O Lv CHEM
-
1 to 13 is solved. please start from q14 onwards. tytyyyy
1)portion of acidified potassium manganate VII solution added into sample of solution X . X decolourised the potassium manganate solution . resulting solution was brown
An orangic solvent Y added to mixture above and left to stand. colourless layer of Y turned violent . What conclusion can be drawn about X ?
a) it is a reducing agent and contain iodide ions
b) it is a reducing agent and contain bromide ions.
2) sample of waste gas passed ino alkaline solution of a compound called pyrograllol . remaining gas burns in air .
gas removed gas remained
1)co2 h2
2)NO2 o2
3) 2 experiment setup to demo diffusion of gases are done. gases in each porous pot are N2 and O2 . In the first exp. gas intro into beaker is CO2 while in the second and Seperate expt , H2 is introduced.
*diagram shows a porous pot in a beaker with the gas introduced(CO2/H2) connected to a U - shaped tube with the side connected to the porous pot and water lvl of P and the other end exposed to atmpsphere pressure in which the water level is labelled Q .
wht changes can be seen in water lvl P and Q in both expt ?
4) Thallium is a metallic elemetn and exits in to isotopic form which have mass number 203 and 205 respectivel, 10g of thallium reacts with excess chlorine to give 15.2g of compound ticl3 . based on the data. calculate Mr or thallium
a) 203.4
b)203.6
c)204.5
d)204.8
my working
10/x=15.2/x+106.5
15.2x=10x+106.5
5.2x= 106.5
x=204.8
but i keep thinking need to do sth with the isotope ?
5)Y is a compound with valencey of 5.
hence, the formula of compound formed between calcium and Y is ?
a) CaY
b) CaY2
c) Ca2Y3
d) Ca3Y2
6) element X has electronic structure 2,8,5 and element Y has electronic strucutre 2,8,7 . what is the likely formula of a compound containing X and Y only?
a) XY3
b)X3Y
c) X5Y7
d) X7Y5
For qn 5 and 6 is there a quik method instead of drawing it out ? ?
7) Metal Z and its compound undergo following reactions
1 X+2Hcl -> ZCl2 + h2
2 ZCo3 -> ZO + CO2
3 ZO+ CO -> Z + CO2
Metal Z could be
calcium
magnesium
copper
zinc
from the above information given i could only be sure that it isnt copper.
for the breakdown of carbonate i cant take away calcium as only sodium and potassium canot be broken down by heat from what i have learn. for the CO reaction i nvr learn b4.
8) X has proton num of 9 (2,7)
Y has proton num of 7 (2,5)
the compound formed by X and Y is
a) total num of 6 shared electrons
b) molecular forumula XY3
c) molecular formula X5Y
9) Which of the folloowing has the most number of electron ?
a)magnesium ion
b) molecular of HCL
10)
11) Z is a non-metal . Oxide of Z has a high melting point. Z is probably
a) carbon
b) silicon
c) posophorus
d) sulfur .
is silicon the answer as it has the biggest negative oxidation state? eg. 2,8,4 . 4+/4- ? ?
12) W has a gaint covalent structure, which 1 of the following description is true ?
a) constituent particles in W are atoms
b) W exsists as discrete molecules
can sum1 explain why answer is A ?
13) which property of Nitrogen gas explained by the covelent bond in its molecules ?
a) acidity of Nitrogen oxides
b) low reactivity of nitrogen
c) low b.p
i choose c . why canot ? x.x
14) 4dm3 of co2 contains
a) 2.4x10^24 atoms
b)3x10^24 atoms
c)10^23 atoms
d) 1.2x10^23 atoms
ans sheet is B , but i choose C , i think ans sheet wrong, can sum1 x2 check ?
15) num. of atoms present in 34g of NH3 is
a) 1.2x10^24
b) 4.8x10^24
ans sheet says B but i choose A . i think ans sheet is wrong, can sum1 check too ?
16) Zinc nitrate decomposes on strong heating acc. to eqn below.
2Zn(NO3)2.6H20 -> ZnO(s)+ 4NO2(g) + O2(g) + 6H20 (g)
14.85 gram of crystal heated strong, what is total vol of gas obtained ?
a) 3 dm3
b) 6.6dm3
c)1.2dm3
d) 7.8dm3
ans sheet say A . but i choose B . can sum1 x2 check answer? i think ans sheet wrong -.-
17) which of the following has the same mass as 6x10^23 atoms of copper ?
a) 6x10^23 atoms of iodine.
b) 12x10^23 atoms of magnesium
c) 12x10^23 of sulfur
d)24x10^23 atoms of nitrogen
i wrote C but ans is A . cant figure out.
