Chemistry: Post your doubts here!

[MiniSacBall]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s05_qp_1.pdf

Q34, Why B and not C Explain? ,
I chose C, because from the graph the reaction still seems to be going on, and it did reach equilibrium, that why i P is not necessary less, it could be.
Of course the reaction is proceeding at a lower rate
And since the reaction couldn't attain equilibrium it could be because the products are being S continuously removed.

 

[DeViL gURl B)]

hey guys,
could someone please help me to get easy detailed notes for all the A LEVEL chapters. if with questions, well appreciated.
thank you
PLEASE URGENT HELP NEEDED! >_<

 

[robinhoodmustafa]

hey guys,
could someone please help me to get easy detailed notes for all the A LEVEL chapters. if with questions, well appreciated.
thank you
PLEASE URGENT HELP NEEDED! >_<

https://www.xtremepapers.com/commun...as-chemistry-notes-more-chapters-added.34599/

 

[robinhoodmustafa]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_1.pdf

Q34, Why B and not C Explain? ,
I chose C, because from the graph the reaction still seems to be going on, and it did reach equilibrium, that why i P is not necessary less, it could be.
Of course the reaction is proceeding at a lower rate
And since the reaction couldn't attain equilibrium it could be because the products are being S continuously removed.

Um! bro

This is what I think after thinking about it.

Experiment 1, shows a curve with decreasing rate until the curve becomes flat (and reaches equilibrium)
so if we look at the options available... Lets take 1) removing P . Well it can be because if you remove an amount of P (from reactant side) , so according to Le Chatlier the equilibrium is disturbed and it would oppose the change so in result the reaction would favor the left hand side (reactant side) because less concentration on reactant so to balance the rate back up , product would convert back to reactant to equalize the concentration.



I just dont know how experiment 2 makes sense with Catalyst
3 seems to get along with Ex 2 but...

I have posted this question on a chemistry forum. *waiting for someone to reply*

 

[Metanoia]

Um! bro

This is what I think after thinking about it.

Experiment 1, shows a curve with decreasing rate until the curve becomes flat (and reaches equilibrium)
so if we look at the options available... Lets take 1) removing P . Well it can be because if you remove an amount of P (from reactant side) , so according to Le Chatlier the equilibrium is disturbed and it would oppose the change so in result the reaction would favor the left hand side (reactant side) because less concentration on reactant so to balance the rate back up , product would convert back to reactant to equalize the concentration.
View attachment 50272


I just dont know how experiment 2 makes sense with Catalyst
3 seems to get along with Ex 2 but...

I have posted this question on a chemistry forum. *waiting for someone to reply*

Statement 1: (TRUE) Less P means less concentration of reactants, so slower rate of production of S and R.
Statement 2: (TRUE) A different catalyst in expt 2 might lower the activation energy to a lesser extent than expt 1, so the rate of achieving eqm and production of S and R is slower.
Statement 3: (FALSE) If S is being removed, the eqm would shift to the right to produce the products (S and R) at a faster rate.

 

[princess Anu]

Why is 2 wrong?View attachment 49827

What's wrong in E and D?View attachment 49817

Why is it C and not B?View attachment 49816

If somebody could explain these please?

 

[Metanoia]

If somebody could explain these please?

Hydrogen bromide: (Organic halogen compound formed) HBr (g) will react and the Br will substitute the OH group.
Alkaline iodine: (No organic halogen compound formed). This is the iodoform test to test for the presence (appearance of yellow ppt) of certain structures of alcohol and carbonyls. There should not be a reaction.
Ethanoyl chloride : (No organic halogen compound formed)Acyl chloride would react with the alcohol groups and amine groups to form ester and amide bonds respectively. HCl (g) would be released.

What's wrong in E and D?View attachment 49817

For D, refluxing with NaOH (aq) would hydrolyze the ester bond -OCOCH3
For E, the phenol group is not reactive enough to react with CH3COOH (and conc H2SO4). CH3COCl is required to form ester bonds with phenol groups.

Why is it C and not B,anyone??View attachment 49816

The clue to refer to is that the compound would form a weakly alkaline solution.
The methoxide CH3CO- will form a strong alkaline solution.
CH3CO- is a much stronger base than H2O, it will readily accept hydrogen atoms from water molecules.
CH3CO- + H2O --> CH3OH + OH-

 

[princess Anu]

Q29 as well please

 

[Metanoia]

Q29 as well please

Rather than me giving away the answer straight away, perhaps you can draw out what you think are the intermediate products, then I'll try to correct you from there.

 

[princess Anu]

Okay,
I think D is incorrect, since the no if c atoms are already seven and Cn will make them 8 which is not the case with the given compound
I think c is
correct for the carbon can replace Br and can make COoH on a corner wheres in the middle it can't make four bonds
and I assume CHBr changes to COOH only and the intermediates will have Br replaced by CN. Correct me if my reasoning is wrong!

