VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Rishi97 on April 07, 2014, 08:46:37 pm
Title: Rishi's Chemistry Thread
Post by: Rishi97 on April 07, 2014, 08:46:37 pm
Welcome to Rishi's Chemistry Thread :) Chemistry is a tough subject so I'll need all the help I can get.
Thanks in advance ;)
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 07, 2014, 08:49:59 pm
Why can glucose be oxidised more rapidly to produce energy, than fat? I don't know if this is a Biology question but it came up while I was doing chem so sorry if I've posted in the wrong section.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 07, 2014, 08:58:07 pm
On pg 186 of the heinemann textbook, it states that the CH2OH groups on adjacent glucose monomers in starch are on the same side of the polymer chain, while in cellulose, they are on alternating sides. What effect does this have? In regards to the strength of the compound?
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on April 07, 2014, 11:01:29 pm
Why can glucose be oxidised more rapidly to produce energy, than fat? I don't know if this is a Biology question but it came up while I was doing chem so sorry if I've posted in the wrong section.
You won't need this for VCE chem, but it probably has something to do with how effective our enzymes are at catalysing each reaction step. Fats and glucose undergo slightly different metabolic processes, but both enter the citric acid cycle at the same place, as acetyl CoA
On pg 186 of the heinemann textbook, it states that the CH2OH groups on adjacent glucose monomers in starch are on the same side of the polymer chain, while in cellulose, they are on alternating sides. What effect does this have? In regards to the strength of the compound?
Enzymes require a very specific substrate for those to fit in the active site. Moving one of the CH2OH groups to the other side changes the shape of the molecule and thus affects the bonding between cellulose and our starch/glycogen hydrolysis enzymes. As it turns out, our enzymes are ineffective here.
As for the strength of the compound, I'm not too sure on this myself; it may well indirectly affect how the glucose monomers bond to form the cellulose polymer. Perhaps with this particular form of glucose, it prefers to form long linear chains, whereas with the form of glucose found in glycogen, the glucose can branch out. I'm not sure if cellulose is any stronger than starch in terms of intermolecular forces, but cellulose is certainly more regular; the uniformity of the hydrogen bonding may allow cellulose to pack together better.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 08, 2014, 12:39:09 pm
Write an equation to show the formation of a dipeptide from serine and cysteine. Thanks
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on April 08, 2014, 05:34:39 pm
Molecular equation? Work out the molecular formula of serine and cysteine, add them, and make sure you leave space for a H2O I'm going to try this from my knowledge of amino acid structures Cysteine has a CH2-SH side group while serine has a CH2-OH side group. And they're joined to H2N-CH-COOH on carbon 2 OK. So cysteine looks like C3H7O2NS and serine looks like C3H7O3N Add them together to get C3H7O2NS + C3H7O3N => C6H12O4N2S + H2O as it's a condensation reaction
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 09, 2014, 10:25:36 am
When a trigylceride is formed, does it undergo an esterification reaction?
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 09, 2014, 04:13:51 pm
Explain why ethanol produced by fermentation is referred to as a ‘biochemical fuel' Thanks
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on April 09, 2014, 09:22:06 pm
Explain why ethanol produced by fermentation is referred to as a ‘biochemical fuel' Thanks
It's produced by biological organisms and it's renewable.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 11, 2014, 12:56:49 pm
A compound has the molecular formula C3H9X. The atom X is most likely to be: a) Chlorine b) Hydrogen c) Nitrogen d) Oxygen I'm confused cause there can't be more than 8 hydrogens so how is there 9? Thanks
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on April 11, 2014, 05:49:07 pm
A compound has the molecular formula C3H9X. The atom X is most likely to be: a) Chlorine b) Hydrogen c) Nitrogen d) Oxygen I'm confused cause there can't be more than 8 hydrogens so how is there 9? Thanks
Nitrogen. Look at C3H8 and replace a hydrogen with an amine NH2 group. Look, extra hydrogen
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 11, 2014, 07:31:27 pm
Nitrogen. Look at C3H8 and replace a hydrogen with an amine NH2 group. Look, extra hydrogen
oh wow, that was easy. I went completely blank but now I get it. Thanks lzxnl
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 14, 2014, 03:06:34 pm
Why is the functioning of the enzyme closely related to its tertiary structure?