18) 10g of CaCO3 just neutralises 100cm3 of hcl acc. to eqn below
CaCO3 + 2HCL - > CaCl2+H2O + CO2
what is conc of HCL ?
a)1mol/dm3
b) 2mol/dm3
ans is A but i wrote B . why sia , suspect ans wrong
19) sample of N2 contains same num of atoms as found in 4g of CH4 .. what is mass of N2 gas ?
a) 7g
b) 17.5g
ans sheet is b but i choose a !
20) which of the following substances has the most number of atoms ?
a) 10g nitrogen
b)10g magnesium
since qn nvr say gas or what.. i assume is diatomic gas.
but ans sheet is a . while my answer is b
21) 6,94g of mixture wwith BaCO3 and CaCO3 is heated. after removing CO2 , 4.74g of mixture with CaO and BaO is obtained. Orginal weight of BaCO3 is. .. ..
a) 1.97g
b)3.94g
c) 5.91g
d)3.06g
seriously dunoe how to do this 1 .
22)gas contains 46.2% of carbon and 53.8% of nitrogen . under r.t.p vol and weight is 6dm3 and 13g . the gas is
a) C2N2
b) CN
ans sheet say is A . i chosoe B . cause i got ratio of 1:1
23)
24) m Fe2+(aq) + n H+ (aq) + MnO4-(aq) -> Mn2+ (aq) + p Fe3+ (aq) + q H2O(l)
what is m,n,p,q
a) 2, 4,2, 2
b)3,6,3,3
c)4,8,4,4
d)5,8,5,4
-
Originally posted by yiha093:
1)portion of acidified potassium manganate VII solution added into sample of solution X . X decolourised the potassium manganate solution . resulting solution was brown
An orangic solvent Y added to mixture above and left to stand. colourless layer of Y turned violent . What conclusion can be drawn about X ?
a) it is a reducing agent and contain iodide ions
b) it is a reducing agent and contain bromide ions.
I will guess b
I reacts too slowly with organic solvents if i didn't remember wrong so mostly its Br that will react.
-
Originally posted by yiha093:
1)portion of acidified potassium manganate VII solution added into sample of solution X . X decolourised the potassium manganate solution . resulting solution was brown
An orangic solvent Y added to mixture above and left to stand. colourless layer of Y turned violent . What conclusion can be drawn about X ?
a) it is a reducing agent and contain iodide ions
b) it is a reducing agent and contain bromide ions.
2) sample of waste gas passed ino alkaline solution of a compound called pyrograllol . remaining gas burns in air .
gas removed gas remained
1)co2 h2
2)NO2 o2
3) 2 experiment setup to demo diffusion of gases are done. gases in each porous pot are N2 and O2 . In the first exp. gas intro into beaker is CO2 while in the second and Seperate expt , H2 is introduced.
*diagram shows a porous pot in a beaker with the gas introduced(CO2/H2) connected to a U - shaped tube with the side connected to the porous pot and water lvl of P and the other end exposed to atmpsphere pressure in which the water level is labelled Q .
wht changes can be seen in water lvl P and Q in both expt ?
4) Thallium is a metallic elemetn and exits in to isotopic form which have mass number 203 and 205 respectivel, 10g of thallium reacts with excess chlorine to give 15.2g of compound ticl3 . based on the data. calculate Mr or thallium
a) 203.4
b)203.6
c)204.5
d)204.8
my working
10/x=15.2/x+106.5
15.2x=10x+106.5
5.2x= 106.5
x=204.8
but i keep thinking need to do sth with the isotope ?
5)Y is a compound with valencey of 5.
hence, the formula of compound formed between calcium and Y is ?
a) CaY
b) CaY2
c) Ca2Y3
d) Ca3Y2
6) element X has electronic structure 2,8,5 and element Y has electronic strucutre 2,8,7 . what is the likely formula of a compound containing X and Y only?
a) XY3
b)X3Y
c) X5Y7
d) X7Y5
For qn 5 and 6 is there a quik method instead of drawing it out ? ?
7) Metal Z and its compound undergo following reactions
1 X+2Hcl -> ZCl2 + h2
2 ZCo3 -> ZO + CO2
3 ZO+ CO -> Z + CO2
Metal Z could be
calcium
magnesium
copper
zinc
from the above information given i could only be sure that it isnt copper.
for the breakdown of carbonate i cant take away calcium as only sodium and potassium canot be broken down by heat from what i have learn. for the CO reaction i nvr learn b4.
