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hardest chem questions (1 Viewer)

nichorowitz

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Now CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

Therefore CaCO3:HCl
1:2
therefore twice as many moles of HCl are needed

Hence there are zero moles of HCl left over

??
 

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The following diagram shows the steps taken in an experiment to find the percentage of pure calcium carbonate in a sample of limestone. The impurities in the limestone are solids that do not react with either the HCL or NaOH.
-Weigh out 10g of limestone
-Add 100mL of 2molL Hydrochloric acid to the limestoneor
-Allow all reaction to cease
-Titrate the reaction mixture against 1molL Sodium Hydroxide solution

1) Calculate the amount of HCl left after the reaction of the Calcium Carbonate with the acid
2) Calculate the amount of HCl that reacted with the Calcium Carbonate in the limestone
3) What is the percentage of pure Calcium Carbonate in the limestone?

Lol tit rate
1) 0.000179838 mols. 0 mols (1 s.f)

2) 0.1999820161 mols, 0.2 mols (1 s.f)

3) u sure they didnt give u the amount of NaOH it was being titrated against?
 

Squishxmishyx

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The following diagram shows the steps taken in an experiment to find the percentage of pure calcium carbonate in a sample of limestone. The impurities in the limestone are solids that do not react with either the HCL or NaOH.
-Weigh out 10g of limestone
-Add 100mL of 2molL Hydrochloric acid to the limestoneor
-Allow all reaction to cease
-Titrate the reaction mixture against 1molL Sodium Hydroxide solution

1) Calculate the amount of HCl left after the reaction of the Calcium Carbonate with the acid
2) Calculate the amount of HCl that reacted with the Calcium Carbonate in the limestone
3) What is the percentage of pure Calcium Carbonate in the limestone?

Lol tit rate
Answer to the hardest Chemistry question.
1) 0.16
2) 0.04
3) 80%
 

Trebla

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A student is trying to work out the formula of an unknown acid he was given. So far he knows that the formula takes the form HnXO3, where n is an unknown positive integer and X is an unknown non-metallic element which can either be Phosphorous (P), Sulfur (S), Carbon (C) or Nitrogen (N). To determine which of the four elements X actually is for the unknown acid, the student performs a series of titrations.

The student decides to prepare a secondary standard solution of NaOH by titrating it with a primary standard solution of 0.1064 mol/L of HCl.

The result was that 25.00mL of NaOH required 25.00mL of HCl with concentration 0.1064 mol/L for an end-point with the indicator colour change.

The student then dissolved 2.569 grams of HnXO3 in water to make a 250.0mL solution. He performed a titration with the unknown acid HnXO3 and the standard NaOH solution.

The result was that 25.00mL of the standard NaOH required 16.31mL of HnXO3 solution for an end-point with the indicator colour change.

Determine the identity of X and the value of n using the given choices. Justify your answer.
 
Last edited:

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A student is trying to work out the formula of an unknown neutral acid he was given. So far he knows that the formula takes the form HnXO3, where n is an unknown positive integer and X is an unknown non-metallic element which can either be Phosphorous (P), Sulfur (S), Carbon (C) or Nitrogen (N). To determine which of the four elements X actually is for the unknown acid, the student performs a series of titrations.

The student decides to prepare a secondary standard solution of NaOH by titrating it with a primary standard solution of 0.1064 mol/L of HCl.

The result was that 25.00mL of NaOH required 25.00mL of HCl with concentration 0.1064 mol/L for an end-point with the indicator colour change.

The student then dissolved 2.569 grams of HnXO3 in water to make a 250.0mL solution. He performed a titration with the unknown acid HnXO3 and the standard NaOH solution.

The result was that 25.00mL of the standard NaOH required 16.31mL of HnXO3 solution for an end-point with the indicator colour change.

Determine the identity of X and the value of n using the given choices. Justify your answer.
firstly how do u have a neutral acid
 

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A student is trying to work out the formula of an unknown neutral acid he was given. So far he knows that the formula takes the form HnXO3, where n is an unknown positive integer and X is an unknown non-metallic element which can either be Phosphorous (P), Sulfur (S), Carbon (C) or Nitrogen (N). To determine which of the four elements X actually is for the unknown acid, the student performs a series of titrations.

The student decides to prepare a secondary standard solution of NaOH by titrating it with a primary standard solution of 0.1064 mol/L of HCl.

The result was that 25.00mL of NaOH required 25.00mL of HCl with concentration 0.1064 mol/L for an end-point with the indicator colour change.

The student then dissolved 2.569 grams of HnXO3 in water to make a 250.0mL solution. He performed a titration with the unknown acid HnXO3 and the standard NaOH solution.

The result was that 25.00mL of the standard NaOH required 16.31mL of HnXO3 solution for an end-point with the indicator colour change.

Determine the identity of X and the value of n using the given choices. Justify your answer.
after a long calculation, Molar mass of HnXO3= 63.0081 g

MHnX= 63.0081 - 48 = 15.0081g

therefore logically X cant be P or S which leaves N and C

but for it to be C the formula of the acid would be H2CO3 which molar mass = 62.026

therefore it has to be HNO3 which molar mass = approx 63

therefore n=1 and X=N

amirite?
 

silence--

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^ i got the same :L

holy shit that is a monster of a question though. haha

guess the chance of that popping up in HSC is extremely low though
 

Trebla

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after a long calculation, Molar mass of HnXO3= 63.0081 g

MHnX= 63.0081 - 48 = 15.0081g

therefore logically X cant be P or S which leaves N and C

but for it to be C the formula of the acid would be H2CO3 which molar mass = 62.026

therefore it has to be HNO3 which molar mass = approx 63

therefore n=1 and X=N

amirite?
Yep :)

How's that for something 'hard'? lol
 

silence--

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^ i suppose but i guarantee that most HSC candidates would probably not even attempt that.

HSC really isnt going to get much harder than that lol
 

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