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Chemistry: Post your doubts here!

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can anyone solve this please and also let me know the reason for it ? please its urgent :)

Apart from B, all C-atoms are coplanar. For C-atoms to be coplanar, there must be less than or equal 3 C-atoms bonded to each other. As soon as there are 4, it changes shape and in most cases become tetrahedral. Hoe this helps.
 
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Ok I have a problems in a questions of Equilibria:
1. When calculating Kp, what are the total number of moles? The number of moles at equilibrium?

2. Finding moles at equilibrium from given data: CO2 + H2 <==> CO + H2O

A mixture containing 0.5 mol of Co2 and H2, with 0.2 moles of H2O and CO was placed in 1.0 dm^3 of container and was allowed to come to equilibrium at 1200k. KC at 1200k is 1.44,

Calculate the moles at equilibrium of each of the substances at 1200k.
Answer: CO2 and H2 = 0.32, CO and H2o= 0.38

Please explain in full detail. Thanks!




3. N2O4 <===> 2NO2 1.0 mole of N2O4 was allowed to reach equilibrium at 400k. At equilibrium the partial pressure of N2O4 was found to be 0.15 atm,
Givent that Kp for this equilibrium is 48 atm, calculate the partial pressure for No2 in the equilibrium mixture.

Answer: 2.68 atm

Please explain in full detail and show all the working.you would help a ton. thanks in advance!
 
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Could someone please answer these

1) discuss the use if the periodic table in the classification of elements as metals and non metals.
2) write down the two ion half equations for the reaction of the IO3 (-1) anion with the iodine ion [I (-1)] in acidic solution
 
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Ok I have a problems in a questions of Equilibria:
1. When calculating Kp, what are the total number of moles? The number of moles at equilibrium?

2. Finding moles at equilibrium from given data: CO2 + H2 <==> CO + H2O

A mixture containing 0.5 mol of Co2 and H2, with 0.2 moles of H2O and CO was placed in 1.0 dm^3 of container and was allowed to come to equilibrium at 1200k. KC at 1200k is 1.44,

Calculate the moles at equilibrium of each of the substances at 1200k.
Answer: CO2 and H2 = 0.32, CO and H2o= 0.38

Please explain in full detail. Thanks!




3. N2O4 <===> 2NO2 1.0 mole of N2O4 was allowed to reach equilibrium at 400k. At equilibrium the partial pressure of N2O4 was found to be 0.15 atm,
Givent that Kp for this equilibrium is 48 atm, calculate the partial pressure for No2 in the equilibrium mixture.

Answer: 2.68 atm

Please explain in full detail and show all the working.you would help a ton. thanks in advance!



CO2 H2 <==> CO + H2O
0.5 : 0.5 == 0.2 : 0.2

Kc = [CO][H2O]/[CO2][H2] = 1.44

[0.2+x][0.2+x]/[0.5-x][0.5-x] = 1.44
(Since v = 1dm, I wont bother with the division step)

0.2+x/0.5-x = 1.2
0.2+x=0.6-1.2x
2.2x = 0.4
x=0.18

Moles of CO + H2O at eq: 0.2+0.18 = 0.38
Moles of CO2 + H2 at eq: 0.5-0.18 = 0.32

Get it?


About the Kp questions, sorry man! Always sucked at those :(
 
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Ok I have a problems in a questions of Equilibria:
1. When calculating Kp, what are the total number of moles? The number of moles at equilibrium?

2. Finding moles at equilibrium from given data: CO2 + H2 <==> CO + H2O

A mixture containing 0.5 mol of Co2 and H2, with 0.2 moles of H2O and CO was placed in 1.0 dm^3 of container and was allowed to come to equilibrium at 1200k. KC at 1200k is 1.44,

Calculate the moles at equilibrium of each of the substances at 1200k.
Answer: CO2 and H2 = 0.32, CO and H2o= 0.38

Please explain in full detail. Thanks!




3. N2O4 <===> 2NO2 1.0 mole of N2O4 was allowed to reach equilibrium at 400k. At equilibrium the partial pressure of N2O4 was found to be 0.15 atm,
Givent that Kp for this equilibrium is 48 atm, calculate the partial pressure for No2 in the equilibrium mixture.

Answer: 2.68 atm

Please explain in full detail and show all the working.you would help a ton. thanks in advance!


For question 3 :
this question is straightforward,recall the Kp equation and insert products in numerator and reactants in denominator,also take care about raising the pressure of the reactant and product to its moles as stated in the equation. So here since NO2 has 2 moles you square it !
Kp = [NO2]^2 /[N2O4]
48 = [NO2]^2/0.15

Now just solve this out and you get the answer for NO2 :)
 
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can anyone send me a level full Chemistry revision notes and worksheet plzzzzzzz
i need it plzz some one help me if you have it send it to [email protected] and if you need any other help in AICT and maths i will help you
but plzzz send me notes guy
 
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Thank you guys for helping out. i have another question of energetics!! Please solve it and explain here.
Its for OCt/NOV 2001 P1.

Water gas is an equimolar gas mixture of H2 and CO, It is formed when Steam is blown through white hot coke in following reaction:

H2O +C ===> H2+ CO

Another gas used in industry is methane.
Enthalpies of combustion is given as:
CH4 = -890
H2= -242

(i) Use the following data to calculate the volume of methane measured at rtp, required to produce 1MJ of heat on burning. (this one was easy and is done)
(ii) Calculate the volume of water gas measured at RTP required to produce same amount of heat. (Answer = 91. This one i cant do)


Please HELPPPP!!!!!!!!!!!! AND solve it for me, Dont tell me to do this and that. Thanks
 
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A 50 cm3 sample of water containing dissolved calcium sulphate was passed through the ionexchange resin. Each calcium ion in the sample was exchanged for two hydrogen ions. The
resulting acidic solution collected in the flask required 25 cm3 of 1.0 × 10–2 mol dm–3 potassium hydroxide for complete neutralisation.
What was the concentration of the calcium sulphate in the original sample?
A 2.5 × 10–3moldm–3
B 1.0 × 10–2moldm–3
C 2.0 × 10–2moldm–3
D 4.0 × 10–2moldm–3

Answer is A
 
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Ca++ and 2H+ -- resin ------------> Ca++--resin and 2H+


50ml is the original volume of solution (we keep this in mind).

(0.01 mole KOH/Liter)*0.025 L = 2.5*10^-4 moles of H+ neutralized

(1 mole Ca++/2 moles H+)*(2.5*10^-4 mole H+) = 1.25*10^-4 mole of Ca++

1.25*10^-4mole CaSO4/0.050L =2.5*10^-3 molar Ca++
A 50 cm3 sample of water containing dissolved calcium sulphate was passed through the ionexchange resin. Each calcium ion in the sample was exchanged for two hydrogen ions. The
resulting acidic solution collected in the flask required 25 cm3 of 1.0 × 10–2 mol dm–3 potassium hydroxide for complete neutralisation.
What was the concentration of the calcium sulphate in the original sample?
A 2.5 × 10–3moldm–3
B 1.0 × 10–2moldm–3
C 2.0 × 10–2moldm–3
D 4.0 × 10–2moldm–3

Answer is A
 
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does Cr2O7/H+ (heat under reflux) react with (CH3)3COH ? What actually is (CH3)3COH? Is it alcohol or an ester? Please help !
 
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