Title: Re: Rishi's Chemistry Thread
Post by: rhinwarr on April 14, 2014, 03:38:34 pm
In order for an enzyme to facilitate reactions, the enzyme's active site must have a complementary shape to the substrates in order to form an enzyme-substrate complex. The tertiary structure determines the shape of the enzyme's active site so it is important for its function because enzymes are specific to their substrates.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 16, 2014, 12:29:56 pm
A student performs two tests on an organic compound. In the first test, 3 mol of the compound was completely reacted with oxygen and 6 mol of carbon dioxide were produced. In the second test a few drops of bromine were added to the compound. The compound did not react rapidly with bromine. The formula of this compound is likely to be: a) C2H4 b) C2H6 c) C3H8 d) C6H14 Pls show working out. Thanks
Title: Re: Rishi's Chemistry Thread
Post by: nhmn0301 on April 16, 2014, 12:58:31 pm
A student performs two tests on an organic compound. In the first test, 3 mol of the compound was completely reacted with oxygen and 6 mol of carbon dioxide were produced. In the second test a few drops of bromine were added to the compound. The compound did not react rapidly with bromine. The formula of this compound is likely to be: a) C2H4 b) C2H6 c) C3H8 d) C6H14 Pls show working out. Thanks
That would be B. So you have 3 ( C x H y ) produces 6 CO2. Hence you can say that 3x = 6 ( Carbon ), x = 2. We can eliminate C and D from now. Next, consider the fact that it DID NOT react rapidly with Bromine, which would be an alkane, since it is a saturated hydrocarbon, which has a formula C2H6. Hope this helps!
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 16, 2014, 01:05:29 pm
That would be B. So you have 3 ( C x H y ) produces 6 CO2. Hence you can say that 3x = 6 ( Carbon ), x = 2. We can eliminate C and D from now. Next, consider the fact that it DID NOT react rapidly with Bromine, which would be an alkane, since it is a saturated hydrocarbon, which has a formula C2H6. Hope this helps!
ohhhh that helps a lot :) I was just confused what bromine had to do with it but now it makes sense since it must be a alkane. Thanks
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 16, 2014, 01:51:25 pm
Can I please have help with q 52 pg 61 of the checkpoints book? Sorry, it's got diagrams so I can't upload all of it. I have identified that it can only be B or C but how do I calculate molecular mass of the bases?
Thanks
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 19, 2014, 04:37:30 pm
What is the name of the following compound ?
CH3CH(CH3)CO2CH2CH3
Thanks
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on April 19, 2014, 05:11:23 pm
So...you have an ester, in which the alcohol used to make this ester is ethanol and it looks like it's reacted with 2-methylpropanoic acid. Something like ethyl 2-methylpropanoate should be fine.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 20, 2014, 10:07:52 am
Describe the experiments you could carry out to distinguish between an alkane and an alkene. Include in your description any potential hazards and how they may be avoided.
Thanks :D
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on April 20, 2014, 10:36:53 am
Describe the experiments you could carry out to distinguish between an alkane and an alkene. Include in your description any potential hazards and how they may be avoided.
Thanks :D
Easiest one is to place samples inside liquid bromine away from the light; the sample that decolourises bromine is an alkene. Potential hazard? Well don't sniff any bromine vapours or let the liquid spill
Title: Re: Rishi's Chemistry Thread
Post by: swagsxcboi on April 20, 2014, 10:44:26 am
Describe the experiments you could carry out to distinguish between an alkane and an alkene. Include in your description any potential hazards and how they may be avoided.
