Bored of Studies 2012 Chemistry Trial Paper (1 Viewer)

[Rawf]

Thanks for the test. However, could you write a section for the option industrial chemistry?

 

[someth1ng]

Thanks for the test. However, could you write a section for the option industrial chemistry?

No.

 

[HeroicPandas]

bump

 

[Menomaths]

bump

Thanks for bumping, never would have found it otherwise.

 

[Frie]

Nice, going to take a look

 

[Frie]

Need help with Question 12 because multiple choice doesn't give explanations.

"5mL of 0.1M HCl is diluted to 100mL with water. What is the new pH of the solution?"

A) 1.0
B) 2.3
C) 4.6
D) 5.3

Answer says B. but isn't the original solution a pH of -log(0.005 x 0.1) = 3.3
So how can diluting the solution cause it to be more acidic.
I thought the answer would be D because it was diluted by a factor of 20, so the hydrogen ion concentration decreases by a factor of 20. A change in hydrogen ion concentration of 20 is equivalent to a change in pH of 2.

Also don't get Question 17 and 20.

For Question 17, is this even in the syllabus? or just a fun question.

Question 20, no idea about question, or how any of the answers affect it.

 

[bedpotato]

Need help with Question 12 because multiple choice doesn't give explanations.

"5mL of 0.1M HCl is diluted to 100mL with water. What is the new pH of the solution?"

A) 1.0
B) 2.3
C) 4.6
D) 5.3

Answer says B. but isn't the original solution a pH of -log(0.005 x 0.1) = 3.3
So how can diluting the solution cause it to be more acidic.
I thought the answer would be D because it was diluted by a factor of 20, so the hydrogen ion concentration decreases by a factor of 20. A change in hydrogen ion concentration of 20 is equivalent to a change in pH of 2.

Also don't get Question 17 and 20.

For Question 17, is this even in the syllabus? or just a fun question.

Question 20, no idea about question, or how any of the answers affect it.

The initial pH is actually 1, not 3.3

The concentration of HCl is 0.1M (i.e. 0.1mol/L).

0.005x0.01 = 0.0005 gives you the moles of HCl.

 

[Frie]

Oh right, got a bad habit of using all information if I get it, regardless if its necessary or not. Thanks.

How about 20?

Question is: "Which of the following can be used to help reduce SO2 emissions from factories?"

A) CH3COOH
B) Mg(OH)2
C) NH4Cl
D) Na

Is it something to do with being basic because A and C are acidic and D is a metal/element.

 

[Menomaths]

Oh right, got a bad habit of using all information if I get it, regardless if its necessary or not. Thanks.

How about 20?

Question is: "Which of the following can be used to help reduce SO2 emissions from factories?"

A) CH3COOH
B) Mg(OH)2
C) NH4Cl
D) Na

Is it something to do with being basic because A and C are acidic and D is a metal/element.

is the answer b?

 

[someth1ng]

It's B - bases can be used in scrubber systems. Basically, it reacts with SO2 to prevent emissions.

 

[Menomaths]

Yeah, the equation is SO2+Mg(OH)2 ---> MgSO3 + H20

 

[Frie]

For Question 26(c),

Will it be alright if we used 14 = pH + pOH to calculate the pH.

Because then, instead of working out the concentration of excess H+, we go straight to the answer, and so we skip a step which was worth 1 mark.

Working out would just be (after the Concentration of excess OH- step)

pH = 14 + log(0.0895)
The answer this gives is still the correct one, 12.95 pH

 

[someth1ng]

For Question 26(c),

Will it be alright if we used 14 = pH + pOH to calculate the pH.

Because then, instead of working out the concentration of excess H+, we go straight to the answer, and so we skip a step which was worth 1 mark.

Working out would just be (after the Concentration of excess OH- step)

pH = 14 + log(0.0895)
The answer this gives is still the correct one, 12.95 pH

Use K_w=[H⁺][OH^-]=1.0×10^-14

 

[Frie]

Use K_w=[H⁺][OH^-]=1.0×10^-14

Will we get marked down if we use 14 = pH + pOH?

It's much simpler.

 

[someth1ng]

You shouldn't use it because it's not a given formula and isn't common sense like C=n/V.