• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

Messages
1,394
Reaction score
1,377
Points
173
i am afraid u are wrong aqueous bromine vapours or gaseous bromine vapours both are brown or orange... gaseous bromine is never purple but yeah aqueous iodine is orange
o right... but then can u plz answer that question sweetninjah asked?? :) and tag me in the answer too :)
 
Messages
1,160
Reaction score
3,858
Points
273
SiCl4 is a covalent compound. The bonding in AlCl3 is more covalent than it is ionic. When the question asks which is ionic which SOME covalent character it is asking about compound where the main bond is ionic.

The question says ionic with some covalent character. this only happens when
- the charge on the cation is large ( Mg > Na )​
- the cation is small ( Mg < Na )​
- the size of the anion is large​
since both anion are the same, and Mg is smaller in size and has a greater charge, it has a higher covalent character then Na.​
The main bonding in MgCl2 and NaCl are ionic bond and MgCl2 has more covalent tendency so the answer is B.​

AlCl3 DOESNOT show Ionic Character , it is an Exception , AlCl3 further Form a Dimer with Al3Cl6
Na is Highly Electropositive in nature ( tendency to give electron , because of being in group 1 and Period 3 ) while Chorine is Highly Electronegative so it is Ionic
SICl4 is purely covalent
and justification for MgCl2 is correct ..

Ionic or Covalent bonding depends on difference in electronegativity between two BONDED atoms
If the Difference is less greater than 1.7 than it is Ionic else covalent ..
flourine have highest electronagativity of 4
 
Messages
1,160
Reaction score
3,858
Points
273
What happens to the volatility of halogens and of hydrogen halides as we go down the group??!!!

volatility decreases down the group coz vanderwaals forces increases as we move down the group (moving down the group the number of electrons increases) so due to these attracted forces volatility trend decreases moving down the group ..........

stability of hydrogen halides decreases down the group ........as we move down the group the bond lenght increases means bond energy decreases therefore less energy is needed to break the bonds so thermal stability is decreased
 
Messages
2,824
Reaction score
3,730
Points
273
volatility decreases down the group coz vanderwaals forces increases as we move down the group (moving down the group the number of electrons increases) so due to these attracted forces volatility trend decreases moving down the group ..........

stability of hydrogen halides decreases down the group ........as we move down the group the bond lenght increases means bond energy decreases therefore less energy is needed to break the bonds so thermal stability is decreased
yes
about halogens, i thought the same thing.. but i had a question in a test, of second MCQ type.. it said that astatine is below Iodine, what r the likely properties of At? one of the sentences was that At forms diatomic molecules which dissociate more readily than Chlorine (1st sentence) i thought its wrong and so automatically marked it C but answer was wrong.. does this has to do with volatility?
 
Messages
2,824
Reaction score
3,730
Points
273
Hassan Ali Abid
one more question.. is At (below I) a reducing agent? i read in one of the notes but according to the question mentioned above, it is not (it was pne of the wrong sentences that said "It is a powerful reducing agent" ...or is it a reducing agent but not a powerful one?
 
Messages
2,824
Reaction score
3,730
Points
273
AlCl3 DOESNOT show Ionic Character , it is an Exception , AlCl3 further Form a Dimer with Al3Cl6
Na is Highly Electropositive in nature ( tendency to give electron , because of being in group 1 and Period 3 ) while Chorine is Highly Electronegative so it is Ionic
SICl4 is purely covalent
and justification for MgCl2 is correct ..

Ionic or Covalent bonding depends on difference in electronegativity between two BONDED atoms
If the Difference is less greater than 1.7 than it is Ionic else covalent ..
flourine have highest electronagativity of 4
*Al2O6 not Al3O6
:D
 
Messages
681
Reaction score
1,731
Points
153
In this question and in similar questions you should consider the outer most shell, ONLY.

Now see in P the outer most electron has 6 electrons
And in Q it has 1

As mentioned in question they form Ionic compound.
So transfer of electron occur.
And this is obviously from Q to P .