Thanks :D
Bromine (or Iodine) Test. - Add a drop of Bromine (liquid at room temperature) into the test tube containing the sample - Observe colour change; from orange to colourless (or pale orange) or orange remaining the same. orange to colourless (or pale orange) means that an addition reaction has taken place between Br2 and the alkene (unsaturated hydrocarbon)
orange remaining the same colour means that no reaction has taken place between Br2 and the alkane (saturated hydrocarbon).
Iodine test is the same except for colour changes. Iodine turns purple to colourless, or remains purple.
Bromine and Iodine is harmful to the skin and eyes so gloves, apron and goggles are a must
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 20, 2014, 04:53:51 pm
Do we need to know what electrophoresis is and how it works?
Title: Re: Rishi's Chemistry Thread
Post by: Yacoubb on April 20, 2014, 05:06:51 pm
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on April 22, 2014, 07:37:55 pm
Instructions for the production of a useful chemical, accompanied by appropriate safety warnings included 1. Take one litre of cooking oil, 3.5 grams of sodium hydroxide and 200mL of methanol. 2. Pour the methanol into a blender 3. Add 3.5g of sodium hydroxide and blend until the sodium hydroxide has completely dissolved. 4. Add the one litre of cooking oil and continue blending until the liquid seperates into two layers The two layer referred to in this procedure would be: a) a fatty acid and water b) biodiesel and glycerol c) an alcohol and a carboxylic acid d) oil and water
I don't have the answer sorry but if you could also give a brief explanation of your answer, it would be great. Thanks :D
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on May 04, 2014, 10:43:22 am
Has anyone had their chem sac on organic chemistry yet? If yes, what did you get tested on? There's so much to remember and I don't know what's important and what I should study. CHEM SAC TOMORROW!!!
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on May 04, 2014, 11:36:41 am
Instructions for the production of a useful chemical, accompanied by appropriate safety warnings included 1. Take one litre of cooking oil, 3.5 grams of sodium hydroxide and 200mL of methanol. 2. Pour the methanol into a blender 3. Add 3.5g of sodium hydroxide and blend until the sodium hydroxide has completely dissolved. 4. Add the one litre of cooking oil and continue blending until the liquid seperates into two layers The two layer referred to in this procedure would be: a) a fatty acid and water b) biodiesel and glycerol c) an alcohol and a carboxylic acid d) oil and water
I don't have the answer sorry but if you could also give a brief explanation of your answer, it would be great. Thanks :D
Oil plus sodium hydroxide => fatty acid (well, conjugate base of fatty acid at least) + glycerol Then plus methanol => methyl ester You have biodiesel and glycerol as your two layers (one is non-polar, one is polar)
Has anyone had their chem sac on organic chemistry yet? If yes, what did you get tested on? There's so much to remember and I don't know what's important and what I should study. CHEM SAC TOMORROW!!!
Each school's SACs will be different. Mine was on aspirin, spectroscopy, biomolecules, naming things and reaction pathways from memory. It wasn't a hard SAC but I completely messed up one of the IR questions (teacher put an amine in the SAC and I hadn't looked over what those absorptions looked like -.-)
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on May 04, 2014, 11:50:30 am
Oil plus sodium hydroxide => fatty acid (well, conjugate base of fatty acid at least) + glycerol Then plus methanol => methyl ester You have biodiesel and glycerol as your two layers (one is non-polar, one is polar)
Each school's SACs will be different. Mine was on aspirin, spectroscopy, biomolecules, naming things and reaction pathways from memory. It wasn't a hard SAC but I completely messed up one of the IR questions (teacher put an amine in the SAC and I hadn't looked over what those absorptions looked like -.-)
Thanks for that :) at least, you still got a 50 :)
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on June 07, 2014, 09:23:00 am
In regards to organic chem, what is meant by cracking?
