H2 Chem Questions
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1 The concentration of the chlorate (I) ions in household bleach can be determined by acidifying with hydrochloric acid to produce chlorine gas.
ClO- (aq) + 2H+ (aq) +Cl- (aq) -> Cl2 (aq) + H2O (l)
In a typical experiment, a 25.0cam3 sample of household bleach was duluted to 250cm3. Excess HCl acid was added to a 10.0cm3 portion of this diluted solution. Excess potassium iodide solution was then added, resulting in the production of a brown solution. The brown solution required 18.50cm3 of 0.200 mol dm-3 sodium thiosulfate, Na2S2O3, for complete reaction
Calculate the concentration of sodium chlorate (I) in the household bleach.
I wrote: 2KI + Cl2 -> 2KCl + I2
I2 + 2 S3O3 2- -> 2I- + S4O6 2-
First, I found the amount of Na2S2O3 to be 0.00370mol
Amount of I2 = 0.5 X 0.00370 = 0.00185mol
Amt of Cl2 = Amt of I2 = 0.00185
Let n mol be mount of ClO- in 10.0cm3
[ClO-] in 10.0cm3 diluted = n / (10.0/1000) = 100n
In 10.0 cm3 (diluted),[ClO-] = 100n
In 250cm3 (diluted) [ClO-] = 2500n
CV (Before dilution) = CV (after dilution)
Concentration before dilition = [(2500 X 0.00185)(250/10000] / (25/1000)
= 46.25 mol dm-3... which is a ridiculously HIGH concentration... anyone can help me out?
2. The lattice energy of sodium chlorate (I) can be determined from a hypothetical Born-Haber cycle using the following date:
Ethalpy change of atomisation of Na = 107 kJ/mol
Cl-O bond energy +269 kj/mol
electron affinity of ClO -506 kJ/mol
enthalpy change of formation of NaClO -564 kJ/mol.
(i) we are asked to find the lattice energy of sodium chlorate using a born-haber cycle. I did draw out the entire steps. The answer obtained is -928 kJ/mol.
Can someone verify my answer? thanks
(ii) Explain how you would expect the numerical magnitude of the LE of sodium chlorate (I) to compare with that of sodium chloride.
2. Phosphoric acid has multiple dissociation constants and can be used to create buffers maintained at a wide range of pH values
H3PO4 <--> H2PO4 - + H+ 2.16
H2PO4- <--> HPO4 2- + H+ 7.21
HPO4 2- <----> PO4 3- + H+ 12.32
The values provide are pKa values.
(i) Explain why the pKa values are increasing down the table.
(ii) Identify the acid/conjugate base pair used to make a phosphate buffer of pH 6.8
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I'll only help out with the non-calculation questions.
Explain how you would expect the numerical magnitude of the LE of sodium chlorate (I) to compare with that of sodium chloride.
The LDE/LFE of NaClO will be less endothermic/exothermic compared to the LDE/LFE of NaCl, because ClO- has lower charge density compared to Cl-.
(i) Explain why the pKa values are increasing down the table.
Successive proton dissociation constants are always smaller, because the increasing magnitude of negative charge on the successive conjugate bases result in decreasing stabilities of the species. Furthermore, (guys only like girls, guys don't like guys) negatively charged bases will repel other negatively charged bases.
(ii) Identify the acid/conjugate base pair used to make a phosphate buffer of pH 6.8
At pH = pKa, we have equal molarities of both members of a conjugate acid-base pair, and we have maximum buffer capacity. Hence we will need a slightly higher molarity of H2PO4 - and a slightly lower molarity of HPO4 2- to achieve a buffer with pH of 6.8 (ie. slightly below 7.21).
When aq. iron (III) chloride is shaken with aq sodium carbonate, no ppt of iron (III) carbonate is formed. Instead, effervescence and a red-brown ppt of iron (III) hydroxide are observed.
2 reasons :
The high charge density Fe3+ ion will polarize and distort the anionic charge cloud of CO3 2-, resulting in it's decomposition (becoming less endothermic).
Simultaneously, the high charge density Fe3+ ion is strongly electron withdrawing by induction, withdrawing electron density from its water ligands through the coordinate dative bonds, weakening the O-H bonds in the water ligands, resulting in proton dissociation from the water ligands (becoming less endothermic).
