This is from the 2010 paper.
Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.
When 0.10 cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of gas(measured at 298K) was produced.
When 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH(aq) caused a further reaction in gas volume of 109cm^3 (measured at 298K).
Use these data to calculate values for x and y in the molecular formula CxHyOH for J.
Someone pls help to answer the above question...(:
Originally posted by jlheartsfly:This is from the 2010 paper.
Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.
When 0.10 cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of gas(measured at 298K) was produced.
When 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH(aq) caused a further reaction in gas volume of 109cm^3 (measured at 298K).
Use these data to calculate values for x and y in the molecular formula CxHyOH for J.
Someone pls help to answer the above question...(:
2010 'A' Level H2 Chemistry Paper 3.
Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.
When 0.10 cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of gas(measured at 298K) was produced.
When 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH(aq) caused a further reaction in gas volume of 109cm^3 (measured at 298K).
Use these data to calculate values for x and y in the molecular formula CxHyOH for J.
TPJC Chem teachers' solution :
In the first experiment, the gas evolved in the reaction with Na metal is hydrogen.
Moles of H2(g) = 10.9/24000 = 0.000454 mol
Þ moles of liquid J present = 0.000454 x 2 = 0.000908 mol
In the 2nd experiment, moles of CO2 evolved = 109/24 000 = 0.00454 mol
Þ moles CO2 : moles J (CxHyOH)
5 : 1 Þ x = 5
Moles of O2 used in the reaction = (109 + 54.4)/24000 = 0.00681 mol
Note that since CO2 was formed, it ‘masked’ the O2 that was used up
Þ Reduction in volume of gas = (vol O2 used up) – (vol CO2 formed)
Þ vol O2 used up = (reduction in vol) + (vol CO2 formed)
moles O2 : moles J (CxHyOH)
7.5 : 1
Þ C5HyOH + 15/2 O2 ® 5CO2 + ((y+1)/2)H2O
16 = 10 + (y+1)/2 Þ y = 11
ii) J is an alcohol that reacts with K2Cr2O7 Þ J is either primary or secondary alcohol.
J can be dehydrated to alkene Þ J contains the -CH2-CH(OH)- structure
Since alkene K is oxidized into ethanoic acid and propanone
Þ K is CH3CH=C(CH3)2
Þ J is CH3CH(OH)CH(CH3)2
SAJC Chem teacher's solution :
When 0.10 cm3 of J is added to sodium metal, hydrogen gas is evolved.
CxHyOH + Na <-----> CxHyO-Na+ + (1/2) H2
No. of moles of hydrogen gas = 10.9 / 24000 = 4.542 x 10-4 mol
No. of moles of alcohol J = 2 x no. of moles of hydrogen = 9.084 x 10-4 mol
When NaOH is added to J, the carbon dioxide gas produced from the combustion is being absorbed.
No. of moles of CO2 = 10.9 / 24000 = 4.542 x 10-3 mol
Since all the C in CO2 is produced from combustion of J, we can find the value of x when finding out the ratio of the no. of moles of CO2 produced to that of alcohol J present initially.
No. of moles of CO2 / No. of moles of J = 4.542 x 10-3 / 9.084 x 10-4 = 5
Therefore, x = 5
When J is combusted in an excess of oxygen, the reduction in the volume of gas is due to the consumption of oxygen to produce water. This reduction in volume of gas is the difference between the volume of oxygen gas consumed in combustion and the volume of carbon dioxide gas produced.
Volume of gas being consumed= 54.4 cm3
No. of moles of gas being consumed = 54.4 / 24000 = 2.267 x 10-3 mol
Taking the ratio between the no. of moles of gas consumed and the no. of moles of alcohol J,
Ratio = 2.267 x 10-3 / 9.084 x 10-4 = 2.5 mol
Since we have determined x = 5 above, we can simplify the equation to,
C5HyOH (l) + ((19+y)/4) O2 (g) 5CO2 (g) + ((y+1)/2) H2O (l)
Hence, difference in no. of moles of gases on combustion = ((19+y)/4) – 5 = 2.5
y = 11
Alcohol J : C5H11OH
Observation : J reacts with acidified K2Cr2O7, J has molecular formula C5H11OH
Deduction : J is oxidised [1/2]. J is a primary or secondary alcohol.
Observation : J dehydrated to form K, K reacts with hot KMnO4 to give equimolar mixture of ethanoic acid and propanone.
Observation : Oxidative cleavage
K has the structure (CH3)2C= and J has the structure CH(CH3)=
K is 2-methylbut-2-ene.
J is 3-methylbutan-2-ol.
thanks so much for your answer! anyways can i know if i would have to know how to draw the secondary structures (alpha helix and beta pleated sheets) of the protein for A levels?
Originally posted by jlheartsfly:thanks so much for your answer! anyways can i know if i would have to know how to draw the secondary structures (alpha helix and beta pleated sheets) of the protein for A levels?
Yes you need to know how to draw both alpha helix and beta pleated secondary structures for H2 Chem.
Hi sorry to trouble again. can you explain the difference in TM and the s-block metals in their reactivity in water to me?thanks.
Originally posted by jlheartsfly:Hi sorry to trouble again. can you explain the difference in TM and the s-block metals in their reactivity in water to me?thanks.
S-block metals are generally more reactive with water (with the reactivity increasing down the group) and react in a redox reaction with water (the metal being oxidized and water being reduced).
D-block metals are generally unreactive with water, as the effective nuclear charge for the valence electrons are stronger (in turn because additional electrons enter into the d orbitals, which are more diffused than s orbitals and hence provide poorer shielding*), and therefore the oxidation potentials for oxidizing d-block metal atoms into their cationic states, are generally not as positive (ie. not as reactive) as that of s-block metals.
* Sometimes you need to say "d-orbitals provide poor shielding", while at other times you need to say "d-orbitals provide effective shielding", much like The Accountant Joke**, depending on the question. For instance, the former statement would be used to explain why the d-block metals generally have a smaller radius than s-block metals (d orbitals provide poorer shielding compared to s and p orbitals of the same shell). The latter statement would be used to explain why the atomic and ionization energies of the d-block metals remain similar (d orbitals provide more effective shielding compared to the s and p orbitals of the next shell). Why do we care about the next shell? Because unlike s-block metals of (say) period 4, for the d-block metals in period 4, together with the additional proton (as you go from left to right of the Periodic Table), the additional electron goes into the 3-d orbital rather than the 4-d orbital (for a period 4 s-block metal, the additional electron goes into the 4s orbital).
** The Accountant Joke :
A businessman was interviewing job applications for the position of manager of a large division. He quickly devised a test for choosing the most suitable candidate. He simply asked each applicant this question, "What is two plus two?"
The first interviewee was a journalist. His answer was, "Twenty-two".
The second was a social worker. She said, "I don't know the answer but I'm very glad that we had the opportunity to discuss it."
The third applicant was an engineer. He pulled out a slide rule and came up with an answer "somewhere between 3.999 and 4.001."
Next came an attorney. He stated that "in the case of Jenkins vs. the Department of the Treasury, two plus two was proven to be four."
Finally, the businessman interviewed an accountant. When he asked him what two plus two was, the accountant got up from his chair, went over to the door, closed it, came back and sat down. Leaning across the desk, he said in a low voice, "How much do you want it to be, boss?" The accountant got the job.
Source :
http://www.accountingcoursesonline.com/accountantjokes.html