Title: Re: Rishi's Chemistry Thread
Post by: Yacoubb on June 07, 2014, 10:03:13 am
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on June 10, 2014, 08:05:11 pm
Lactic acid (HC3H5O3) is a product of anaerobic respiration in muscle cells. It accumulates in muscle tissue during exertion, causing pain. In a 0.15M solution of lactic acid, the acid is 3.0% ionised. i) Determine the concentration of the C3H5O3- anion in a 0.15M aqueous solution of lactic acid
Thanks heaps :)
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on June 10, 2014, 08:43:03 pm
3.0% ionised => concentration of ionised lactate = 3% * 0.15 M = 3*10^-2 * 1.5*10^-1 M = 4.5*10^-3 M
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 01, 2014, 02:32:00 pm
A 10mL solution of HCl has a pH of 2. WHat volume of water,in mL, must be added to it to change the pH to 3? a) 10 b) 90 c) 99 d) 100
Thanks :)
Title: Re: Rishi's Chemistry Thread
Post by: brightsky on July 01, 2014, 02:41:41 pm
Need to dilute by a factor of 10, i.e. turn 10 mL into 100 mL. So need to add 90 mL. B.
Title: Re: Rishi's Chemistry Thread
Post by: Yacoubb on July 03, 2014, 12:42:36 am
Back to the stem of the question, it asks you to find the amount of water needed to change the pH to 3. 100mL - 10mL = 90mL. Hence, the answer is B.
Thanks for that Yacoubb :)
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 03, 2014, 10:57:38 am
Would different low resolution proton NMR spectra be observed for C3H6O2(l) as opposed for C3H6O2(aq) ? Give a reason for your answer
Many thanks :-)
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 07, 2014, 10:06:40 am
Citric acid, C6H12O6, the main acid in lemon juice, is a tripotic acid. When 20.00ml aliquots of lemon juice were fully titrated with 0.350M NaOH, the volume of the average titre was 37.09mL. What was the concentration, in g/L, of citric acid in the lemon juice? a) 0.831 b) 2.50 c) 41.5 d) 374
Thanks :)
Title: Re: Rishi's Chemistry Thread
Post by: Rod on July 07, 2014, 01:35:24 pm
n(NaOH) = 0.350 X 0.03709 = 0.01298 MOL n (citric acid) = 1/3 x 0.00432 mol c (Citric acid) + 0.00432 /0.02 = 0.216 mols per litre c (citric acid) = 0.216 x 192 = 41.5 = C Assuming it's the reaction I have given you.
I think you made some mistakes on the parts I coloured in red. Your final answer is correct though, you just typed some of your steps in wrong I guess.
Firstly, the equation should be C6H8O7 + 3NaOH --> Na3C6H5O7 + 3H2O. Note how the sodium gets the 3 from the citric acid when it citric acid loses/donates it's 3 protons. Also, you typed in the second line of your working incorrectly. It should read 1/3 * 0.01298. The rest of your steps are correct (:
Title: Re: Rishi's Chemistry Thread
Post by: Rod on July 07, 2014, 05:15:00 pm
I think you made some mistakes on the parts I coloured in red. Your final answer is correct though, you just typed some of your steps in wrong I guess.