This results in an acidic solution whereby the CO3 2- ions are protonated to generate H2CO3, which exists in equilibrium with, and hence can decompose into, CO2 and H2O.
The water ligands that were deprotonated earlier constitutes the OH- ions that now combine with the Fe3+ ions (when ionic product Qsp exceeds solubility product Ksp) to generate the Fe(OH)3 ppt.
Iron (II) iodide can be crystallised from an aq mixture of iron (II) sulfate and potassium iodide but iron (III) iodide cannot be produced by mixing aq iron (III) sulfate and potassium iodide.
Fe3+ will oxidize I- to I2, and I- will reduce Fe3+ to Fe2+.
Cell potential = Reduction potential @ Cathode + Oxidation potential @ anode.
Use Data Booklet values to show that standard cell potential is positive, and therefore you cannot obtain FeI3(aq) under standard conditions.
Suggest an advantage of using nickel carbonyl complex in the mond process.
CO complexes with nickel readily and reversibly, allowing convenient extraction from nickel oxides. On heating, nickel tetracarbonyl decomposes to give the desired nickel metal, and the carbon monoxide gas, which can hence be easily separated.
0.382 g of the nikel carbonyl complex was completely vaporised in a gas syringe at 120 deg cel and 1 atm. 72.4 cm3 of vapor was produced. calculate the % by mass of CO in the complex.
Note that the final vapour is purely CO, as nickel would present as a solid residue. Using PV=nRT, find the moles of CO. Hence find sample mass of CO. Therefore find % by mass of CO in the complex.
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Thank you Ultimaonline for providing the clear explanations, as always! At least I have a brief understanding on some questions now.
If anyone can, feel free to help me check on the calculation questions. Importantly, question 1 (Which I am afraid of making a mistake due to conceptual misunderstanding)
Thank you all!
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The mistake (I think) you made:
Concentration of 10 cm3 dilute solution = Concentration of 250 cm3 dilute solution.
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How you distinguish between phenol and methoxy benzene?
I think Adding neutral FeCl3 (aq) will distinguish. Only phenol will give a violet-color complex while methoxy benzene will have no reaction.
Please correct me.
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Originally posted by hoay:
How you distinguish between phenol and methoxy benzene?
I think Adding neutral FeCl3 (aq) will distinguish. Only phenol will give a violet-color complex while methoxy benzene will have no reaction.
Please correct me.
Correct.Methoxybenzene, unlike phenol, does not have an acidic proton which can be lost, which is a prerequisite for its function as a ligand in ionic coordination complex formation presenting as a visible violet coloration, as only the phenoxide ion (rather than phenol or methoxybenzene) is sufficiently electron-rich to function effectively as the ligand.
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Three compounds
A is an aromatic ester and also has a phenol group
B is an aromatic ester and also has alcohol group
C has ether group and carboxylic acid attached to benzene ring.
Based on their relative acidities, suggest how samples of A, B and C could be
distinguished from each other by the use of NaOH(aq) and Na2CO3(aq).Ans..... The increasing order of acid strength is B,A,C.
C will react with both NaOH and Na2CO3. while A and B will react with only NaOH(aq).
Is this sufficient explanation?
Secondly both A and B contain ester group so how we are going to distinguish which is more acidic as the patter suggest.
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Originally posted by hoay:
Three compounds
A is an aromatic ester and also has a phenol group
B is an aromatic ester and also has alcohol group
C has ether group and carboxylic acid attached to benzene ring.
Based on their relative acidities, suggest how samples of A, B and C could be
distinguished from each other by the use of NaOH(aq) and Na2CO3(aq).Ans..... The increasing order of acid strength is B,A,C.
C will react with both NaOH and Na2CO3. while A and B will react with only NaOH(aq).
Is this sufficient explanation?
Secondly both A and B contain ester group so how we are going to distinguish which is more acidic as the patter suggest.
It'll help if you can specify the condensed structural formula of the 3 compounds, eg. C6H5COOCH3, etc.
Assuming B's alcohol is an aliphatic alcohol, regardless of the aromatic benzene ring present (further away from the OH group), then the aliphatic alcohol cannot be deprotonated by OH-. Even if the ester group has an acidic alpha proton, it'll take a stronger base such as NH2- to deprotonate.
