will always be the bane of my existence! I always feel quit stupid that I'm a uni student who struggles with GCSE stuff.
Has anyone seen any papers recently on making petroleum products from CO2 and Water? I understand it would be a complex and energy laden process but I've been quite interested in it recently
i've been sat staring at this all day and i just don't get it.
can anyone explain how to use this equation?
[img]http://preparatorychemistry.com/images/Ka_expression_simpler_CS.gif[/img]
i'm trying to work out the hydrogen concentration of a weak acid, HCOOH in a 4.6x10-3 mol dm−3 solution. Ka = 1.8x10^4
........HCOOH ==> H+ + HCOO-
I.......4.6x10-3..........0.......0
C..........-x................x.......x
E......4.6x10-3-x........x.......x
"Substitute the E line into the Ka expression and solve for x = H^+." (this is what i don't get how do i work out x? )
[QUOTE=download;39549008]Has anyone seen any papers recently on making petroleum products from CO2 and Water? I understand it would be a complex and energy laden process but I've been quite interested in it recently[/QUOTE]
I'm fairly sure that CO + 2H[SUB]2[/SUB] -> H[SUB]3[/SUB]COH would work under a catalyst and be quite exothermic, but thats not really a petroleum product
CO + 3H[SUB]2[/SUB] -> CH[SUB]4 [/SUB]+ H[SUB]2[/SUB]O should also occur under the right catalysts
[editline]11th February 2013[/editline]
[url]http://www.ems.psu.edu/~elsworth/courses/egee580/2010/Final%20Reports/co2_electrochem.pdf[/url]
Googling this I found basicaly what I was thinking about but done electrochemically
[QUOTE=Tobba;39549780]I'm fairly sure that CO + 2H[SUB]2[/SUB] -> H[SUB]3[/SUB]COH would work under a catalyst and be quite exothermic, but thats not really a petroleum product
CO + 3H[SUB]2[/SUB] -> CH[SUB]4 [/SUB]+ H[SUB]2[/SUB]O should also occur under the right catalysts
[editline]11th February 2013[/editline]
[url]http://www.ems.psu.edu/~elsworth/courses/egee580/2010/Final%20Reports/co2_electrochem.pdf[/url]
Googling this I found basicaly what I was thinking about but done electrochemically[/QUOTE]
I imagine from methane you could create longer hydrocarbons. I was thinking synthetic petrol myself
[QUOTE=joe588;39549757]i've been sat staring at this all day and i just don't get it.
can anyone explain how to use this equation?
[img]http://preparatorychemistry.com/images/Ka_expression_simpler_CS.gif[/img]
i'm trying to work out the hydrogen concentration of a weak acid, HCOOH in a 4.6x10-3 mol dm−3 solution. Ka = 1.8x10^4
........HCOOH ==> H+ + HCOO-
I.......4.6x10-3..........0.......0
C..........-x................x.......x
E......4.6x10-3-x........x.......x
"Substitute the E line into the Ka expression and solve for x = H^+." (this is what i don't get how do i work out x? )[/quote]
I think you have to use the fact that the concentration [H+] equals the concentration [A-], which makes sense, as one molecule of HCOOH dissociates to HCOO- and H+. The other assumption you need is that the concentration of [HA] before dissociation is the same after, which is roughly true here because of the tiny Ka value - [url=http://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text12/Tx126/tx126.html]only a tiny bit dissociates[/url], so you're cool.
From there you get the equation Ka = ([H+]^2)/[HA], and you know all those values, so you can work out [H+] from that
[QUOTE=Turnips5;39550007]I think you have to use the fact that the concentration [H+] equals the concentration [A-], which makes sense, as one molecule of HCOOH dissociates to HCOO- and H+. The other assumption you need is that the concentration of [HA] before dissociation is the same after, which is roughly true here because of the tiny Ka value - [url=http://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text12/Tx126/tx126.html]only a tiny bit dissociates[/url], so you're cool.
From there you get the equation Ka = ([H+]^2)/[HA], and you know all those values, so you can work out [H+] from that[/QUOTE]
thanks
still don't get it though. where does ka = 1.8x10^4 come in? is it just simply [H+] = 0.0046?
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