Firstly, the equation should be C6H8O7 + 3NaOH --> Na3C6H5O7 + 3H2O. Note how the sodium gets the 3 from the citric acid when it citric acid loses/donates it's 3 protons. Also, you typed in the second line of your working incorrectly. It should read 1/3 * 0.01298. The rest of your steps are correct (:
Thanks but didn't really make any difference, I used 0.01298 mol when I multiplied by a third
mehs all this for a multi choice question xD
Title: Re: Rishi's Chemistry Thread
Post by: alchemy on July 07, 2014, 06:28:39 pm
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 10, 2014, 01:45:34 pm
What is the concentration in mol/L, of I- (aq) in a solution that is 5.00% KI, by mass, if the solution has a density of 1.038 g/ml? a) 0.0301 b) 0.0313 c) 0.313 d) 0.625
Thanks :)
Title: Re: Rishi's Chemistry Thread
Post by: nhmn0301 on July 10, 2014, 01:59:31 pm
What is the concentration in mol/L, of I- (aq) in a solution that is 5.00% KI, by mass, if the solution has a density of 1.038 g/ml? a) 0.0301 b) 0.0313 c) 0.313 d) 0.625
Thanks :)
5.00% of KI meaning 5g of KI in 100g of solution. Density of solution = m/V 1.038= 100/V V = 96.3 mL. Hence, there are 5g of KI in 96.3mL solution. n (KI) = m/M = 5 / 166 = 0.0301 (mol) = n(I-) Hence, there are 0.0301 mol of I- in 96.3mL => 0.0301 mol in 0.0963 L => 0.313 mol in 1L So I guess the answer is C. Let me know if there is any errors!
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 13, 2014, 11:32:06 am
Consider the following equation:
2H2O2(l) --> 2H2O (l) + O2 (g) In this reaction: a) H2O2 is undergoing oxidation only b) H2O2 is undergoing reduction only c) H2O2 is undergoing both oxidation and reduction d) H2O2 is not undergoing either reduction or oxidation
Title: Re: Rishi's Chemistry Thread
Post by: nhmn0301 on July 13, 2014, 11:49:11 am
2H2O2(l) --> 2H2O (l) + O2 (g) In this reaction: a) H2O2 is undergoing oxidation only b) H2O2 is undergoing reduction only c) H2O2 is undergoing both oxidation and reduction d) H2O2 is not undergoing either reduction or oxidation
Normally, when you assign oxidation number, you let O = -2. But in here, we have H2O2, so the order to the rules becomes important, if you have a look back to the order, the rule which states that H is a +1 is above O is a -2. Hence, in H2O2, we assign H as +1 and O as -1. Hence, you have O goes from -1 -> -2, which means it's being reduced. So I reckon it's B.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 13, 2014, 12:11:42 pm
Normally, when you assign oxidation number, you let O = -2. But in here, we have H2O2, so the order to the rules becomes important, if you have a look back to the order, the rule which states that H is a +1 is above O is a -2. Hence, in H2O2, we assign H as +1 and O as -1. Hence, you have O goes from -1 -> -2, which means it's being reduced. So I reckon it's B.
The answer says D
Title: Re: Rishi's Chemistry Thread
Post by: nhmn0301 on July 13, 2014, 12:22:51 pm
Sorry about that, my bad. But even if I need to choose again, I would choose C. The half equation I think goes like this: H2O2 -> O2 + 2H+ + 2e^- oxidation 2e- + 2H+ + H2O2 -> 2H2O reduction That's what I think but according to the question, it's wrong so please correct me.
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on July 13, 2014, 01:52:55 pm
Question must be wrong then. You've given the equation for the spontaneous (albeit slow if uncatalysed) decomposition of hydrogen peroxide, which is interesting as it's a disproportionation reaction, when H2O2 is both oxidised and reduced simultaneously. The answers given must be wrong. O2 has oxidation state zero for oxygen, while H2O has oxidation state -2 and H2O2 has oxidation state -1
In weird cases like these, don't just go by rules. Draw out the structure, which is H-O-O-H Each hydrogen is bonded to an oxygen => oxidation state convention says each hydrogen has 'lost' an electron as oxygen is more electronegative and has oxidation state +1. Then, each oxygen has only gained one electron each from the hydrogen as the O-O bond is completely symmetric => oxygen gains one electron, oxidation state -1. No need for 'dodgy' application of rules
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 21, 2014, 01:30:19 pm
When methane reacts with chlorine in the presence of uv light, a) an addition reaction occurs to form CH3Cl b) a substitution reaction occurs to form CH3Cl c) the products are CH3Cl and HCl only d) the products include CH3Cl, CH2Cl2, CHCl3, CCl4 and HCl
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on July 21, 2014, 02:48:08 pm
When methane reacts with chlorine in the presence of uv light, a) an addition reaction occurs to form CH3Cl b) a substitution reaction occurs to form CH3Cl c) the products are CH3Cl and HCl only d) the products include CH3Cl, CH2Cl2, CHCl3, CCl4 and HCl
Troll question. b can occur, but d is more accurate because of all the possible products. Note that HCl IS a product of the substitution, no matter if it's the first, second or third stage.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 26, 2014, 05:51:58 pm
From the following statement, is it possible to tell the direction in which the rate of reaction is greater? ---> The concentration fraction is greater than the value of the equilibrium constant
Title: Re: Rishi's Chemistry Thread
Post by: jgoudie on July 26, 2014, 06:11:06 pm
Yes. The concentration factor is the Q value (ie. [products]/[reactants]) at any time during an equation. The equlibrium constant is the K value (ie. [products]/[reactants]) at equilibrium.