Likely answer :
B cannot be deprotonated by either OH- or CO3 2-
A will only be deprotonated by OH- and cannot be deprotonated by CO3 2-.
C will be deprotonated by both.
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(i) SnO2 reacts with concentrated sulfuric acid to form a colourless solution with no
evolution of gasSnO2 + 2H2SO4 Sn(SO4)2 + 2H2O
(ii) PbO2 reacts with concentrated sulfuric acid to give a white solid, B, and oxygen gas.
PbO2 + H2SO4 PbO + PbSO4 + O2 + H2O
B
(iii) (a) PbO2 reacts with cold concentrated hydrochloric acid to give a yellow solution
containing the [PbCl 6]2– ion, with no evolution of gas. (b) Warming this yellow solution causes the evolution of Cl 2 gas, leaving a colourless solution which on cooling in ice precipitates a white solid, C.(a) PbO2 + 6HCl [PbCl6]2- + H2O
(b) [PbCl6]2- Cl2 + [PbCl4]2-
C
Please make any corrections.
thank you.
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Originally posted by hoay:
(i) SnO2 reacts with concentrated sulfuric acid to form a colourless solution with no
evolution of gasSnO2 + 2H2SO4 Sn(SO4)2 + 2H2O
(ii) PbO2 reacts with concentrated sulfuric acid to give a white solid, B, and oxygen gas.
PbO2 + H2SO4 PbO + PbSO4 + O2 + H2O
B
(iii) (a) PbO2 reacts with cold concentrated hydrochloric acid to give a yellow solution
containing the [PbCl 6]2– ion, with no evolution of gas. (b) Warming this yellow solution causes the evolution of Cl 2 gas, leaving a colourless solution which on cooling in ice precipitates a white solid, C.(a) PbO2 + 6HCl [PbCl6]2- + H2O
(b) [PbCl6]2- Cl2 + [PbCl4]2-
C
Please make any corrections. Thank you.
Interesting inorganic chemistry deductive elucidation questions.Your answers all appear to be fine, but the very last question warrants further discussion.
Heating provides the activation energy for a redox reaction to occur, in which lead(IV) (in the dinegative hexachloroplumbate(IV) ion) is reduced to lead(II) (thus generating the dinegative tetrachloroplumbate(II) ion), and 2 of the chloride ligands are oxidized to chlorine gas.
However, because the question specified that the resulting product is a "colourless solution which on cooling in ice precipitates a white solid, C", it is unlikely to be the soluble dinegative tetrachoroplumbate(II) ion, which as a dinegative ionic species will experience thermodynamically favourable ion-permament dipole bonding with strongly polar protic water.
The white solid C is therefore more likely to be relatively insoluble lead(II) chloride.
As described earlier, heating provides the activation energy for the transfer of electrons from 2 of the chloride ligands to reduce the lead(IV) (in the dinegative hexachloroplumbate(IV) ion) to lead(II) (thus generating the dinegative tetrachloroplumbate(II) ion), and the 2 chloride ligands are oxidized to chlorine gas, and so we get the reaction exactly as you (correctly) described :
[PbCl6]2-(aq) <---> Cl2(g) + [PbCl4]2-(aq)
However, that is probably not the final equation that generates C.
Bearing in mind that "cold concentrated hydrochloric acid" was used to generate the yellow complex ion hexachloroplumbate(IV) (which btw, is yellow due to other electron transitions other than d-d*, since lead is not a d-block transition metal and does not possess partially filled d-orbitals, a possible common exam trick question that many 'A' level students fall for), therefore the coordination compound containing the yellow complex ion, is dihydrogen hexachloroplumbate(IV). So there are plenty of protons to serve as the positive counter ion where necessary.
