Chemistry: Post your doubts here!

[KateCake]

Hey are you also writing chem tomorrow?

yes you too right?

 

[MdaraGoat]

yes you too right?

Yes

 

[Rid@Ahmed]

Answer is C but How?
Anyone?

 

[Mazeeen mohd]

Guys can anyone explain me what's the difference between pi and sigma bonds?

 

[Kiara_S]

Re: All Chemistry help here!! Stuck somewhere? Ask here!

tonnes of CHEMISTRY WORKSHHETS
http://www.chemactive.com/a_level_chemi ... tions.html

So I wanted to ask a question from periodicity... It says in the book that the oxidation number increases across a period because the elements in period 3 till phosphorus use all their electrons in bonding with chlorine. And their oxidation numbers are positive as chlorine has the highest electronegatavity...
I didn't get a word of this! What does this mean??

 

[Ayesha abid]

no matter how much i study chemistry i just cant get the hang of MCQs,what should i do?

 

[Zash Riyash]

Can someone explain why the answer is C ?

 

[KashishV]

Can someone explain why the answer is C ?

If you draw the propene molecule, it will help. so the molecule looks something like this: CH2=CH-CH3. Now we have Br2 which is (i) Br-Br and HBr which is (ii) H-Br. In case (i) the nucleophile will attack. the double bond will break and one extra bond will be available on C1 and one more on C2. a Br ion can attach to each one of these free bonds, thus forming option D. in (ii) H can bond to first free bond or the second. Br will attach to the one left. Through this, the process forms option A and B.

Option C is impossible because it implies that both the bromine ions went to the first carbon, but that is not possible because if you revisit molecular bonding, a pi bond is basically an unhybridised p orbital from each atom interacting. SO the breaking of a double bond will automatically free up one unhybridised p orbital on either C atom. You just can't have option C. Others are possible, as described above.

Hope this helps

 

[Zash Riyash]

If you draw the propene molecule, it will help. so the molecule looks something like this: CH2=CH-CH3. Now we have Br2 which is (i) Br-Br and HBr which is (ii) H-Br. In case (i) the nucleophile will attack. the double bond will break and one extra bond will be available on C1 and one more on C2. a Br ion can attach to each one of these free bonds, thus forming option D. in (ii) H can bond to first free bond or the second. Br will attach to the one left. Through this, the process forms option A and B.

Option C is impossible because it implies that both the bromine ions went to the first carbon, but that is not possible because if you revisit molecular bonding, a pi bond is basically an unhybridised p orbital from each atom interacting. SO the breaking of a double bond will automatically free up one unhybridised p orbital on either C atom. You just can't have option C. Others are possible, as described above.

Hope this helps

Thank you so much!!!!

 

[ZubairXD]

I got 1/root of 2

How?

 

[KashishV]

How?

yeah you're actually supposed to get 1/2

 

[raeee111]

Can anyone pls explain why is the is answer C? Thank you

 

[KashishV]

Can anyone pls explain why is the is answer C? Thank you

So the graph is characteristics for an ENDO thermic reaction which is deltaH +ve value. For the reverse reaction, energy needs to be supplied to increase the potential energy of products to the energy level of intermediate(which has a greater potential energy). So this also requires an INPUT of energy (i.e, +ve sign). This makes C the right answer.

 

[EXPERTS]

any one ? plz help

 

[KashishV]

any one ? plz help
View attachment 65104

Is the answer C? It's a very simple working, Mr=200 is extra information given to throw you off.
number of molecules= Avagadro's number* number of moles=6.02*10^23 x 3*10^(-21) = 1806.

 

[A*(a*)]

Is there any whatsapp group for this group and if yes can someone kindly add me over there
Should I tell my whatsapp number

 

[KashishV]

Hey guys, anyone wrote paper 52 today? How'd it go?

 

[Ramlaa]

Hi can any one explain this??
You have 100.000g of a 60% by mass sulfuric acid mixture of a calculated density of 1.370gcm–3. Calculate the mass of water you would need to add to the 100.000g in order to give a final calculated density of 1.154 g cm–3.?

 

[Jean Lim]

OK so first you need to know the reactions of the period 3 oxides with water. AL2O3 does not react so optionA b are wrong. Next we react 2NaOH (which is a basic oxide and the product of reaction between oxide and H2O) with corresponding acidic oxide & balance the equations.
NOTE The Q states"neutralised by exactly 1mol....so D is correct as it balances perfectly with 2NaOH.

Al2O3 does react with acid and alkali since it is amphoteric.
a) when Y is added to water H3PO4 will form and if you balance the equation when reacted with Al2O3, you wont get one mole for each of the reactants.
b) when SO3 is added to water H2SO4 will form and if you balance the equation when reacted with Al2O3, you wont get one mole for each of the reactants.
c) same goes to when Na2O reacts with H2SO4. (wont get one mole)
d) the equation is Na2O + H2SO4 = Na2SO4 + H2O. X and Y both have one mole.
Therefore the ans is D.

 

[TJ Lok]

compressing at constant temp will never liquify any gas... you have to cool the cylinder as well for liquifying... else the cylinder will blast away....

Hey brother, I don’t think this statement is correct. Gas can be liquified by increasing pressure at constant T. Just look at phase diagram.
Reason 2 is wrong is bcause using the relationship of PV=PV, volume of gas at 6000k pa should be 19cm3.
However, the measure volume is 20.5cm3.
This shows that the gas is less compressible than expected (what we call the positive deviation from ideal gas behaviour)
Option 2 cannot be valid. Gas can be liquify if they are compressible i.e. stronger intermolecular attraction force (a negative deviation from ideal gas behavior should be observed.
Meaning volume should be LOWER than ideality when gas is liquified not higher.