Predictions for Chemistry 2014 HSC? (9 Viewers)

[Queenroot]

D

 

[enigma_1]

haha do the question, it's a multiple choice.

Lel
I hate doing these questions for practice, they waste so much time. If you think the answers are wrong they probably are.

 

[faisalabdul16]

D

Yeah i also got D, however answers say A for some reason...

 

[Queenroot]

Yeah i also got D, however answers say A for some reason...

Maybe they just meant without changing the sign in front of the potential. Who knows, the HSC is dumb. Is it a HSC question?

 

[faisalabdul16]

Maybe they just meant without changing the sign in front of the potential. Who knows, the HSC is dumb. Is it a HSC question?

Nope it was a question from 2007 Independent paper

 

[Queenroot]

Nope it was a question from 2007 Independent paper

I guess they fucked up

 

[faisalabdul16]

2M Nitric Acid.

 

[QZP]

2M Nitric Acid.

Shoulda added some bases for funs. OMG IM back from the BOS trials I feel so good to be in the chemistry realm again

 

[enigma_1]

Shoulda added some bases for funs. OMG IM back from the BOS trials I feel so good to be in the chemistry realm again

how was it????

 

[QZP]

produce ester: let's make ethyl butanoate

1) Add 30 ml Ethanol and 20 ml butanoic acid to a round bottomed flask. Excess of the ethanol is added coz it's cheaper #cheapos

2) Add 5ml of concentrated sulphuric acid (as a dehydrating agent to shift the equilibrium in the forward direction --> increas yield of ester AND it's a catalyst so it increases the rate of the reaction making it go faster) and add some ceramic boiling chips (anti bumping granules to allow the even heating of the liquid and to increase surface area for vaporisation to occur on)

3) Attach a condenser to the the round bottomed flask with the reactants and stuff in it (used because the system is in equilibrium and if the reactants evaporate then like the equilibrium will shift in the reverse direction and reduce yield of ester which is bad). The round bottomed flask is placed in a water bath to allow even heating of water BUT a better thing to use is the heat plate because the reactants are volatile. I think in hsc we should write that we used a heat plate lol. cheap schools.

4) Anyway next put the Bunsen Burner on top of a heat mat and put it under the reaction vessel. (or use the heat plate)

5) Pass the cooling water through the condenser so it condenses the reactants, preventing evaporation and preventing the equilibrium from shifting to the left. Turn on Bunsen burner and the reaction occurs for half an hour at 100 degrees Celsius.

6) then after 30 mins, turn off the Bunsen burner and let the cooling water pass through for a while. after it's cooled, take the round bottomed flask and pour the ester mixture into a separating funnel. 20 ml of NAHCO3 was added to the separaring funnel, this is because NaHCO3 is amphiprotic and it acts as a base :. neutralising the excess butanoic acid which may remain unused in the reaction. The mixture was shaken and left to stand. The lower aqueous layer was drained off.

7) then add 20ml of DISTILLED WATER to the separating funnel. Since water is a polar substance, it would dissolve the excess polar ethanol if there is any because likes dissolve likes. Again shake it and drain off the lower aqueous layer leaving only the ester and you've successfully purified the ester ethyl butanoate (I hope )

Pretend I drew the diagram and the equilibrium equation for refluxing

k done

now mark it

Hi guys serious question regarding all prac Q's: How do you know the volumes of liquids to add? Do you make it up??

how was it????

destroyed

 

[faisalabdul16]

Hi guys serious question regarding all prac Q's: How do you know the volumes of liquids to add? Do you make it up??



destroyed

You make up suitable volumes.

Do not Say add 1234234234L of ethanol.

 

[SuchSmallHands]

Hi guys serious question regarding all prac Q's: How do you know the volumes of liquids to add? Do you make it up??

For esterification, you have to have more of the alkanol than the alkanoic acid. For some of the other pracs, you use small amounts because you're working with noxious gasses/carcinogens/both (eg the bromine water prac). It's mostly the ratio you have to get right, most of our pracs don't have a set amount of each reactant you have to use, but often you need to have more of reactant x than reactant y.

 

[faisalabdul16]

Solutions containing nickel (II) ions were analysed by AAS. A standard solution of

5 ppm nickel had an absorbance of 0.200. A second solution of unknown

concentration was found to have an absorbance of 0.500.

100 mL of this second solution was reacted with excess sodium carbonate solution.

The precipitate formed was weighed and dried.

What mass of precipitate formed?