P needs 2 more electrons to form full outershell ( tht is 8 in total)
And Q can only give 1
thus, we need 2 from Q to satisfy one P
So the formula will be
PQ2 which is the answer A
 
Messages
249
Reaction score
419
Points
73
In this question and in similar questions you should consider the outer most shell, ONLY.

Now see in P the outer most electron has 6 electrons
And in Q it has 1

As mentioned in question they form Ionic compound.
So transfer of electron occur.
And this is obviously from Q to P .

P needs 2 more electrons to form full outershell ( tht is 8 in total)
And Q can only give 1
thus, we need 2 from Q to satisfy one P
So the formula wilL be
PQ2 which is the answer A

ThanKs ALOT!!!
This really helped :)
JAZAK'ALLAH
 
Messages
887
Reaction score
466
Points
73
when u get the answers to the rest of the questions... plz tag me 2 bcuz they were mighty DIFICULT O__O x.x
for q8 mj2010
just do the mole ratio: Mass/Relative molecular mass
and check which satisfies the equation
2.92/RMM= 5.287/RMM
like for example A
2.92/137 =5.287/(137+35.5*2)
l.h.s does not equal R.H.S

for q 12
If u acidify the pool, the equlibrium would shift to RHS to make more OH-
so conc. of HOCl would be high
adding hydroxide ions would actually shift equilibrium to lhs thus reducing the yield
since second equation is not reversible so any change in chloride or oxygen would have no effect to produce hocl.
 
Messages
681
Reaction score
1,731
Points
153
For question 2,
Chlorine has 2 isotopes 35 and 37
And they are asking how many peaks are formed with Cl2+

Now, look at the formula again, it 2 of Cl...
so consider the possible arrangements
35-35
35-37 (which is same as 37-35)
37-37

So the number of peaks shall be 3, answer is B

For question 10, (according to me way)
Now you have an equation
1 -> 2 ( tht is the equation showing the mole only )
50% of the reactant dissociate.
Since mole ratio is 1:2 so for every 50% dissociation 2 x 50% is formed
thus 50% give 100% and the total thing is 100 + 50 = 150

to find Kp u need to know the formula
P product/ P reactant (P is partial pressure)

Now to find P of product u take ratio of product that is 100% and divide by total percentage that is 150%
this gives 2/3 as answer

Now for the P of reactant u take ratio of reactant that is 50% and divide by total percentage that is 150%
this gives 1/3 as answer

Now place them in formulea

(2/3) / (1/3)

BUT this is wrong, since i forgot to square the product. ( Remember the mole ratio ! )

so it should be

(2/3) ^2/ (1/3)

Do this in you calculator, and you get 4/3.

I am sure there is an easy way. But sorry that is the way I do it.
 
Messages
887
Reaction score
466
Points
73
when u get the answers to the rest of the questions... plz tag me 2 bcuz they were mighty DIFICULT O__O x.x
for w10 q7
At constant T, PV is also constant. In a mixture, partial pressures add to be the total pressure.
the total volume will be 15 dm^3. This is what you get by adding the volume of gas 1 to the volume of gas 2. Since the flasks are connected, the overall volume will be the sum of the volume of the two flasks. To get the pressures, I just used P1V1 = P2V2 and figured out the new pressure of each gas.

Gas X; (12kPa)(5 dm^3) = P2(15 dm^3). P2 = 4 kPa.

Gas Y; (6kPa)(10 dm^3) = P2(15 dm^3). P2 = 4 kPa.

So the partial pressure of each gas is 4 kPa. The total pressure is the sum of the partial pressures, so you just add the two partial pressures and get 8 kPa.

for w10 q12
thermal stability increases down the group so strontrium would have higher value of delta T .
change in ph depends on dissolving of these metal hydroxides into water or in other words
reaction of metal hydroxides with water.
u know solubility of metal hydroxides increases down the group so strontrium would also give higher value of delta ph

for q13
oxidation number of iodine in HIO is -1
in I2 is 0
and in HIO3 is +5
now balancing left hand side with r.h.s
-1*5=0+5
both equal 5 signs are different but what is important is the number of ions in both side
so u now know that
m=5
since there are 5 iodine atoms in left.h.s there should be 5 iodine atoms in r.h.s
so n=2
p=1

for q18
bromine is soluble in nonpolar solvents.
bromine will vaporize significantly as it has only weak van der walls so If you have a large volume of bromine liquid, it will begin to vaporize at room temperature and pressure.
any element having mass more than 32 is considered denser than air so bromine is denser than air

for q40
SN2 reactions take place in one step and take place only if halogen is attached to that carbon which is primary.
SN1 reactions take place in two steps and take place only if halogen is attached to tertiary carbon .
the diagram shows reactions taking place in two steps so only SN1 reactions are correct in this question.