8) X has proton num of 9 (2,7)
Y has proton num of 7 (2,5)
the compound formed by X and Y is
a) total num of 6 shared electrons
b) molecular forumula XY3
c) molecular formula X5Y
9) Which of the folloowing has the most number of electron ?
a)magnesium ion
b) molecular of HCL
10)
11) Z is a non-metal . Oxide of Z has a high melting point. Z is probably
a) carbon
b) silicon
c) posophorus
d) sulfur .
is silicon the answer as it has the biggest negative oxidation state? eg. 2,8,4 . 4+/4- ? ?
12) W has a gaint covalent structure, which 1 of the following description is true ?
a) constituent particles in W are atoms
b) W exsists as discrete molecules
can sum1 explain why answer is A ?
13) which property of Nitrogen gas explained by the covelent bond in its molecules ?
a) acidity of Nitrogen oxides
b) low reactivity of nitrogen
c) low b.p
i choose c . why canot ? x.x
2) Hmm, O2 is generally the 1 that burns, H2 is quite explosive. Not very sure i would say... Both acidic gases can react with alkaline.
3) P will drop and Q will rise for CO2 since it diffuses slower than N2 and O2 so the 2 gases will diffuse out and increase the pressure thus pushing P down. Its the reverse for H2.
4) Isotope part is just to tell you answer is between 203-205.
Simpler method i feel using more chemistry than maths in ur working
Take 15.2-10g gives you mass of Cl.
From that, calculate moles of the element.
Use mass/moles gives u Mr5) correct
Just use the powers to X. Ca2+ and Y3-, bring the charges to the oppersite elements give u Ca3Y2 if you get what i mean.
6) a (I'm surprise you can't do this)
In this case, you know that X a +5 needs 3 more electrons to form octet, 1 way is to share 3 electrons. Y is a -1 so just share 1 electron. Combine the 2 together gives you a XY3
7) Not too sure but Ca and Mg might too reactive for them to react similar to Na and K since they are also high up in the reactivity series.
8) a. If it was period 3 onwards with the d-orbitals, c can also be correct due to violation of octet structure but in O's i don't think you all learn that.
9) b. Cl itself already have more electrons than Mg
10) a. Only N, P and B can form 3 bonds so B is out. For Y, only P can form 5 bonds due to violation of octet. C and N strictly obeys octet rule and thus will never form 5 bonds.
11) Si is correct. SiO2 which is common sand has a very high melting point. The other 3 forms oxides which are mostly gaseous in nature.
12) Everything is made up of atoms regardless whether its a macro or micro molecule.
13) Nothing wrong with c. Is answer a and c? If a is correct then it could be related to the fact that all the acids are covalently bonded compounds while all the alkalines are generally ionic bonded.
-
QN updated ! 14 to 24 please ! ~.~
-
You post new reply la, i no need scroll so much...
14) My answer would be neither of a-d. I would pick 3X10^23. 10^23 is the number of moles but in terms of atoms, you need X 3 since there are 3 atoms in each molecule of CO2
15) Same explaination as qn 14. Read the question ask for ATOMS not MOLECULES.
16) I got answer similar to yours. Maybe someone else can try.
17) I agree with you
18) Agree with you
19) Similiar with Qn 14 and 15
20) Similar with the top few i believe. Never try it but all qns that say atoms u seem to treat as molecules
21) unsure and lazy to really go think.
22) 6dm3 means 1/4 mole of gas. 1/4 mole of gas weights 13g, 1 mole weighs 52g which is the mass of ans a
23) QN 57 b and 58 a since the mass difference is ratio 1:2
24) You need to calculate number of electrons needed then try to balence the equation. Getting old le, do so many qn feeling tired liao so leaving this to u :p
-
new qn~~ i have 500 mroe to go... xD
*typical electrolysis setup*
1) at 1st, the ammeter showed tat current is flowing. some dilute sulfuric acid added to liquid Z and current flowing greatly reduced.
what could be Z ?
a) sodium hydroxide
b)calcium hydroxide
ans sheet say b but i wrote a , x2 chck please?
2) when i electroyse dilute H2SO4 with plat. electrodes, does the concentration of h2so4 remain the same or decrease ?
3) which of the following solutions would produce bubbles of gas with the ratio 1:2 with plat. electrode ?
a) 1 mol/dm3 AGNO3
b) 5MOL/DM3 KCL
c)1mol/dm3 NA2SO4
d) 5mol/dm3 MGCL2
-
Originally posted by yiha093:
new qn~~ i have 500 mroe to go... xD
*typical electrolysis setup*
1) at 1st, the ammeter showed tat current is flowing. some dilute sulfuric acid added to liquid Z and current flowing greatly reduced.
what could be Z ?
a) sodium hydroxide
b)calcium hydroxide
ans sheet say b but i wrote a , x2 chck please?
