Chemistry: Post your doubts here!

[SpadeAce61]

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!

 

[Snowysangel]

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

 

[AbbbbY]

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

 

[scouserlfc]

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

 

[johnjase0]

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

 

[AbbbbY]

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

http://www.scribd.com/doc/140353885/A-Level-Chemistry-Factsheets

Give it a read. Excellent stuff.

 

[SpadeAce61]

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

 

[Alice123]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_42.pdf
Q7-b (i) why is it -CH3 and not R-CH2-R for peak at 1.26ppm?

someone please help!!!

 

[Suchal Riaz]

Anyone please tell me which books are best for chemistry and why they are best.

 

[Dr.MMM]

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

 

[Jaspreet Dhaliwal]

Hey Guys, sorry ive got a lot of questions to ask, mcq if you guys dont mind answering them really struggling with as chemistry!
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Question 2
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf Question 13
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_12.pdf Question 1
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_11.pdf Question 5
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_11.pdf Question 2
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Question 31
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf Question 33
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_12.pdf Question 40

 

[Jaspreet Dhaliwal]

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

 

[ShreeyaBeatz]

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

 

[Jaspreet Dhaliwal]

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

It's 2-methylpropan-2-ol so yes it will be oxidised, as its an alcohol

 

[Vounn Rose]

Hey Guys, sorry ive got a lot of questions to ask, mcq if you guys dont mind answering them really struggling with as chemistry!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Question 2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_11.pdf Question 13
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf Question 1
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_11.pdf Question 5
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf Question 2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf Question 31
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_11.pdf Question 33
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf Question 40

I will try.

Your 1st question: The answer is "D", simply divide the number of moles of oxygen by the volume of bleach (in dm^3 unit), as the number of moles of oxygen correspond to the number of moles of NaClO, provided that hydrogen peroxide is in excess

2nd: To answer this, try to write out a balanced equation for the reaction to determine the ratio of the amount of reactant to that of product, but it is ok even if you choose not to write out. Just have to know that the aluminium will end up forming aluminium chloride (AlCl3) in a complete reaction. So 1 mol of aluminium will form 1 mol of AlCl3 with 3 mol of chlorine. For hydrochloric acid (HCl), 1 mol of HCl contains 1 mol of chlorine, so 3 mol of HCl is needed to form 1 mol of AlCl3 to completely react with 1 mol of aluminium.

In this case, 0.02 mol of aluminium is used, so 0.02 x 3 mol of HCl is needed, which is 0.06 mol of HCl. Here it says the concentration of HCl is 2.00 mol dm-3, so to provide 0.06 mol of HCl, the volume needed is (0.06 divided by 2.00) dm3. Convert the volume in dm3 to cm3 and you will get the answer which is "C", 30 cm3.

3rd: To answer this question, one way is to find out how many number of moles of carbon is contained in 35.2 gram of carbon dioxide and how many moles of hydrogen are there in 14.4 gram of water. Do this by dividing 35.2 by molar mass of carbon dioxide, whose quotient is about 0.765 mol, and dividing 14.4 by molar mass of water, whose quotient is about 0.8 mol. This is not yet over. 0.765 mol of carbon dioxide means 0.765 mol of carbon, whereas 0.8 mol of water means 1.6 mol of hydrogen atoms.

With the information so far, it is informed that 0.2 mol of hydrocarbon contains 0.765 mol of carbon and 1.6 mol of hydrogen atoms. The formula of hydrocarbon can be obtained by knowing how many carbon atoms and hydrogen atoms are in there in one molecule of hydrocarbon. No. of carbon atoms = 0.765 / 0.2 = 3.825 rounded off to 4. No. of hydrogen atoms = 1.6 / 0.2 = 8. So the formula of hydrocarbon is C4H8, a butane. The answer is D.

4th: First, know the mass of nickel in the coin, which is 10.0 g x 20/100 = 2g. Second, divide the mass of nickel by its molar mass to get its no. of moles, which is 2/58.7 = 0.0341 mol. Third, multiply the mol by Avogadro's Constant, which is 0.0341 x 6.02 x 10^23 = 2.05 x 10^22. The answer is A.

5th: If one Pb 4+ ion is reduced to Pb 2+ ion, it requires 2 Br - ions to do that. So 6.98 g of lead (IV) chloride contains 0.02 mol of lead (IV) chloride molecules （By dividing the mass by its molar mass). So it contains 0.02 mol of lead or therefore lead (IV) ions. So it needs 0.02 x 2 = 0.04 mol of bromide ions. 0.04 mol of bromide ions will be reduced to form 0.04 mol of bromine atoms. So 0.04 mol of bromine atoms has mass of 0.04 x its molar mass = 3.196 gram. The answer is C.

6th: It asks about the properties of chlorine atoms, which are always the same in whichever isotope of chlorine. So the isotopic mass is definitely not the answer because isotopic mass is different between different isotopes. As different isotopes of one element have different number of neutrons, so is the nucleon number, so this answer is out too. All is left is the atomic radius, which is the same for all isotopes of one element, because the number of protons and electrons is the same for every isotope, so is the structure of orbitals. This is just a simplified explanation, there are more reasons to account for the fact. So the answer is D.

7th: Simply determine how many moles of zinc are there in each statement, and how many moles of HCl in 0.1 dm3 of 1.00 mol dm-3 of the acid, and how many moles of hydrogen gas molecules (H2) in 9.6 dm3 of the gas ( the molar volume of gas varies in different temperature, but in this case try both 22.5 dm3 and 24.0 dm3 ). Then compare the number of moles of the reactant and that of product, using the balanced chemical equation. The answer is D, as only the first statement is correct.

8th: If 1 mol of monomers is fully polymerised, the most possible number of polymers is 0.5 mol if every polymer formed is a dimer, whereas the least possible number of polymers is 1/(6.02 x 10^23) mol if all monomers bond together into one (Note: 1 mol of particles consists of 6.02 x 10^23 of the particles). So the possible number of polymers formed is an integer lying in the range between the maximum and minimum numbers. So the answer is C, statement 2 and 3.

Hope they help, and correct me for any point mistaken.

 

[snowbrood]

http://www.scribd.com/doc/140353885/A-Level-Chemistry-Factsheets

Give it a read. Excellent stuff.

yar can u upload this document scribd does not give the access to download this

 

[Jinosupreme]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf

Help guys! :0

Q2, 30, 33.

2B
30D
33A

 

[aminah_jn]

Can someone plz xplain this question:
A student mixed 25cm3 of 0.10mol dm–3 sodium hydroxide solution with 25cm3 of 0.10mol dm–3 hydrochloric acid and noted a temperature rise of 2.5°C.
What is the enthalpy change of the reaction per mole of NaOH?
A –209 kJ mol–1
B –104.5 kJ mol–1
C –209 J mol–1
D –522.5 J mol–1

 

[AbbbbY]

yar can u upload this document scribd does not give the access to download this

I've downloaded several files from scribd by uploading stuff (if you don't have anythign, uploading bogus files and taking them down)

Anyhow. From phone right now. WIll upload on dropbox or 4shared etc and send.

 

[AbbbbY]

yar can u upload this document scribd does not give the access to download this

I've downloaded several files from scribd by uploading stuff (if you don't have anythign, uploading bogus files and taking them down)

Anyhow. From phone right now. WIll upload on dropbox or 4shared etc and send.