Thinking about this. If the Q value is larger than the K value that means there is more products than there would be at equilibrium. These means that the reaction would proceed in the reverse direction.
From the following statement, is it possible to tell the direction in which the rate of reaction is greater? ---> The concentration fraction is greater than the value of the equilibrium constant
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 26, 2014, 07:23:12 pm
An aqueous solution was prepared by dissolving 0.0200g of NaOH in 500.0mL of deionized water. The pH of this solution would be closest to: a) 11.0 b) 3.0 c) 3.3 d) 10.7
Thank you heaps :)
Title: Re: Rishi's Chemistry Thread
Post by: alchemy on July 26, 2014, 10:49:20 pm
An aqueous solution was prepared by dissolving 0.0200g of NaOH in 500.0mL of deionized water. The pH of this solution would be closest to: a) 11.0 b) 3.0 c) 3.3 d) 10.7
Thank you heaps :)
n=0.02/(23+16+1)=0.0005 c(OH-)=[OH-]=n/V=0.0005/0.5=0.001M [H+]=(10^-14)/0.001=10^-11 pH=-log10[H+]=-log10(10^-11)=11. So the answer is A.
Title: Re: Rishi's Chemistry Thread
Post by: Yacoubb on July 27, 2014, 08:48:01 am
An aqueous solution was prepared by dissolving 0.0200g of NaOH in 500.0mL of deionized water. The pH of this solution would be closest to: a) 11.0 b) 3.0 c) 3.3 d) 10.7
Thank you heaps :)
Before you do any calculations, just keep in mind that sodium hydroxide is a strong base, eliminating B and C. Then, calculations as alchemy has shown. :)
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 27, 2014, 09:17:25 am
The addition of some solid sodium propanoate to a 0.10M aqueous solution of propanoic acid resulted in the equilibrium concentration of the propanoate ion being 5.2 x10-3. The pH of this solution would be:
a) 3.6 b) 2.9 c) 2.3 d) 6.2
Title: Re: Rishi's Chemistry Thread
Post by: alchemy on July 27, 2014, 07:00:05 pm
The addition of some solid sodium propanoate to a 0.10M aqueous solution of propanoic acid resulted in the equilibrium concentration of the propanoate ion being 5.2 x10-3. The pH of this solution would be:
a) 3.6 b) 2.9 c) 2.3 d) 6.2
Use the Henderson-Hasselbalch equation. pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration) =-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6. So the answer is A.
Title: Re: Rishi's Chemistry Thread
Post by: Blondie21 on July 27, 2014, 07:39:38 pm
Use the Henderson-Hasselbalch equation. pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration) =-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6. So the answer is A.
Hey alchemy, which textbook do you use? I have never heard of that equation before.