(A) 2.5 x 10^-3

(B) 1.3 x 10^-3

(C) 1.7 g

(D) 2.5 g

 

[QZP]

For esterification, you have to have more of the alkanol than the alkanoic acid. For some of the other pracs, you use small amounts because you're working with noxious gasses/carcinogens/both (eg the bromine water prac). It's mostly the ratio you have to get right, most of our pracs don't have a set amount of each reactant you have to use, but often you need to have more of reactant x than reactant y.

Why?? :S

 

[enigma_1]

Why?? :S

because alkanol is cheaper than alkanoic acis and schools are cheap (sorry it's the sad reality lol)
You really just need more of one reactant than another

 

[zhertec]

2M Nitric Acid.

20mL of 15M H2SO4
50mL of 2M HNO3
100mL of 0.00001M HCl

H2SO4 -->2H+ + SO4 2-
0.02L (20mL) x 15M = 0.3 moles of H2SO4
H2SO4 ratio with H+ 1:2
Hence 0.6 moles of H+ produced (theoretically, though ionisation of HSO4- is highly dependent on concentration and rarely occurs).

HNO3 --> H+ + NO3-
0.05L (50mL) x 2M = 0.1 moles of HNO3
Since 1:1 0.1 moles of H+ produced.

HCl --> H+ + Cl-
0.1L (100mL) x 0.00001M = 1x10^-6 moles of HCl
again 1:1 hence 1x10^-6 moles of H+ is produced.

Adding all the moles of H+ produced:
0.6 + 0.1 + 1x10^-6 = 0.700001 moles of H+ in total.
0.700001 moles/total volume (0.17L) or 170mL = 4.118 M of H+ in the solution

pH=-log(base 10) (H+ concentration)
Thereforce -log(base 10) (4.118M)
pH = -0.615

Did I get it right? lol

 

[faisalabdul16]

20mL of 15M H2SO4
50mL of 2M HNO3
100mL of 0.00001M HCl

H2SO4 -->2H+ + SO4 2-
0.02L (20mL) x 15M = 0.3 moles of H2SO4
H2SO4 ratio with H+ 1:2
Hence 0.6 moles of H+ produced (theoretically, though ionisation of HSO4- is highly dependent on concentration and rarely occurs).

HNO3 --> H+ + NO3-
0.05L (50mL) x 2M = 0.1 moles of HNO3
Since 1:1 0.1 moles of H+ produced.

HCl --> H+ + Cl-
0.1L (100mL) x 0.00001M = 1x10^-6 moles of HCl
again 1:1 hence 1x10^-6 moles of H+ is produced.

Adding all the moles of H+ produced:
0.6 + 0.1 + 1x10^-6 = 0.700001 moles of H+ in total.
0.700001 moles/total volume (0.17L) or 170mL = 4.118 M of H+ in the solution

pH=-log(base 10) (H+ concentration)
Thereforce -log(base 10) (4.118M)
pH = -0.615

Did I get it right? lol

Tbh idk, ill check right now since i made the question lolol

 

[seventhroot]

20mL of 15M H2SO4
50mL of 2M HNO3
100mL of 0.00001M HCl

H2SO4 -->2H+ + SO4 2-
0.02L (20mL) x 15M = 0.3 moles of H2SO4
H2SO4 ratio with H+ 1:2
Hence 0.6 moles of H+ produced (theoretically, though ionisation of HSO4- is highly dependent on concentration and rarely occurs).

HNO3 --> H+ + NO3-
0.05L (50mL) x 2M = 0.1 moles of HNO3
Since 1:1 0.1 moles of H+ produced.

HCl --> H+ + Cl-
0.1L (100mL) x 0.00001M = 1x10^-6 moles of HCl
again 1:1 hence 1x10^-6 moles of H+ is produced.

Adding all the moles of H+ produced:
0.6 + 0.1 + 1x10^-6 = 0.700001 moles of H+ in total.
0.700001 moles/total volume (0.17L) or 170mL = 4.118 M of H+ in the solution

pH=-log(base 10) (H+ concentration)
Thereforce -log(base 10) (4.118M)
pH = -0.615

Did I get it right? lol

yeah it's right

from the same guy who gave this answer:

so they don't combust too much

 

[QZP]

because alkanol is cheaper than alkanoic acis and schools are cheap (sorry it's the sad reality lol)
You really just need more of one reactant than another

To shift equilibrium fwd you mean!!

 

[faisalabdul16]

Solutions containing nickel (II) ions were analysed by AAS. A standard solution of

5 ppm nickel had an absorbance of 0.200. A second solution of unknown

concentration was found to have an absorbance of 0.500.

100 mL of this second solution was reacted with excess sodium carbonate solution.

The precipitate formed was weighed and dried.

What mass of precipitate formed?

(A) 2.5 x 10^-3

(B) 1.3 x 10^-3

(C) 1.7 g

(D) 2.5 g

. do this guys.