HSC 2012-2015 Chemistry Marathon (archive) (1 Viewer)

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AnimeX

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re: HSC Chemistry Marathon Archive

Most pH indicators are weak acids or bases which change colour at a certain pH.

Explain how pH indicators work.
In your answer, include relevant chemical equations.
(4 marks)
I'll give this a go..

pH indicators are weak acids or bases which change colours depending on their pH, phenolphthalein is a colourless indicator which is an acid, it changes its colour at the 8.2-10 pH range. The following reaction is a neutralisation reaction:
phenolphthalein, an acid, (colourless) + NaOH(colourless) --> Naphenolphthaleite (pink) + H2O.
[would this get me any marks, no idea what else to write besides more examples]

edit: how did people know to use Hin+ and H in-? is it a standard thing to do?
 
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abdog

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re: HSC Chemistry Marathon Archive

HCO3- is amphiprotic though, so it can act as either a base or an acid by either accepting or donating a proton.

Acting as an acid:
HCO3- + OH- --> CO32- + H2O

Acting as a base:
HCO3- + H3O+ --> H2CO3 + H2O

Don't really understand what you're asking :s
(a) Identify whether the salt, sodium hydrogen carbonate is acidic, basic or neutral. Write a balanced equation to explain its acidic, basic or neutral nature in water.
- Looking at the above equations, this salt can be both acidic and basic. How do we know which one to use?
 

superSAIyan2

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re: HSC Chemistry Marathon Archive

It is slightly basic as it is the salt of a strong base and a weak acid. In solution the salt dissociates as NaHCO3 ---> Na+ + HCO3-
The sodium cation has no tendency to hydrolyse, whereas the hydrogen carbonate ion will hydrolyse in water to produce hydroxide ions
HCO3- + H20 ---> H2CO3 + OH-

Although the hydrogen carbonate ion is amphiprotic (i.e able to react with acids and bases) it has a higher tendency to act as a base with water.

Edit: Question; Explain the effect of taking the salt bridge out of a galvanic cell
 
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someth1ng

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re: HSC Chemistry Marathon Archive

It is slightly basic as it is the salt of a strong base and a weak acid. In solution the salt dissociates as NaHCO3 ---> Na+ + HCO3-
The sodium cation has no tendency to hydrolyse, whereas the hydrogen carbonate ion will hydrolyse in water to produce hydroxide ions
HCO3- + H20 ---> CO3)2- + H30+.

Although the hydrogen carbonate ion is amphiprotic (i.e able to react with acids and bases) it has a higher tendency to act as a base with water.

Edit: Question; Explain the effect of taking the salt bridge out of a galvanic cell
I would say that's wrong - it does have a tendency to hydrolyse but it is extremely small...but that's not zero.
 

bedpotato

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re: HSC Chemistry Marathon Archive

It is slightly basic as it is the salt of a strong base and a weak acid. In solution the salt dissociates as NaHCO3 ---> Na+ + HCO3-
The sodium cation has no tendency to hydrolyse, whereas the hydrogen carbonate ion will hydrolyse in water to produce hydroxide ions
HCO3- + H20 ---> CO3)2- + H30+.

Although the hydrogen carbonate ion is amphiprotic (i.e able to react with acids and bases) it has a higher tendency to act as a base with water.

Edit: Question; Explain the effect of taking the salt bridge out of a galvanic cell
HCO3- + H2O --> H2CO3 + OH-

Is this the equation?
 

abdog

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re: HSC Chemistry Marathon Archive

It is slightly basic as it is the salt of a strong base and a weak acid. In solution the salt dissociates as NaHCO3 ---> Na+ + HCO3-
The sodium cation has no tendency to hydrolyse, whereas the hydrogen carbonate ion will hydrolyse in water to produce hydroxide ions
HCO3- + H20 ---> CO3)2- + H30+.

Although the hydrogen carbonate ion is amphiprotic (i.e able to react with acids and bases) it has a higher tendency to act as a base with water.

Edit: Question; Explain the effect of taking the salt bridge out of a galvanic cell
A salt bridge is responsible for allowing the movement of ions between the half cells and for maintaining neutrality of the half cells. For example, if one half cell, say Zn electrode in Zn2+ electrolyte(let the other half cell be anything under Zn). When the reaction is occurring, Zn is oxidised and produces Zn2+. The salt bridge will bring negative ions from the other half cell to the zinc half cell, to maintain electrical neutrality.

