Chemistry P1 Help

[beacon_of_light]

Nov06: q5 pi bond is formed by sideways overlapp of P orbitals only!!!

 

[beacon_of_light]

Nov09///q3: angle X is 104.5 just like in water due to 2 lone pairs..y is 109 whilez is 107 due to 1 lone pair....so arrange them nw...

 

[beacon_of_light]

june09:q29 the alcohol must be a secondary or primary as the product suggests but D has -OH attached to tertiary alcohol !! if u don't get this try dehydrating all the options and come up to the right answer!!!

 

[flyingcrayons]

THANK YOU! *BOWS* <3

 

[flyingcrayons]

Beacon, could you help out with my Physics too? Under the topic "Physics P2 Help". Sorry for the bother

 

[sizbeauty]

hey can u plz help me???
june08
q2,6,15,27 and
nov08
q1,5(why is it D and nt A)
q7(why nt hydrogen since it has less mass thn all oder givn optionz??)
q14!!!!
plz help out!!!

 

[beacon_of_light]

june08: q2

15+30+15 =60
mass of N in 100g = 15/60 x 100 = 25g

mass of N in 14g = 25x14/100 = 3.5g

moles of N = 3.5/28=0.125

n=cv c= n/v = 0.125/5 = 0.03mol/dm^3

q15: equation is...

CaCO3 ------> CaO + CO2
Total mass of lime = 1000+200 = 1200
moles of lime = 1200/100=12
moles of CO2 = 12

mass of CO2 = 12x44= 527

q27:the alcohol before dehydration must be a tertiary one...So if you add OH and H to double bond in D, you get a tertiary alcohol.

 

[beacon_of_light]

nov08:
q1: equation.......FeTiO3 ------> FeO + TiO2

now carry out ur calculations...
q5:
hydrogen bonds are intermolecular forces which only effect physical properties of compounds like mp or bp....and enthalpy change of vapourisation is when these bonds will break and convert liquid to a gas....

while A refers to intramolecular forces( forces b/w molecules or atoms) which result in chemical changes....

q7: Hydrogen is diatomic gas...while He is monoatomic ..so obviously it will have weak intermolecular forces!!!

q14: ammonia reacts with SO2 to form a sulphite....and the condition is until just alkaline...not excess ......

 

[kugs]

Can anyone explain these

j05
13, 17, 18

n05
12, 17, 21, 24

j06
8, 10

 

[British Council]

@ Beacon: dude please help me with June 2005: MCQ 2,21, 25, 33 AND 34............... PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

 

[beacon_of_light]

q2: see that both equations result in Nitrogen...calculate it like this...

2NaN3 give 3 moles of N
1 mole will give 3/2 moles of N

2NaN3 give 2 moles of Na
1mole will give 1 mole of Na...

now in the second eq:
10Na give 1 mole of N
1 mole of Na will give 1/10 moles of N2....
Total moles 1/10 + 3/2 =1.6

q21: the reaction is isomerisation just c before n after reaction no of carbon, chlorine and fluorine is same!!!

q25: in step 1, OH oxidises to CHO...while in step 3 secondary alcohol oxidises to ketone....

q33: hess's cycle is used to calculate enthalpy changes which are impossible to calculate directly....so 1 and 2 are impossible...du think burning C in hydrogen will give us methane...lol...while 3 is possible since if u remember we can calculate this enthalpy using calorimeter.....

q34: rate of reaction reduces or less product is formed....1 & 2 cause this.....since a different catalyst makes the reaction proceed by a diff. route...which has a dif. activation eenergy...

 

[lotto]

dude ur supposed to be the BRITISH Council!!! y dont u freaking use ur contacts to get the papers??!!! !! LMAO!!

 

[lotto]

well i had problemz in Q25 from the above mentioned!!! thanks beacon of light!! btw cool name!!

 

[beacon_of_light]

Can anyone explain these

j05
13, 17, 18

n05
12, 17, 21, 24

j06
8, 10

june05:
q13: since Al has a similar size to that of beryllium, so its electronegativity is similar to that of beryllium...

q17: D....its a fact that sulphuric acid corodes limestone work....

n05: q12...i too couldn't do that question an amazingly read its examiner report, this question was withdrawn from P1!!!
q17 is not in the syllabus...
q21: when one R group and one H is attached to each carbon of double bond...u get the corresponding COOH...here it would be CH3COOH!!!
q24: its butan-2-ol...the alkenes are but-1-ene ....and cis-but-2-ene...and trans-but-2-ene...

j06: q8: since the reaction is 2 step reaction so obviously two curves or two peaks would be formed on the graph..nw since second reaction is faster as it is exothermic...it will have lower activation energy and therefore the second peak will be low...

q10: ive explained it in detail in another topic buh lemme brief it out!!!

assume the initial moles of N2O4 as 1....since 50% dissociated to NO2, so 50% are 0.5 moles dissociated and 1 moles of NO2 formed....

nw calculate the partial pressurre of each gas and then finally Kp!

 

[lotto]

dude nice!! i still cant believe i guessed no5 q 12 and got it rite miraculously!!! where az had problemz in q17!!!

oh and btw the j06 q10 could u tel where i could find it!! the link to ur post would be very helpful!!! and Thanks ur great help!!!

 

[beacon_of_light]

N204 ---> 2NO2
initial moles: 1 0
moles reacted: 0.5 0
equilibrium moles
or moles left: ( 1-0.5) = 0.5 1( since 1 mole of n2o4 forms 2 moles of no2 so 0.5 moles will form 0.5 x 2)

Nw find the partial pressure of each gas by dalton's law of partial pressure ka formula...
PP would be 0.5/1.5 1/1.5

so Kp = (1/1.5)^2 / (0.5/1.5) = 4/3 n u get the answer!!!

 

[kugs]

Thank you

 

[mz23]

please help in june 07, 27

 

[mz23]

also, november 06, 31 and 32 and 39

 

[beacon_of_light]

please help in june 07, 27

K2Cr2O7 would oxidise both alcohol and aldehyde...

DNPH and tollen's reagent would only react with aldehyde...sodium wuld only react with alcohol ...