-
I solved this differently and used acidity constants (Rishi it's on page 289 of Heineman)
CH3CH2COOH <-> CH3CH2COO- + H+
Ka = [CH3CH2COO-] [H+ or H3O+.. same thing] / [CH3CH2COOH]
Sub in the info we now. NB: the acidity constant for propanoic acid is in our data books
1.3 x 10-5 = [5.2x10-3] x [H+] / 0.10
H+ = 0.00025M
pH = -log(0.00025) pH = 3.6
(My class did this last week ;))
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 27, 2014, 07:43:22 pm
Use the Henderson-Hasselbalch equation. pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration) =-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6. So the answer is A.
Never seen this before..hmm.. something new for me:) yay Thanks Alchemy
Title: Re: Rishi's Chemistry Thread
Post by: Blondie21 on July 27, 2014, 07:59:55 pm
Use the Henderson-Hasselbalch equation. pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration) =-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6. So the answer is A.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on July 27, 2014, 08:32:08 pm
A student titrates sulfuric acid against a 0.200 M sodium hydroxide solution. 20.00 ml aliquots of sodium hydroxide are used How do I know if the acid or the base is in the burette?
Title: Re: Rishi's Chemistry Thread
Post by: alchemy on July 27, 2014, 08:42:19 pm
A student titrates sulfuric acid against a 0.200 M sodium hydroxide solution. 20.00 ml aliquots of sodium hydroxide are used How do I know if the acid or the base is in the burette?
think about it Rishi.. If there are 20mL ALIQUOTS of NaOH hmmmm where does this go? In the burette or in the conical flask?
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on August 04, 2014, 09:15:53 pm
A 20mL aliquot of sodium hydroxide solution, NaOH, is added to a flask from a stock solution and is titrated against 0.66M sulfuric acid. The average titre is 16 mL
1) Another 20mL aliquot of NaOH is taken from the same stock solution and is diluted to 100mL. The volume of 0.66M sulfuric acid required in the titration this time, should be, in mL
Thanks :)
Title: Re: Rishi's Chemistry Thread
Post by: Blondie21 on August 04, 2014, 10:06:14 pm
A 20mL aliquot of sodium hydroxide solution, NaOH, is added to a flask from a stock solution and is titrated against 0.66M sulfuric acid. The average titre is 16 mL
1) Another 20mL aliquot of NaOH is taken from the same stock solution and is diluted to 100mL. The volume of 0.66M sulfuric acid required in the titration this time, should be, in mL
Thanks :)
Remember this: Diluting something will NOT change the amount of mol. Think about it like this... If I had 3 balls in a container of 50mL of water and I added another 100mL, would this change the amount of balls? No. Same with molecules. Keep this in mind.
n(NaOH) = 0.011mol (For the reaction with the 20mL aliquot) Once this is diluted, the mol remains at the value. However, the concentration changes. It is now: c(NaOH) = 0.11M
We know the concetration of sulphuric acid remains at .66M so c1V1=c2V2 (0.11)x(0.1) = (.66)V2 V= 0.016L = 16mL
Orrrrr you could have skipped all of those calculations as the mol of H2SO4 remains the same. Because the mol and concentration is the same, the volume must be the same too! :P
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on August 24, 2014, 02:44:29 pm
Is the following statement correct regarding electrochemical series?
The more negative the E0 value is, the more readily oxidation occurs and the stronger the reducing agent
Thanks
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on August 24, 2014, 05:56:41 pm
List two factors that need to be considered when selecting an appropriate substance for use in a salt bridge
Thanks :)
Some off the top of my head;
- Whether it is soluble or insoluble - Salt chosen cannot react with cell contents - Whether it provides the appropriate ions to balance charges generated in each half cell
Title: Re: Rishi's Chemistry Thread
Post by: Yacoubb on August 31, 2014, 10:06:31 am
List two factors that need to be considered when selecting an appropriate substance for use in a salt bridge
Thanks :)
This may be one reason, but really it covers everything. The salt being used should contain soluble ions (this is so that no precipitate forms in the half-cells, which would disrupt the experiment).