If the salt bridge is removed, electrical neutrality cannot be maintained and movement of ions cannot occur, affecting the reaction.

Someone correct me if I'm wrong

For my previous question, so we have to memorise that HCO3- has a tendency to act as a base?
 
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bangladesh

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re: HSC Chemistry Marathon Archive

A salt bridge is responsible for allowing the movement of ions between the half cells and for maintaining neutrality of the half cells. For example, if one half cell, say Zn electrode in Zn2+ electrolyte(let the other half cell be anything under Zn). When the reaction is occurring, Zn is oxidised and produces Zn2+. The salt bridge will bring negative ions from the other half cell to the zinc half cell, to maintain electrical neutrality.

If the salt bridge is removed, electrical neutrality cannot be maintained and movement of ions cannot occur, affecting the reaction.

Someone correct me if I'm wrong

For my previous question, so we have to memorise that HCO3- has a tendency to act as a base?
Correct answer. However I don't think an example is required unless the question asks you specifically to use an example. Apart from that everything is good.
 

bedpotato

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re: HSC Chemistry Marathon Archive

A student wished to determine the percentage of calcium carbonate in a shell found at the beach. The clean dry shell, which weighed 1.306g, was placed in a small beaker and 10mL of 5mol/L of hydrochloric acid was added. When the shell had completely dissolved, the resulting solution was transferred to a volumetric flask and the volume made up to 25mL with distilled water. A 10mL sample from this solution required 11.2mL of 1mol/L sodium hydroxide for complete neutralisation.

a) Write a balanced equation for the reaction of calcium carbonate with hydrochloric acid. [1]
b) Calculate the number of moles of NaOH present in the 11.2mL of 1mol/L NaOH solution. [1]
c) How many moles of acid remained in the beaker after the reaction with the shell (before the dilution was made)? [2]
d) How many moles of acid reacted with the shell? [1]
e) What mass of calcium carbonate was present in the shell? [2]
f) What was the percentage of calcium carbonate in the shell? [1]​
 

abdog

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re: HSC Chemistry Marathon Archive

A student wished to determine the percentage of calcium carbonate in a shell found at the beach. The clean dry shell, which weighed 1.306g, was placed in a small beaker and 10mL of 5mol/L of hydrochloric acid was added. When the shell had completely dissolved, the resulting solution was transferred to a volumetric flask and the volume made up to 25mL with distilled water. A 10mL sample from this solution required 11.2mL of 1mol/L sodium hydroxide for complete neutralisation.

a) Write a balanced equation for the reaction of calcium carbonate with hydrochloric acid. [1]
b) Calculate the number of moles of NaOH present in the 11.2mL of 1mol/L NaOH solution. [1]
c) How many moles of acid remained in the beaker after the reaction with the shell (before the dilution was made)? [2]
d) How many moles of acid reacted with the shell? [1]
e) What mass of calcium carbonate was present in the shell? [2]
f) What was the percentage of calcium carbonate in the shell? [1]​
a)CaCO3 + 2HCL ---> CaCl2 + CO2 + H20
b)Concentration = No.of moles/Volume of solution(C=n/V, becomes n=CV), so n=1 x 0.0112L = 0.0112 moles of NaOH
c)0.036 moles?(not sure so I left the working out)
d)0.014 moles?(again, not sure so I left the working out)
e)0.701g?(confidence wavering....)
f))0.701/1.306 x 100 = 53.7%....? (Confidence collapsed....)

I GIVE UP!

Can anyone help?
 
