Chem p22

[Saad (سعد)]

It was the Enthalpy Change of Reaction, but the Reaction was the Formation of CH3OH.

 

[Amy farvin]

@aaditya menon
are u sure that the empirical formula is C6H8O7??? How did u get it?? I got CHO...

 

[Pals_1010]

It was burning in Chlorine. Its not confirmed that Sulfur burns in a blue flame in Chlorine, I'm 90% sure that Sulfur burns in a yellow flame in a chlorine atmosphere; producing a dark yellow liquid (S2Cl2).

Sodium was more correct, and safer.

I screwed up this question big time. I lost like at least 2 marks here just because of sheer stupidity, even though I knew the correct answers. I wrote 'Light yellow flame' for Aluminum's reaction in Chlorine, even though I could have written 'yellow color fades' (I had that in my mind then) as I wasn't sure of Aluminum's flame color - but I didn't. I made up something entirely new 'light yellow flame'.

Bleh.

Anyways, the correct observations were; white sparks, white solid produced, or yellow color of gas disappears.

Then I turned Aluminum chloride's reaction with water as Basic. >_< I wrote it as Acidic (ph 3), then I crossed it and wrote pH8. >_< How idiotic of me.

Question 1: The table in which the reactions of chlorides with water were asked;- was it worth 3 marks or 6?

Question 2: When writing the pH of SiCl4's reaction with water, I wrote 3-4. In the Book it says 2. Will mine be accepted or not?

Question 3: What was the enthalpy change of formation of Methanol? I got -129 kJ mol-1.

Question 4: How do we know that, one of those organic compounds we were told to draw the structure of; that contained 2 Alcohol groups?

Question 5: What was the empirical formula they asked us to calculate? I got C6H8O7 or something...

1) 6 marks

2) I would go with pH 2 as HCl is produced when SiCl4 reacts with water.

3) -129 kJmol-1

4) You found the no of moles of H atoms in the previous part right?

That means 0.30 g of G releases 6.67 x 10^-3 mol of H atoms with Na
90 g (Mr of G) , which is also 1 mol of G, releases by calculation 2 mol of H atoms with Na

1 mol (6.02 x 10^23 molecules) of G releases (2 x 6.02 x 10 ^23 )atoms of H with Na

1 molecule of G releases hence 2 atoms of H with Na... So it contains 2 OH groups.... This is how I did it anyway... I am not sure about the method though. There may be a shorter one...

5) C6H8O7

 

[aaditya menon]

^^ Are you sure it's -129? Coz i'm pretty sure it's +129. Confirmed it with my friends too

 

[Amy farvin]

did u ppl keep the massES as a % and itself calculated the empirical formula?? coz the mass was in % form.. i made it as a just mass and got CHO.

 

[gary221]

I kinda gt tht...bt i multiplied it by 5( tht ws wrong i think) 2 gt a whole no...

For the empirical formula part:
C was 37.5 %, H was 4.17% and O was 58.3% so,
C:H:O = 37.5/12 : 4.17/1 : 58.3/16
= 3.125 : 4.17 : 3.644 then divide all values by lowest value 3.125 so,
= 1 : 1.334 : 1.166 then multiply sll values by 6 to obtain simplest whole no. ratio
= 6 : 8 : 7

 

[Amy farvin]

u hav probably multiplied it by 6... but u dont hav to do as such simply had to round it up.. is my ans correct wat den...

 

[Pals_1010]

^^ Are you sure it's -129? Coz i'm pretty sure it's +129. Confirmed it with my friends too

Yeah, am pretty sure...

 

[Pals_1010]

u hav probably multiplied it by 6... but u dont hav to do as such simply had to round it up.. is my ans correct wat den...

In fact, I think you had to multiply. You round off your figures only when the difference between the whole number is 0.1 (like 1.1 becomes 1 or 8.9 becomes 9) and it's very very close (7.9888888...) .... In this case, the values were too far to simply round off

 

[gary221]

u hav probably multiplied it by 6... but u dont hav to do as such simply had to round it up.. is my ans correct wat den...

In fact, I think you had to multiply. You round off your figures only when the difference between the whole number is 0.1 (like 1.1 becomes 1 or 8.9 becomes 9) and it's very very close (7.9888888...) .... In this case, the values were too far to simply round off

Yeah..wsnt sure whtr i shud round up the values i gt...or multiply it 2 make a whole no...
2 be honest m nt very sure even now whtr m ans is correct or nt!!!