The tetrachloroplumbate(II) ion itself exists in equilibrium with its constituent species, as illustrated in the equation for the formation constant (Kf) a.k.a. stability constant (Kstab) as follows :
Pb2+(aq) + 4Cl-(aq) <---> [PbCl4]2-(aq)
and that all relatively insoluble precipitates nonetheless exist in equilibrium with their constituent species, as illustrated in the equation for the solubility product, Ksp :
PbCl2(s) <---> Pb2+(aq) + 2Cl-(aq)
Because C is rather insoluble at low temperatures (as specified in the question), it's conceivable that, should the forward reaction (for either or both Kf-Kstab and Ksp equations above) be endothermic, lowering the temperature will shift the position of equilibrium to the left, generating PbCl2(s), with the protons (available as mentioned earlier) acting as counter ions for the Cl- ions, in effect generating HCl(g), which in gaseous state leaves the reaction mixture, pulling the position of equilibrium over to the right as predicted by Le Chatelier's principle, with a favourable positive entropy change as a thermodynamic bonus, too.
H2[PbCl4] <---> PbCl2(s) + 2HCl(g)
To summarize in chronological order, the equations would be :
Coordination complex formation :
PbO2(s) + 6HCl(l) <---> [PbCl6]2-(aq) + H2O(l)
Redox reaction :
[PbCl6]2-(aq) <---> Cl2(g) + [PbCl4]2-(aq)
Coordination complex equilibria :
[PbCl4]2-(aq) <---> Pb2+(aq) + 4Cl-(aq)
Ionic solubility equilibria :
Pb2+(aq) + 2Cl-(aq) <---> PbCl2(s)
Overall chemical equation :
PbO2(s) + 6HCl(l) ---> H2[PbCl6](aq) + H2O(l) ---> Cl2(g) + H2[PbCl4](aq) ---> Cl2(g) + Pb2+(aq) + 2Cl-(aq) + 2HCl(g) ---> Cl2(g) + PbCl2(s) + 2HCl(g)
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Wow!! Excellent explanation. Thank you very much Sir.
But PbO is yellow so what about my equation of PbO2 with H2SO4.
PbO2 + H2SO4 PbO + PbSO4 + O2 + H2O.
Is it correct?
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Originally posted by hoay:
Wow!! Excellent explanation. Thank you very much Sir.
But PbO is yellow so what about my equation of PbO2 with H2SO4.
PbO2 + H2SO4 PbO + PbSO4 + O2 + H2O.
Is it correct?
Oops, yes thanks for pointing that out.Lead(IV) compounds are unstable, and thus often functions as oxidizing agents as they act to reduce themselves to more stable lead(II) compounds.
Here, sulfur already has a maximum OS of +6, and thus cannot be oxidized further.
Therefore, it is oxygen that is oxidized : one (mole) of the oxide anion loses (two moles of) electrons to (one mole of) the eager I-cant-wait-to-be-reduced lead(IV) ion which is reduced to (one mole of) the more stable lead(II), and the oxide anion itself becomes oxidized to (half mole of) molecular dioxygen.
The other oxide anion, being a strong base (due to its dinegative high charge density) is protonated by the strong sulfuric(VI) acid, generating water.
When ionic product exceeds solubility product, the lead(II) ion combines with the sulfate(VI) anion, generating the white solid B, ie. PbSO4(s) precipitate.
PbO2(s) + H2SO4(l) --> PbSO4(s) + H2O(l) + 1/2 O2(g)
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Are the reagetnts and conditions for the hydrolysis of ester, amides and proteins are same i.e dil HCl or dil NaOH and heat under reflux? Is there any particular concentration to be used such as 2M HCl.
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Originally posted by hoay:
Are the reagetnts and conditions for the hydrolysis of ester, amides and proteins are same i.e dil HCl or dil NaOH and heat under reflux? Is there any particular concentration to be used such as 2M HCl.
Yes, they are the same. Hydrolysis of esters and amides can be acid-catalyzed or base-promoted (with pros and cons in regards of activation enthalpies for the various stages in the hydrolysis mechanism).'A' level students are not required to memorize any specific concentration of acid or alkali used for hydrolysis. If Cambridge asks the student to suggest a molarity, then 2M is about fine.
'A' level students are expected to be aware that, due to the much larger number of covalent bonds required to be broken, the complete hydrolysis of proteins will naturally require a significantly longer time, compared to the hydrolysis of simple esters and amides.
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I did not pay any attention to the comparative rate of hydrolysis of proteins and amides. Indeed proteins would require a longer time than ester and amides. Thank you very much Sir.
The use of acid halides as halogentaing agent is common. But there is confusion regarding their conditions. I am aware of the following reagents and conditions.