Rather than me giving away the answer straight away, perhaps you can draw out what you think are the intermediate products, then I'll try to correct you from there.

..

 

[princess Anu]

Why is F-F bond weaker than Cl-Cl bond? where as H-F bond is stronger than H-Cl bond?

 

[MafaldaC]

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 ÅãC. 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

 

[princess Anu]

Why not C

 

[Wkhan860]

Why is F-F bond weaker than Cl-Cl bond? where as H-F bond is stronger than H-Cl bond?

F-F is an exception in the general trend of decreasing energy of bonds
The reason is tht Fluorine is an exception due to its extremely small size. The F-F bond length is so short that the lone pairs of electrons on the fluorine atoms repel each other and weakens the F-F bond.

 

[Metanoia]

Okay,
I think D is incorrect, since the no if c atoms are already seven and Cn will make them 8 which is not the case with the given compound
I think c is
correct for the carbon can replace Br and can make COoH on a corner wheres in the middle it can't make four bonds
and I assume CHBr changes to COOH only and the intermediates will have Br replaced by CN. Correct me if my reasoning is wrong!


..

You are on the right track by first counting the number of carbons. Actually, both C AND D start with seven carbons, so we eliminate both C and D.
Therefore, we focus on A and B.
Step 1: Replace the Br atom with CN
Step 2: hydrolyze the CN to COOH

Why is F-F bond weaker than Cl-Cl bond? where as H-F bond is stronger than H-Cl bond?

Generally, the smaller the atoms orbitals that overlap, the stronger the bonds. This trend is seem by the bond strength decreasing from Cl-Cl>Br-Br>I-I.
H-F being stronger than H-Cl can also be explained in that manner.

However although F atoms are really small and thus expected to have the most overlap it is actually an exception to the general rule. This is because the two nuclei (each 9 protons) will repel each other if there are too close together (lone pair of electrons will also repel each other). Thus the bond length ends up longer than expected.

 

[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 ÅãC. 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 released by the reaction = mass of water x specific heat capacity of water x temperature rise = 50 x 4.2 x 2.5 = 525 kJ

We have 0.00875 moles of NaOH and 0.00875 moles of HCl, so we can take either reactant as limiting reactant.

heat of reaction = heat/moles of limiting reactant = - 525 kJ/0.00875 mol = - 60 kJ mol–1

TIP: If we need to guess an option without calculation, the heat of neutralization of a strong acid and strong alkali should be in the region of ~58 kJ mol–1.

 

[Metanoia]

Why not C

C is obtained by reacting methylbenzene and bromine in UV light (free radical substitution).
Halogen carrier will attach bromine to the ring, and not the side chain.

 

[MafaldaC]

heat released by the reaction = mass of water x specific heat capacity of water x temperature rise = 50 x 4.2 x 2.5 = 525 kJ

We have 0.00875 moles of NaOH and 0.00875 moles of HCl, so we can take either reactant as limiting reactant.

heat of reaction = heat/moles of limiting reactant = - 525 kJ/0.00875 mol = - 60 kJ mol–1

TIP: If we need to guess an option without calculation, the heat of neutralization of a strong acid and strong alkali should be in the region of ~58 kJ mol–1.

thank you!! and this one
10 The equilibrium constant, Kc, for the reaction H2(g) + I2(g) 2HI(g), is 60 at 450 ÅãC.

What is the number of moles of hydrogen iodide in equilibrium with 2 mol of hydrogen and 0.3 mol

of iodine at 450 ÅãC?

paper may june 2014 11 question 10

 

[Metanoia]

thank you!! and this one
10 The equilibrium constant, Kc, for the reaction H2(g) + I2(g) 2HI(g), is 60 at 450 ÅãC.

What is the number of moles of hydrogen iodide in equilibrium with 2 mol of hydrogen and 0.3 mol

of iodine at 450 ÅãC?

paper may june 2014 11 question 10

Form an equation between the Kc and the respective concentrations,then solve for the concentration of HI

 

[princess Anu]

Answer is A .
I seriously need to get a stylus
View attachment 50002

See..
You know that group no in periodic table represents the no of valence electron. so X+ shows group 1 , X3+ shows group 3 such as
so Just find the difference between two ionization energy and compare
You can see that the first one is 510 but the second one is way too high 3800. If energy shows slight increase it is in the same shell but second ionization energy (3800) shows it is in an another shell...
I.E always start by pulling the valence electrons.. and 510 shows its the valence shell with 1 electron so it is fact that it is in GROUP 1

But isn't the second and third I.E's difference the greatest? and so, the 3rd electron can be said to be in a new shell not the 2nd one?