 

[Pals_1010]

Yeah..wsnt sure whtr i shud round up the values i gt...or multiply it 2 make a whole no...
2 be honest m nt very sure even now whtr m ans is correct or nt!!!

You got which answer?

 

[TahaJamshed]

It was burning in Chlorine. Its not confirmed that Sulfur burns in a blue flame in Chlorine, I'm 90% sure that Sulfur burns in a yellow flame in a chlorine atmosphere; producing a dark yellow liquid (S2Cl2).

Sodium was more correct, and safer.

I screwed up this question big time. I lost like at least 2 marks here just because of sheer stupidity, even though I knew the correct answers. I wrote 'Light yellow flame' for Aluminum's reaction in Chlorine, even though I could have written 'yellow color fades' (I had that in my mind then) as I wasn't sure of Aluminum's flame color - but I didn't. I made up something entirely new 'light yellow flame'.

Bleh.

Anyways, the correct observations were; white sparks, white solid produced, or yellow color of gas disappears.

Then I turned Aluminum chloride's reaction with water as Basic. >_< I wrote it as Acidic (ph 3), then I crossed it and wrote pH8. >_< How idiotic of me.

Question 1: The table in which the reactions of chlorides with water were asked;- was it worth 3 marks or 6?

Question 2: When writing the pH of SiCl4's reaction with water, I wrote 3-4. In the Book it says 2. Will mine be accepted or not?

Question 3: What was the enthalpy change of formation of Methanol? I got -129 kJ mol-1.

Question 4: How do we know that, one of those organic compounds we were told to draw the structure of; that contained 2 Alcohol groups?

Question 5: What was the empirical formula they asked us to calculate? I got C6H8O7 or something...

DUDE aluminium does burn with light yellow flame to give rise to pale yellow aluminium chloride you are correct and you dont loose marks here ..... official revision guide for A levels says this aluminium chloride is yellow and white too both http://en.wikipedia.org/wiki/Aluminium_chloride

 

[gary221]

You got which answer?

C6H8O7...or smth like tht...bt loads of m frnds gt CHO...

 

[Pals_1010]

C6H8O7...or smth like tht...bt loads of m frnds gt CHO...

I got that as well... Let's hope for the best then...

 

[TahaJamshed]

^^ Are you sure it's -129? Coz i'm pretty sure it's +129. Confirmed it with my friends too

i got positive 129 too :\ some of my friends got positive some got negative i dont know which ones correct thought but this is the only part where i will loose marks thank God rest waas pretty good

 

[TahaJamshed]

I got that as well... Let's hope for the best then...

PEOPLE RELAX it was C6H8O7 you can calculate the percentage by mass and they gave the EXACT numbers as in the question i got C6H8O7 too

 

[Amy farvin]

ok watevr ppl gone is gone now leave thinking about this and wasting the tym try hard for the next paper... best of luck..
go to the chem p4 thread post ur doubts and study...
my mom always says for eg: "dont show your hand for the bus which went, warmly welcome the other ones..."

 

[Pals_1010]

ok watevr ppl gone is gone now leave thinking about this and wasting the tym try hard for the next paper... best of luck..
go to the chem p4 thread post ur doubts and study...
my mom always says for eg: "dont show your hand for the bus which went, warmly welcome the other ones..."

Nice saying

 

[TahaJamshed]

1) 6 marks

2) I would go with pH 2 as HCl is produced when SiCl4 reacts with water.

3) -129 kJmol-1

4) You found the no of moles of H atoms in the previous part right?

That means 0.30 g of G releases 6.67 x 10^-3 mol of H atoms with Na
90 g (Mr of G) , which is also 1 mol of G, releases by calculation 2 mol of H atoms with Na

1 mol (6.02 x 10^23 molecules) of G releases (2 x 6.02 x 10 ^23 )atoms of H with Na

1 molecule of G releases hence 2 atoms of H with Na... So it contains 2 OH groups.... This is how I did it anyway... I am not sure about the method though. There may be a shorter one...

5) C6H8O7

LOL you actually calculated the MOLES OF H2 gas the volume of H2 gas was given so you had to change it into number of moles of atoms of H only this way did the ratio of alchol and H atoms came 1:2 meaning two H atoms were realsed per one mole of that compund meaning that each mole had two OH groups so two H atoms were liberated

 

[Amy farvin]

thanx..