Chemistry: Post your doubts here!

[Metanoia]

My teacher says you can conclude the answer is C simply because Al3+ and C4- can only form Al4C3 based on their valency.

However I think there must be a reason behind the other numbers given, if there is a method which utilises the given information please help. Thanks.

 

[The Sarcastic Retard]

Yes, I got the same answers too. Where did you get these questions from?

http://www.rewardinglearning.org.uk/common/includes/microsite_doc_link.aspx?docid=14322-1

 

[princess Anu]

Hey, I don't think halogenoalkanes even make Hydrogen bonds :/ there is no H attached to Fluorine, Nitrogen Or Oxygen.. ?

Option 2: Halogenoalkanes have less H atoms because they are replaced by halogens, so they form hydrogen bonds less readily.

Option 3: Alkanes cannot undergo addition polymerisation.

Therefore, they are both wrong.

 

[princess Anu]

can somebody help me with M/J 08 Q 6 please

 

[Mathemagical]

Hey, I don't think halogenoalkanes even make Hydrogen bonds :/ there is no H attached to Fluorine, Nitrogen Or Oxygen.. ?

You see, hydrogen bonds are very much an intermolecular force as they are an intramolecular one. This is still a grey area, but for simplicity's sake we can ignore the possibility of it forming hydrogen bonds with water.

 

[Mathemagical]

can somebody help me with M/J 08 Q 6 please

Mass of ice = Volume*Density = 1.00*1 = 1g
Number of moles = 1/18

From here on, apply the formula pV=nRT to calculate V.

 

[Studydayandnight]

A student mixed 25.0 cm3 of 0.350 mol dm–3 sodium hydroxide solution with 25.0 cm3 of 0.350 mol dm–3 hydrochloric acid. The temperature rose by 2.50 C. Assume that no heat was lost to the surroundings.

The final mixture had a specific heat capacity of 4.20 J cm–3K–1.

What is the molar enthalpy change for the reaction?

A –150 kJ mol–1

B –60.0 kJ mol–1

C –30.0 kJ mol–1

D –0.150 kJ mol–1

 

[Metanoia]

A student mixed 25.0 cm3 of 0.350 mol dm–3 sodium hydroxide solution with 25.0 cm3 of 0.350 mol dm–3 hydrochloric acid. The temperature rose by 2.50 C. Assume that no heat was lost to the surroundings.

The final mixture had a specific heat capacity of 4.20 J cm–3K–1.

What is the molar enthalpy change for the reaction?

A –150 kJ mol–1

B –60.0 kJ mol–1

C –30.0 kJ mol–1

D –0.150 kJ mol–1

heat of reaction x moles of water formed = mass of solution heated x temperature change x specific heat capacity of solution

Are you able to figure out the values to use for the following?

moles of water formed =
mass of solution heated =
temperature change =
specific heat capacity of solution =

 

[fantastic girl]

A student mixed 25.0 cm3 of 0.350 mol dm–3 sodium hydroxide solution with 25.0 cm3 of 0.350 mol dm–3 hydrochloric acid. The temperature rose by 2.50 C. Assume that no heat was lost to the surroundings.

The final mixture had a specific heat capacity of 4.20 J cm–3K–1.

What is the molar enthalpy change for the reaction?

A –150 kJ mol–1

B –60.0 kJ mol–1

C –30.0 kJ mol–1

D –0.150 kJ mol–1

Q=VxCxdeltaT
total volume is 25+25 = 50 cm3
= 50 x 4.3 x (2.5)
=525
now we shud find enthalpy change fore one mole
number of moles = vol. x conc.
25/1000 x 0.35 = 8.75x10^-3

525 Joules was for 8.75x10^-3 moles
525 J = 8.75x10^-3 moles
? = 1 mole
525/(8.75x10^-3) = 60000 Joules/mole

i hope u got it

 

[Studydayandnight]

Q=VxCxdeltaT
total volume is 25+25 = 50 cm3
= 50 x 4.3 x (2.5)
=525
now we shud find enthalpy change fore one mole
number of moles = vol. x conc.
25/1000 x 0.35 = 8.75x10^-3

525 Joules was for 8.75x10^-3 moles
525 J = 8.75x10^-3 moles
? = 1 mole
525/(8.75x10^-3) = 60000 Joules/mole

i hope u got it

Thanks!!