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khoavo12

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re: HSC Chemistry Marathon Archive

I don't know if I'm right but here's my result:

a)CaCO3 + 2HCL ---> CaCl2 + CO2 + H20
b) 0.0112 mol
c) 0.0388 mol
d) 0.0112 mol
e) 0.56 g
f) 43%
 

babberz

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re: HSC Chemistry Marathon Archive

A student wished to determine the percentage of calcium carbonate in a shell found at the beach. The clean dry shell, which weighed 1.306g, was placed in a small beaker and 10mL of 5mol/L of hydrochloric acid was added. When the shell had completely dissolved, the resulting solution was transferred to a volumetric flask and the volume made up to 25mL with distilled water. A 10mL sample from this solution required 11.2mL of 1mol/L sodium hydroxide for complete neutralisation.

a) Write a balanced equation for the reaction of calcium carbonate with hydrochloric acid. [1]
b) Calculate the number of moles of NaOH present in the 11.2mL of 1mol/L NaOH solution. [1]
c) How many moles of acid remained in the beaker after the reaction with the shell (before the dilution was made)? [2]
d) How many moles of acid reacted with the shell? [1]
e) What mass of calcium carbonate was present in the shell? [2]
f) What was the percentage of calcium carbonate in the shell? [1]​
this is what i got hope im right hahahaha

a)CaCO3(s) + 2HCL(aq) ----> CaCl2(aq) + CO2(g) + H2O(l)
b) n(NaOH)= 0.0112x1
=0.0112 moles of NaOH
c) NaOH+HCL are in a 1:1 so no need to write equation
since 10mL of the diluted sample required 11.2mL of 1mol/L NaOH
we see that there are 0.0112 moles of HCL in the 10mL sample
therefore in a 25mL sample there are
0.0112x2.5
=0.028 moles of HCL
(before and after dilution there are still the same amount of moles of HCL)
d)n(HCL reacted) = n(HCL initially added) - n(HCL excess)
therefore 0.05-0.028
=0.022 moles HCL reacted
e)from the equation at the top n(CaCO3)=2n(HCL)
therefore n(CaCO3)=0.011
then M(CaCO3)= 0.011x100.1
=1.1011g
f) % CaCO3= (1.1011/1.306) x 100
=84.31%
 
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bedpotato

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re: HSC Chemistry Marathon Archive

this is what i got hope im right hahahaha

a)CaCO3(s) + 2HCL(aq) ----> CaCl2(aq) + CO2(g) + H2O(l)
b) n(NaOH)= 0.0112x1
=0.0112 moles of NaOH
c) NaOH+HCL are in a 1:1 so no need to write equation
since 10mL of the diluted sample required 11.2mL of 1mol/L NaOH
we see that there are 0.0112 moles of HCL in the 10mL sample
therefore in a 25mL sample there are
0.0112x2.5
=0.028 moles of HCL
(before and after dilution there are still the same amount of moles of HCL)
d)n(HCL reacted) = n(HCL initially added) - n(HCL excess)
therefore 0.05-0.028
=0.022 moles HCL reacted
e)from the equation at the top n(CaCO3)=2n(HCL)
therefore n(CaCO3)=0.011
then M(CaCO3)= 0.011x100.1
=1.1011g
f) % CaCO3= (1.1011/1.306) x 100
=84.31%
You got it, well done :)
 
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bedpotato

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re: HSC Chemistry Marathon Archive

What is the difference between the terms 'strong' and 'concentrated' when referring to acids?
 

abdog

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re: HSC Chemistry Marathon Archive

What is the difference between the terms 'strong' and 'concentrated' when referring to acids?
When an acid is defined as strong, it means the acid ionises completely in solutions, such as HCL. Concentrated is used when an acid has a high concentration, such as 2molL-1, regardless of the degree of ionisation.
 

bangladesh

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re: HSC Chemistry Marathon Archive

When an acid is defined as strong, it means the acid ionises completely in solutions, such as HCL. Concentrated is used when an acid has a high concentration, such as 2molL-1, regardless of the degree of ionisation.
Adding a diagram to this answer would guarantee you full marks.
 

Menomaths

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When an acid is defined as strong, it means the acid ionises completely in solutions, such as HCL. Concentrated is used when an acid has a high concentration, such as 2molL-1, regardless of the degree of ionisation.
I'm not sure, but is it okay to say that? Because relative to something else that may not be concentrated?
 

abdog

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I'm not sure, but is it okay to say that? Because relative to something else that may not be concentrated?
Well, 2molL-1 of anything is concentrated, no?
 

Menomaths

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Well, 2molL-1 of anything is concentrated, no?
No idea, relative to a 18 mol/L sulfuric acid solution used in esterification, it's not really concentrated.
 
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