PCl3 heat
PCl5 room temperature
SOCl2 room temperature
Or P + Cl2 + heat
In case of Br:
PBr3 heat
P + Br2 + heat
I think you cannot quote PBr5 directly as it does not exist.
I am not aware of use of iodine.
please make any corrections.
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Originally posted by hoay:
I did not pay any attention to the comparative rate of hydrolysis of proteins and amides. Indeed proteins would require a longer time than ester and amides. Thank you very much Sir.
The use of acid halides as halogentaing agent is common. But there is confusion regarding their conditions. I am aware of the following reagents and conditions.
PCl3 heat
PCl5 room temperature
SOCl2 room temperature
Or P + Cl2 + heat
In case of Br:
PBr3 heat
P + Br2 + heat
I think you cannot quote PBr5 directly as it does not exist.
I am not aware of use of iodine.
please make any corrections.
Yes, your stated conditions are all correct.
However, for generating phosphorous halide in situ, it is advised that the 'A' level student specify the use of red phosphorous, as opposed to white phosphorous ('A' level students should be able to explain why; see Wikipedia on phosphorous).
There are at least 2 different reasons why PCl5 exists, but not PI5. Cambridge may ask the 'A' level student to "suggest" reasons for this.
The first reason has to do with oxidizing power. Iodine has much weaker oxidizing strength compared to chlorine ('A' level students should be able to explain this, based on redox potentials as well as atomic structure), and thus while Cl has the capacity to oxidize P to an OS of +5 in phosphorous(V) chloride, iodine only has the capacity to oxidize P to an OS of +3 in phosphorous(III) iodide.
The second reason has to do with steric considerations. Iodine has a much larger atomic radius compared to chlorine, and therefore an analogous PI5 would experience too much steric crowding and van der Waals repulsion (specifically, lone pair - lone pair repulsions) between the 5 iodine atoms in the trigonal bipyramidal geometry, destabilizing the molecular structure.
Being an intermediate between chlorine and iodine, bromine is still able to form PBr5, but it is an unstable species and students should just quote the use of PBr3 instead.
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In an electrochemical cells why a high resistance voltmeter is attached? Specially when measuring the emf of system. I am with the word "high resistance" what we mean exactly by that?
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Originally posted by hoay:
In an electrochemical cells why a high resistance voltmeter is attached? Specially when measuring the emf of system. I am with the word "high resistance" what we mean exactly by that?
This part is more physics than chemistry. The idea being, the higher the resistance, the lower the current, and therefore the higher the voltage registered. This allows for the maximum potential difference to be registered, which is then taken to be the reduction or oxidation potential of the other half-cell (ie. other than the SHE half-cell). -
I have some chemistry MCQ questions and I will greatly appreciate if anyone could enlighten me on how to get about solving them. Thank you very much!
Q1:
Equation: CCl2CHCl + (HF) -> CF3CH2Cl
• In addition to electrophilic addition, I'm curious as to what is the other reaction that converts some of the Cl (but not all, why) to F?
Q2: Aluminium chloride conducts electricity in its aqueous state but not in its molten state.
• Ans: True, but why (got to do with it being a covalent compound?)? Is this under the topic of periodicity?
Q3: Which substances have a higher boiling point than its functional group isomer?
1. CH3CH(OH)CH2CH3
2. CH3COOCH2CH3
3. CH3CH2CH2CHO
Q4: Chlorine in the presence of sunlight reacts with gaseous hydrocarbon to produce a mixture of mono-substituted alkanes and HCl. Which of the following molecules could represent that hydrocarbon?A. Ethane
B. Propane
C. 2,2-dimethylpropane
D. Cyclohexane
• Seems like all of them can be multi-substituted in the presence of Cl..
Q5: In an experiment 20cm3 of 0.1moldm-3 Mn2+ was acidified and completely oxidised to MnO4- by 25cm3 of 0.2moldm-3 solution of Na2S2O8. Which of the following is the sulphur-containing product of this reaction?