 

[nehaoscar]

Option 3 is wrong because it could either be an alcohol or a carboxylic acid. There is no observation which confirms either of these, so we cannot assume that it is an alcohol definitely.

In one of the papers, I remember there being a table asking with which each alcohol and carboxyllic acid reacts with;
and there was i think Na metal and NaHCO3 and i think that it said that the Na metal doesn't react with carboxyllic acid??
So I'm confused on which of the Na metal or compounds react with what respectively (OH and COOH groups)
So could you please summarize on which ones react with which?

 

[Mathemagical]

In one of the papers, I remember there being a table asking with which each alcohol and carboxyllic acid reacts with;
and there was i think Na metal and NaHCO3 and i think that it said that the Na metal doesn't react with carboxyllic acid??
So I'm confused on which of the Na metal or compounds react with what respectively (OH and COOH groups)
So could you please summarize on which ones react with which?

Both alcohols and carboxylic acids react with sodium metal. Follow these links for further information:
Alcohols: http://tinyurl.com/bawcd94
Carboxylic acid: http://tinyurl.com/5r72ms

However, only carboxylic acids react with sodium carbonate in a neutralisation reaction, not alcohols. The above link has further information about this. Hope this helps!

 

[fantastic girl]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_12.pdf
Can someone tell me how to find the answer for Q6 ? :/
A is the answer
is it like we see if the equation is balancing wid each hydrocarbon ? xP
in that case its only A i think.

 

[Mathemagical]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_12.pdf
Can someone tell me how to find the answer for Q6 ? :/
A is the answer
is it like we see if the equation is balancing wid each hydrocarbon ? xP
in that case its only A i think.

Yes, the question is on balancing reactions. Option A is the only hydrocarbon that can balance the reaction.

 

[fantastic girl]

Yes, the question is on balancing reactions. Option A is the only hydrocarbon that can balance the reaction.

Thanks alot
i thought i was being stupid xP
so there's no other way to find out ryt?

 

[nehaoscar]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_12.pdf
Can someone tell me how to find the answer for Q6 ? :/
A is the answer
is it like we see if the equation is balancing wid each hydrocarbon ? xP
in that case its only A i think.

Yes:
The equation will be
Al4C3 + 4NaOH + 4H2O = 4NaAlO2 + 3CH4
This requires a bit of trial and error

firstly writing without balancing will give
Al4C3 + NaOH + H2O = NaAlO2 + CH4

Now pop a 4 in front of NaAlO2 so the Al's balance and a 4 at NaOH so the Na balances and a 4 at H2O so the O's balance too.
Al4C3 + 4NaOH + 4H2O = 4NaAlO2 + ?

Now we see that there are 3C's and 12H's left over
C3H8 is not possible as then H's don't balance
C2H6 is also not possible
So the only option is 3CH4

 

[nehaoscar]

Both alcohols and carboxylic acids react with sodium metal. Follow these links for further information:
Alcohols: http://tinyurl.com/bawcd94
Carboxylic acid: http://tinyurl.com/5r72ms

However, only carboxylic acids react with sodium carbonate in a neutralisation reaction, not alcohols. The above link has further information about this. Hope this helps!

Alright! Thanks!
And which of the two react with NaOH?

 

[Mathemagical]

Alright! Thanks!
And which of the two react with NaOH?

NaOH is basic, so carboxylic acids would react with it. However, alcohols would not.

 

[nehaoscar]

Does an ester react with HCN? If yes then please can you give an example with structural or displayed formula?

 

[Mathemagical]

Does an ester react with HCN? If yes then please can you give an example with structural or displayed formula?

No, they do not react.