2) when i electroyse dilute H2SO4 with plat. electrodes, does the concentration of h2so4 remain the same or decrease ?
3) which of the following solutions would produce bubbles of gas with the ratio 1:2 with plat. electrode ?
a) 1 mol/dm3 AGNO3
b) 5MOL/DM3 KCL
c)1mol/dm3 NA2SO4
d) 5mol/dm3 MGCL2
Wah u not tired ah, i am lo :p
1) Current flow decrease cos reaction with Ca(OH)2 forms CaSO4 which is a solid and thus reduce the numbre of ions in the solution to facilitate electrolysis.
2) The H2SO4 qn i remember very clearly answering before and i think my answer differs from ultimaonline's answer and we had a 'debate' or something over it. Try look at ultimaonline's O's/A's thread and you should find it somewhere :p
3) I suck at such qns so not going to try
-
~ when i electroyse dilute H2SO4 with plat. electrodes, does the concentration of h2so4 remain the same or decrease? ~ Neither. ~ which of the following solutions would produce bubbles of gas with the ratio 1:2 with plat. electrode ? ~ What does a 2:1 ratio remind you of? Find a setup in which water would be preferentially reduced and oxidized at both electrodes.
-
Originally posted by UltimaOnline:
~ when i electroyse dilute H2SO4 with plat. electrodes, does the concentration of h2so4 remain the same or decrease?
Neither.
which of the following solutions would produce bubbles of gas with the ratio 1:2 with plat. electrode ? ~
What does a 2:1 ratio remind you of? Find a setup in which water would be preferentially reduced and oxidized at both electrodes.
Oi dun be lazy, i went through all his 20+ questions, you should too and verify my answers :p There are a few questionss that i am not 100% sure and at least 1 i skip i think waiting for you to answer :)
-
can you explain to me why the concentration of H2SO4 did not decrease? i thought that the hydrogen ions will be discharged. thanks!
-
-
more coming out. stay tuned. LOL
-
ok end of thread. will be updating nearer to O lv.
thx for ur help guys.
confident of 38/40 .
heng heng just 2 marks above A1 . XD
-
~ can you explain to me why the concentration of H2SO4 did not decrease? i thought that the hydrogen ions will be discharged. thanks! ~ That's at the cathode. What gets oxidized ("discharged" means removal of charge, notice the product is a neutral molecule) at the anode is OH- ions, which means in effect water is being electrolysed away as hydrogen gas (at the cathode) and oxygen gas (at the anode). So the sulfuric(VI) acid gets more concentrated as time goes by.
-
~ 21) 6,94g of mixture wwith BaCO3 and CaCO3 is heated. after removing CO2 , 4.74g of mixture with CaO and BaO is obtained. Orginal weight of BaCO3 is? ~ Let x and y be the moles of each carbonate present. Before heating, write an equation in x and y. After heating (ie. removal of CO2), write an equation in x and y (notice that no. of moles of metal oxide = no. of moles of metal carbonate). Solve for x and y using simultaneous equations. ~ m Fe2+(aq) + n H+ (aq) + MnO4-(aq) -> Mn2+ (aq) + p Fe3+ (aq) + q H2O(l) ~ First write reduction half-equation. MnO4- is reduced to Mn2+. Second write oxidation half-equation. Fe2+ is oxidized to Fe3+. Third, balance these half-equations within themselves. To balance oxygens, add water molecules. To balance hydrogens, add protons (H+). (If it were in alkaline conditions, you need to remove all H+ by adding OH- on both sides.) To balance charges, add electrons (e-). Fourth, ensure that the no. of electrons gained (in reduction) is the same as the no. of electrons lost (in oxidation). Do this by multiplying across one or both half equations (eg. lowest common multiple). Fifth, add up both half-equations together to obtain the balanced redox. The electrons cancel out perfectly (if you did everything right). Anything that's common on both sides, cancel them out as well. (Eg. 16 water on LHS and 10 water on RHS, becomes 6 water on LHS and 0 water on RHS). Have a bit of practice with balancing redox equations. It's useful to understand these processes better, even at 'O' levels (where it's not strictly required).
-
Originally posted by UltimaOnline:
~ 21) 6,94g of mixture wwith BaCO3 and CaCO3 is heated. after removing CO2 , 4.74g of mixture with CaO and BaO is obtained. Orginal weight of BaCO3 is?
Let x and y be the moles of each carbonate present. Before heating, write an equation in x and y. After heating (ie. removal of CO2), write an equation in x and y (notice that no. of moles of metal oxide = no. of moles of metal carbonate). Solve for x and y using simultaneous equations.
Thought of simultaneous but lazy to try it out :p
-
ok . will look thru it.