(ie. ratio of Mn:S = 2:5)
A. S
B. S2O3 2-
C. SO3 2-D. SO4 2- (Ans)
• I worked out that the oxidation state of the sulphur containing product to be +5 but the oxidation state of S in SO42- is +6
Q6: To identify an oxide of nitrogen, 0.1mol of the oxide was mixed with 10dm³ of hydrogen gas and passed over a heated catalyst. At the end of the reaction, 0.4dm³ of hydrogen gads remained. The ammonia produced required 125cm³ of 0.8moldm-3 H2SO4 for neutralisation. All gaseous volmes were measured at rtp.What is the formula of the oxide of nitrogen?
A. NO
B. N2O (Ans)
C. NO2
D. N2O4• I worked out NxOy +4H2 -> 2NH3, but what is the full equation for the reaction in order for me to be sure of the values of x&y?
Q7:
The compressibility factor, pV/RT of one mole of any real gas approaches 1.0 when p approaches 0atm.• True, but why?
Q8: NOCl decomposes to form NO and Cl2 as follows:
2NOCl(g) <=>2NO(g)+Cl2(g)
At 35 degrees, the Kc=1.6x10^-5moldm-3.
Which relationship must be correct at equilibrium at 35 degrees?
A. 2 [NOCl] = 2[NO]+[Cl2]
B. 2[NO]=[Cl2]
C. [NO]<[NOCl] (Ans)
D. [NOCl]<[Cl2]
Is the answer C because Kc is of a very small value, so barely any product is formed relative to the reactant? -
In future, please post a few questions at a time. It'll help your chances of getting a reply sooner. For instance, if I've only got a few mins to spare when dropping by this forum, I would be more inclined to immediately reply a post asking about 2 or 3 qns, but might decide not to reply (if busy, perhaps for a couple days or even more) a post asking about 8 qns.
Q1. It's nothing more than a typo error in the question.
Q2. AlCl3 doesn't conduct in the molten state because it remains covalent, but undergoes hydrolysis in aqueous state to generate ions. Topic : both chemical bonding and periodicity.
Q3. Option 1 only. Alcohols are capable of both donating and accepting H bonds, but ethers can only accept H bonds.
Q4. Only propane has two different possible mono-halogenated products. The rest only have 1 possible mono-halogenated product.
Q5. It's a trick question. It's not the S atoms that accept electrons in peroxodisulfate, it is the peroxo O atoms that accept electrons.
Q6. Based on NH3 generated, both N2O and N2O4 are possible. Based on H2 used up to generate H2O, only N2O4 is possible.
Q7. As pressure tends towards zero, a real gas tends towards ideal gas behaviour (due to reduced interactive forces between gaseous particles), where PV=nRT holds true, and hence = PV/RT = n =1.
Q8. Correct. But if the question is made tougher by having a Kc value closer to 1, then a mathematical proof (using algebra and ICE table) is required to discern the best answer.
Originally posted by gohby:I have some chemistry MCQ questions and I will greatly appreciate if anyone could enlighten me on how to get about solving them. Thank you very much!
Q1:
Equation: CCl2CHCl + (HF) -> CF3CH2Cl
• In addition to electrophilic addition, I'm curious as to what is the other reaction that converts some of the Cl (but not all, why) to F?
Q2: Aluminium chloride conducts electricity in its aqueous state but not in its molten state.
• Ans: True, but why (got to do with it being a covalent compound?)? Is this under the topic of periodicity?
Q3: Which substances have a higher boiling point than its functional group isomer?
1. CH3CH(OH)CH2CH3
2. CH3COOCH2CH3
3. CH3CH2CH2CHO
Q4: Chlorine in the presence of sunlight reacts with gaseous hydrocarbon to produce a mixture of mono-substituted alkanes and HCl. Which of the following molecules could represent that hydrocarbon?A. Ethane
B. Propane
C. 2,2-dimethylpropane
D. Cyclohexane
• Seems like all of them can be multi-substituted in the presence of Cl..
Q5: In an experiment 20cm3 of 0.1moldm-3 Mn2+ was acidified and completely oxidised to MnO4- by 25cm3 of 0.2moldm-3 solution of Na2S2O8. Which of the following is the sulphur-containing product of this reaction?
(ie. ratio of Mn:S = 2:5)
A. S
B. S2O3 2-
C. SO3 2-D. SO4 2- (Ans)
• I worked out that the oxidation state of the sulphur containing product to be +5 but the oxidation state of S in SO42- is +6
Q6: To identify an oxide of nitrogen, 0.1mol of the oxide was mixed with 10dm³ of hydrogen gas and passed over a heated catalyst. At the end of the reaction, 0.4dm³ of hydrogen gads remained. The ammonia produced required 125cm³ of 0.8moldm-3 H2SO4 for neutralisation. All gaseous volmes were measured at rtp.What is the formula of the oxide of nitrogen?