only reaction 1 is SN1
so D is correct
 
Messages
887
Reaction score
466
Points
73
when u get the answers to the rest of the questions... plz tag me 2 bcuz they were mighty DIFICULT O__O x.x
for W11 q30
there are three different isomers of alcohol possible
1. 1-butanol
2. 2-butanol
3 trimethyl ethanol.

for 1 u will get ch3ch2=ch2
for 2 u will get ch3ch=chch3
for 3 u will get CH3)2-C=CH2
so only 2 wil get u cis trans isomers so only 1 is correct which is option A.
 
Messages
887
Reaction score
466
Points
73
for q11 s11
Experimentally, moles of C present = 3x10^-3
Molar mass of AlxCy = 27x + 12y
Moles of AlxCy present = sample mass / molar mass = 0.144 / (27x + 12y)
look if we want to find the number moles of element in a compound we multiply the number of moles of that compound and number of moles of that element present in it for example
H20
if i have 2 moles of h20 how many moles of hydrogen are there simple 2*2
so
By stoichiometry, moles of C present = moles of (AlxCy) x (y) = y (0.144 / (27x + 12y))
Which can be equated to 3x10^-3 found earlier.
Simplifying the mathematical expression, we obtain x/y = 4/3.
Hence the answer is C, aluminium carbide.

for q23
In chemistry, mole fraction x is a way of expressing the composition of a mixture hence the mole fractions are
50% or 1/2 are ethene,
25% or 1/4 are methane,
and 25% or 1/4 are propene.
we need whole number so multiply by 4
we get 200 percent and 100 percent 100percent respectively.
this means 2 moles of ethene and 1,1 moles of methane and propene

for 40
 
Messages
887
Reaction score
466
Points
73
for q11 s11
Experimentally, moles of C present = 3x10^-3
Molar mass of AlxCy = 27x + 12y
Moles of AlxCy present = sample mass / molar mass = 0.144 / (27x + 12y)
look if we want to find the number moles of element in a compound we multiply the number of moles of that compound and number of moles of that element present in it for example
H20
if i have 2 moles of h20 how many moles of hydrogen are there simple 2*2
so
By stoichiometry, moles of C present = moles of (AlxCy) x (y) = y (0.144 / (27x + 12y))
Which can be equated to 3x10^-3 found earlier.
Simplifying the mathematical expression, we obtain x/y = 4/3.
Hence the answer is C, aluminium carbide.

for q23
In chemistry, mole fraction x is a way of expressing the composition of a mixture hence the mole fractions are
50% or 1/2 are ethene,
25% or 1/4 are methane,
and 25% or 1/4 are propene.
we need whole number so multiply by 4
we get 200 percent and 100 percent 100percent respectively.
this means 2 moles of ethene and 1,1 moles of methane and propene

for 40
Ca + 2CH3COOH ----> Ca2+ + 2CH3COO- + H2.
as u know that ch3coo- u need two ch3coo to balance Ca
so u get 2*(CH3COO)Ca
c4h6O4
use first equation to find the empirical formula for other two equations u will find only first one is correct.
 
Messages
957
Reaction score
6,644
Points
503
Ca + 2CH3COOH ----> Ca2+ + 2CH3COO- + H2.
as u know that ch3coo- u need two ch3coo to balance Ca
so u get 2*(CH3COO)Ca
c4h6O4
use first equation to find the empirical formula for other two equations u will find only first one is correct.
thanks a lot snowbrood and daredevil;):)snowbrood u r a genius!!
Sorry actually i dont know how 2 tag.
 
Top