Answers To Difficult Questions (2 Viewers)

[joshuajspence]

REASON FOR NOT DOING FLAME TEST:

The reason for not conducting the flame test is the mixing of the cation and anions. Normally when conducting a flame test you use a variety of metallic ions which have a common salt base such as Cl. This allows you to distinguish between the metals as the Cl remains constant in all of them. Since the compounds had a differing anion, NO3, Cl, CO3, there was no way of identifying them correctly.

i agree... i carefully noticed that the question said IDENTIFY these compounds... not DISTINGUISH them...

if you use a flame test on a solution of sodium carbonate, you can deduce that it contains sodium, but not carbonate.

 

[Steth0scope]

My example of a model from Charles Sturt .. i said we did this in class and a risk was that we could clash into each other ...

The example of people dancing at a party can be used to model equilibrium.
People around the room represent reactants, they come together to form products (they join to dance). Sometimes products decompose and form reactants again (dancing partners break up and become single again).

Equilibrium is reached when the doors are closed, so no-one else can enter or leave the room, and the number of couples dancing is constant. At any time, a couple may sit down, as long as another couple replaces them. This represents microscopic changes. Changes are occurring, but the concentration stays constant.

If the doors are opened briefly and people are allowed in or out, (perhaps the footy match is over and more people arrive to celebrate) this change in concentration disturbs the equilibrium and a new equilibrium position is eventually established (with different numbers of people sitting and dancing).

Closing off part of the room, decreasing volume and thus increasing pressure, will disturb the equilibrium and push more people together to make couples and a new equilibrium is established.


Limitation - when people move faster they become tired and slow down, this doesn't happen for particles. Can you think of any other limitations

 

[simplistic]

c₁ V₁ = c₂ V₂

(NOTE: You need pH 2 for c₂, to convert:
pH 2 = 10^-2 = 0.01)

0.1 mol L^-1 x 0.09 L = 0.01 x V₂
V₂ = 0.1 mol L^-1 x 0.09 L / 0.01 mol L^-1
V₂ = 0.9 L
V₂ = 900 mL

I don't know if I'm right ... or it should be 900 mL - 90 mL = 810 mL

coz ur adding it to the 90mL its gna be 810mL

 

[Forbidden.]

coz ur adding it to the 90mL its gna be 810mL

I know, right method but wrong answer.
It's too late to change things now, 0 marks for attempt and wrong answer.

 

[Fish Sauce]

coz ur adding it to the 90mL its gna be 810mL

Yeah that's what I got too.

 

[amyg238]

How's this?


7. Neutralisation is exothermic, thus ΔH must be negative. D is the only graph where ΔH is negative.

Why can't it be A?
I put D first.. then changed it to A at the end. Daaaamn

Actually yeah its D..

 

[Forbidden.]

Why can't it be A?
I put D first.. then changed it to A at the end. Daaaamn

For A, after the enthalpy the system will continue to absorb energy but less than the energy it was given for activation.

For D, after the enthalpy, the system releases more energy than the activation energy given to it.

 

[Kmara2nv]

i put 0.6 ml for the for the h2so4 one ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh i dont no why though loll i full went thru working out and shittt and got 0.6

 

[esther]

dang XD for that sulfur one i compelteli misread
and thought the ppm they were giving us was forthe so2 content
i have no idea wat my answer is though
cant rmemeber
and for the enthalpy multiple choice
LOL totalli forgot how to read that enthalpy graph.
stupidity at its peak (A)

-edit-
damn i loggedin with my sisters and notmine

 

[jayel89]

hey does any1 remember wat the actual question for the adding of KOH to sulfuric acid was?? it was the one after diluting the sulfuric acid and the red cabbage indicator.......???

 

[dietjelly]

does anyone remember the mark value of the chloride ion ppm question? totally screwed it up

 

[Crofty0]

i said it was an initiator because it sets off the chain reaction. come on guys it was Q 15 as if it would be that easy- everyone would have assumed it was a catalyst

 

[jayel89]

does anyone remember the mark value of the chloride ion ppm question? totally screwed it up

2/3 marks............think it was 3

 

[aja152]

hello~

the wikipedia site is very unreliable!
i think for question15: the answers c) initiator
but everywhere i read it seems like most people agree with b) catalyst.

man... i tripped in all of the calculations questions..
i hate calculations....T_T;;

 

[Kmahal1990]

i said it was an initiator because it sets off the chain reaction. come on guys it was Q 15 as if it would be that easy- everyone would have assumed it was a catalyst

I thought it was an initiator too at the start. But i thought of polymerisation, and a initiator works by FORMING radicals by splitting, THEN causing the reaction to happen. In this case, i thought CL- was the catalyst, and the HCFC that the chlorine radical broke off from would be the initiator. On a side not, any reason why it wouldn't be a catalyst?

 

[Armenikum]

Boo to the Sodium Hydrogen Carbonate neutralisation question, i lost at LEAST 3 marks for that.
​

 

[drynxz]

well im fked..hopefully ill gett marks for working..

For the KOH diluting with H2S04 i considered it was diprotic..
everyone got 3 ml...i got 6x 10-3...but i convert that to 60ml -___-"

For the ppm question. i got 0.51 mol/L i converted it to 18050...but then i thought, that number is too big..so i moved it to 18 ppm -____-"

and industrial chem.
misread my 7 for a 2..got a k value of 4 instead of 1.33

 

[yoakim]

Cl radical is an initiator. Just like how an organic peroxide initiates an addition reaction to produce poly.E, in initiation and propagation...

 

[Armenikum]

Cl radical is an initiator. Just like how an organic peroxide initiates an addition reaction to produce poly.E, in initiation and propagation...

LDPE only though, right?

 

[yoakim]

Yeah, I think?.

Kmahal1990, from my knowledge, a catalyst is a thing that affects both the reactant and product side. From what I saw in the Cl/O3 reactions was that the activation energy was not reduced (the equations didn't show that) and the reaction rate on both sides of the equation was not increased. I'm not positive.

It's funny how we're all trying to justify our claims and reason them with our minute 'stage 6' knowledge. Ah well.