Title: Re: Rishi's Chemistry Thread
Post by: Phenomenol on August 31, 2014, 10:25:55 am
- Whether it is soluble or insoluble - Salt chosen cannot react with cell contents - Whether it provides the appropriate ions to balance charges generated in each half cell
I'm not sure if the third point is necessary. There was a terrific question on this topic in the VCAA 2013 exam.
Title: Re: Rishi's Chemistry Thread
Post by: lucas.vang on August 31, 2014, 05:11:29 pm
Hi everyone, I have a question regarding the products of electrolysis
When it is something like potassium nitrate? How would I predict it when nitrate isnt on the electrochemical series? and what about the others like sulphate and others?
thanks in advance
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on August 31, 2014, 09:19:52 pm
Hi everyone, I have a question regarding the products of electrolysis
When it is something like potassium nitrate? How would I predict it when nitrate isnt on the electrochemical series? and what about the others like sulphate and others?
thanks in advance
There is always water to electrolyse. Never forget that.
VCAA won't ask you about the others. You can reduce sulfate and nitrate but VCAA won't ask you for any reaction that's not in the data book unless they give you the reaction.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on September 23, 2014, 11:07:03 pm
Which of the following compounds contain the highest percentage by mass of oxygen? a) Propyl propanoate b) 1-propanol c) Propanoic acid d) 1- aminopropane
I've ruled put b and d but I can't decide between a and c
Thanks guys :)
Title: Re: Rishi's Chemistry Thread
Post by: lzxnl on September 24, 2014, 12:07:10 am
Which of the following compounds contain the highest percentage by mass of oxygen? a) Propyl propanoate b) 1-propanol c) Propanoic acid d) 1- aminopropane
I've ruled put b and d but I can't decide between a and c
Thanks guys :)
Propanoic acid is C3H6O2 Propyl propanate is the same as the above, except you're substituting a hydrogen for a CH3CH2CH2 group So it has the same number of oxygens with a higher molecular mass => lower percentage by mass of oxygen
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on October 13, 2014, 11:30:08 am
The pH of water is 7.0 at 25 degrees and it is 7.5 at 0 degrees. This means that: A. water is not neutral at 0 degrees B. the self-ionization of water is an endothermic process C. water becomes more alkaline as the temperature drops D. the percentage of ionization increases as the temperature drops
I always get these sorts of questions wrong. Is there like a basic theory that I'm missing?
Title: Re: Rishi's Chemistry Thread
Post by: nhmn0301 on October 13, 2014, 11:57:12 am
The pH of water is 7.0 at 25 degrees and it is 7.5 at 0 degrees. This means that: A. water is not neutral at 0 degrees B. the self-ionization of water is an endothermic process C. water becomes more alkaline as the temperature drops D. the percentage of ionization increases as the temperature drops
I always get these sorts of questions wrong. Is there like a basic theory that I'm missing?
A neutral solution (water in this case) has [OH-] = [H+]. So if at 25C, you have pH = 7, hence, [H+]=[OH-]= 10^(-7) => Kw= 10^(-14). At 0 C, pH= 7.5, hence, [H+]=[OH-]= 10^(-7.5) => Kw= 10^(-15). This means that as temperature decrease, Kw decreases, indicating that the self-ionization of water is endothermic. So I think the answer is B.
Title: Re: Rishi's Chemistry Thread
Post by: Rishi97 on October 13, 2014, 12:11:32 pm
A neutral solution (water in this case) has [OH-] = [H+]. So if at 25C, you have pH = 7, hence, [H+]=[OH-]= 10^(-7) => Kw= 10^(-14). At 0 C, pH= 7.5, hence, [H+]=[OH-]= 10^(-7.5) => Kw= 10^(-15). This means that as temperature decrease, Kw decreases, indicating that the self-ionization of water is endothermic. So I think the answer is B.