A. NO
B. N2O (Ans)
C. NO2
D. N2O4• I worked out NxOy +4H2 -> 2NH3, but what is the full equation for the reaction in order for me to be sure of the values of x&y?
Q7:
The compressibility factor, pV/RT of one mole of any real gas approaches 1.0 when p approaches 0atm.• True, but why?
Q8: NOCl decomposes to form NO and Cl2 as follows:
2NOCl(g) <=>2NO(g)+Cl2(g)
At 35 degrees, the Kc=1.6x10^-5moldm-3.
Which relationship must be correct at equilibrium at 35 degrees?
A. 2 [NOCl] = 2[NO]+[Cl2]
B. 2[NO]=[Cl2]
C. [NO]<[NOCl] (Ans)
D. [NOCl]<[Cl2]
Is the answer C because Kc is of a very small value, so barely any product is formed relative to the reactant? -
Originally posted by UltimaOnline:
In future, please post a few questions at a time. It'll help your chances of getting a reply sooner. For instance, if I've only got a few mins to spare when dropping by this forum, I would be more inclined to immediately reply a post asking about 2 or 3 qns, but might decide not to reply (if busy, perhaps for a couple days or even more) a post asking about 8 qns.
Q1. It's nothing more than a typo error in the question.
Q2. AlCl3 doesn't conduct in the molten state because it remains covalent, but undergoes hydrolysis in aqueous state to generate ions. Topic : both chemical bonding and periodicity.
Q3. Option 1 only. Alcohols are capable of both donating and accepting H bonds, but ethers can only accept H bonds.
Q4. Only propane has two different possible mono-halogenated products. The rest only have 1 possible mono-halogenated product.
Q5. It's a trick question. It's not the S atoms that accept electrons in peroxodisulfate, it is the peroxo O atoms that accept electrons.
Q6. Based on NH3 generated, both N2O and N2O4 are possible. Based on H2 used up to generate H2O, only N2O4 is possible.
Q7. As pressure tends towards zero, a real gas tends towards ideal gas behaviour (due to reduced interactive forces between gaseous particles), where PV=nRT holds true, and hence = PV/RT = n =1.
Q8. Correct. But if the question is made tougher by having a Kc value closer to 1, then a mathematical proof (using algebra and ICE table) is required to discern the best answer.
My apologies, UltimaOnline. Thank you for your help and I will take note of that!
Q3, is there an efficient schematic way of ensuring that all different functional isomers have been drawn, before one compares the intermolecular forces of attraction to compare boiling points?
Q4: Doesn't all the compound have the ability to form a mono-substituted compound with HCl too?
Q5: Could you write down the equation to illustrate?
For Q6, the answer is N2O, what's the full equation, and where do I find the general equation for the reaction between nitrogen oxide and hydrogen?
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No problem at all, gohby.
Q3. Unfortunately, no. This tests the candidates' knowledge of various functional groups, for instance, that alcohols are functionally isomeric with ethers, etc. Ethers can only accept H bonds, but cannot donate H bonds, hence there are no intermolecular hydrogen bonds between ethers.Q4. Yes, all of them except one of the options, can only generate one particular monohalogenated product. The singled out option, can generate two different monohalogenated products. This is due to, for instance, an internal plane of symmetry, such that no matter which H atom you substitute, you end up with exactly the same (ie. has the same structure and same IUPAC name) monohalogenated product. The phrasing "a mixture of monohalogenated products" in the qn implies that the correct answer can form more than one monohalogenated product.
Q5. The reduction half equation can be balanced easily, that's not the trick. The trick is in the OS (OS = formal charge + electronegativity consideration) of the peroxo / peroxy group O atoms, which have OS of -1, and thus will gain electrons to have an OS of -2. The sulfur atoms are neither oxidized nor reduced, in redox reactions involving the peroxodisulfate(VI) ion S2O8 2-.
Q6. I didn't actually work out the answer (so ignore my earlier post about the answer being N2O4 vs 2N2O). The underlying concept is this : once you've got the moles of NH3 generated, you can find the moles of N atoms present per mole of the oxide. Presumably this would be 2. So the answer is hence N2O or N2O4. Once you got the moles of H2 in excess, you then know the moles of H2 used up, and you can then determine how many moles of O atoms are present, per mole of the oxide. Presumably this would be 1, hence the answer is N2O. If the value is 4 (ie. 4 moles of O atoms present to combine with the H2 gas reacted), then the answer is N2O4.
The candidate is expected to deduce that hydrogenating oxides will ultimately generate H2O as the only stable oxygen & hydrogen containing product, and hydrogenating inorganic nitrogen compounds will produce NH3 as the only stable nitrogen & hydrogen containing product.
Originally posted by gohby:My apologies, UltimaOnline. Thank you for your help and I will take note of that!
Q3, is there an efficient schematic way of ensuring that all different functional isomers have been drawn, before one compares the intermolecular forces of attraction to compare boiling points?
Q4: Doesn't all the compound have the ability to form a mono-substituted compound with HCl too?
Q5: Could you write down the equation to illustrate?
For Q6, the answer is N2O, what's the full equation, and where do I find the general equation for the reaction between nitrogen oxide and hydrogen?
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Thank you very much for your lucid explanation, UltimaOnline, they're of great help.
For Q3, in order to analyse if CH3CH2CH2CHO has a higher bp than its functional group isomer, I believe we are looking between the bp of aldehyde and ketone right? I've read up that ketone has a higher bp than its corresponding aldehyde as it is more polar (since the presence of 2 electron-donating alkyl groups (instead of 1 in aldehydes) will intensify the δ- charge on the C=O bond). However, why doesn't the electron-donating alkyl groups lessen the δ+ charge on the C instead, thereby lessening the polarity of the C=O bond in ketones (as compared to aldehydes)?
Another qn:
Beryllium forms covalent compounds which are said to be electorn deficient. In many ways, beryllium resembles aluminium.
BeF2 reacts with NAF
Since BeF2 only has 4 valence electrons, it needs 2 dative bonds right? How can the ionic compound of NaF form dative bonds with BeF2?
Once again, thank you! :)
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Originally posted by gohby:
Thank you very much for your lucid explanation, UltimaOnline, they're of great help.
For Q3, in order to analyse if CH3CH2CH2CHO has a higher bp than its functional group isomer, I believe we are looking between the bp of aldehyde and ketone right? I've read up that ketone has a higher bp than its corresponding aldehyde as it is more polar (since the presence of 2 electron-donating alkyl groups (instead of 1 in aldehydes) will intensify the δ- charge on the C=O bond). However, why doesn't the electron-donating alkyl groups lessen the δ+ charge on the C instead, thereby lessening the polarity of the C=O bond in ketones (as compared to aldehydes)?
Another qn:
Beryllium forms covalent compounds which are said to be electorn deficient. In many ways, beryllium resembles aluminium.
BeF2 reacts with NAF
Since BeF2 only has 4 valence electrons, it needs 2 dative bonds right? How can the ionic compound of NaF form dative bonds with BeF2?
Once again, thank you! :)
Ketones are less reactive than aldehydes due to both electronics (as you mentioned) and sterics. For 'A' levels, aldehydes and ketones with the same number of electrons and same molecular size are considered to have similar boiling points, which are only due to permanent dipole - permanent dipole Keesom van der Waals (and no hydrogen bonds involved).The solution to the riddle is : because two moles of NaF react with one mole of BeF2.
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Originally posted by UltimaOnline:
Ketones are less reactive than aldehydes due to both electronics (as you mentioned) and sterics. For 'A' levels, aldehydes and ketones with the same number of electrons and same molecular size are considered to have similar boiling points, which are only due to permanent dipole - permanent dipole Keesom van der Waals (and no hydrogen bonds involved).The solution to the riddle is : because two moles of NaF react with one mole of BeF2.
Thank you for your reply, so for an MCQ question like
Q3: Which substances have a higher boiling point than its functional group isomer?
1.
2.
3. CH3CH2CH2CHO,how do students know if option 3 is correct/wrong?
As for the BeF2 question, how does the NaF bond with BeF2?