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July 02, 2020, 03:31:37 pm

Author Topic: HSC Chemistry Question Thread  (Read 595647 times)  Share 

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milie10

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Re: Chemistry Question Thread
« Reply #3810 on: January 30, 2020, 10:37:40 pm »
+1
Hey!You are right! 10L is the answer, so you choose the answer that is closest to this value, which is C!
Here is my working out of the question (click on the image to make it larger):
(Image removed from quote.)
When you use \( n=c \times v \), it only finds n(HCl). I find limiting reagent questions confusing, so I like to write statements as I did in my working. In those statements, I calculated how many moles a substance will react with/ use up (according to stoichiometric ratios). I compare the two statements and pretty quickly find out which substance is the limiting reagent.
I am not sure of a quicker way to do this question (hopefully someone can chip in!).

Let me know if you need any more help!

Thanks heaps, that clears things up!! Writing the limiting reagent statements really helps- I'll start doing that. :D
* Noticed a trivial mistake in your working- HCl leftover = 0.002- 0.0008 =0.0012

r1ckworthy

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Re: Chemistry Question Thread
« Reply #3811 on: January 31, 2020, 09:22:34 am »
+1
Thanks heaps, that clears things up!! Writing the limiting reagent statements really helps- I'll start doing that. :D
* Noticed a trivial mistake in your working- HCl leftover = 0.002- 0.0008 =0.0012

Oops! Not surprised though, did rush a lot ;D Glad it was of help to you!
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milie10

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Re: Chemistry Question Thread
« Reply #3812 on: February 02, 2020, 12:58:59 am »
+1
Hi!

I'm a bit confused about why CaCO3 is a basic salt instead of neutral. Could someone explain this please?

Thanks! :)


r1ckworthy

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Re: Chemistry Question Thread
« Reply #3813 on: February 02, 2020, 11:34:02 am »
+2
Hi!

I'm a bit confused about why CaCO3 is a basic salt instead of neutral. Could someone explain this please?

Thanks! :)
Hey!

It all has to do with how \( \text{CaCO}_3 \) dissociates in water. \( \text{CaCO}_3 \) dissociates in water to produce \( \text{Ca}^\text{2+} \) and \( \text{CO}_3^\text{2-} \) ions. The  \( \text{CO}_3^\text{2-} \) ions react with water to produce \( \text{HCO}_3^- \) and hydroxide ions, which leads the solution to be basic.

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BlackFrost

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Re: Chemistry Question Thread
« Reply #3814 on: February 15, 2020, 04:06:49 pm »
0
In an experiment of separating sand from salt water (which involves filtration and evaporation), what could be the sources of errors that could eventually cause substances to be lost ?

milie10

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Re: Chemistry Question Thread
« Reply #3815 on: February 15, 2020, 07:56:38 pm »
+6
In an experiment of separating sand from salt water (which involves filtration and evaporation), what could be the sources of errors that could eventually cause substances to be lost ?

Hi!
- Some of the salt water could be left on the filter paper or the filter funnel while filtrating
- Over evaporating the water with a Bunsen Burner can result in the salt 'spitting out' from the evaporating basin

There can also be errors that causes an increased mass:
- Under evaporating the solvent could result in leftover water in the resultant mass of salt
- Evaporating the water overnight (to try and eliminate the error of over evaporating with a Bunsen Burner) can lead to dust particles and other contaminants getting into the evaporating basin.

Hope that helps :)

Einstein_Reborn_97

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Re: Chemistry Question Thread
« Reply #3816 on: March 02, 2020, 09:35:30 pm »
+1
If the equivalence point of a reaction is between the pH range of colour change for the indicator in use, does the change in colour happening at the maximum or minimum pH within that range or at the equivalence point? E.g. phenolphtalein has a pH range of colour change of around 8.3 to 10.0; if the equivalence point of the reaction is 9.1, will you observe the colour change from colourless to red-pink at about a pH of 8.3 or 9.1?
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milie10

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Re: Chemistry Question Thread
« Reply #3817 on: March 08, 2020, 12:26:14 am »
0
hi!





Did I do q2 correctly? I'm also not sure what the difference between q2 and q3 is.

thanks :D

Erutepa

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Re: Chemistry Question Thread
« Reply #3818 on: March 08, 2020, 05:21:03 pm »
+5
If the equivalence point of a reaction is between the pH range of colour change for the indicator in use, does the change in colour happening at the maximum or minimum pH within that range or at the equivalence point? E.g. phenolphtalein has a pH range of colour change of around 8.3 to 10.0; if the equivalence point of the reaction is 9.1, will you observe the colour change from colourless to red-pink at about a pH of 8.3 or 9.1?
Sorry for the late reply!
The colour change of the indicator occurs somewhat gradually within its range. An indicator doesn't change instantly at a precise pH, but rather changes over a pH interval. For your example, phenolphtalein will begin turning slightly pink at a pH of 8.3 and will become darker as you reach pH of 10 If you want to read more about this you can do so here  :)

hi!

(Image removed from quote.)

(Image removed from quote.)

Did I do q2 correctly? I'm also not sure what the difference between q2 and q3 is.

thanks :D
For question 2, the c value you are using is 4.18J/g/K, meaning your mass should be expressing in grams, however it seems you might have expressed your mass in kg. Also note that you are calculating the temperature change of the final (100ml) solution, as such the mass in your energy calculation should be 100g (since 1ml of water = 1g of water).
When you are calculating the change in temperature you don't need to be converting it to kelvin as the change in degrees celcius is equal to the change in kelvin. If you do want to convert to kelvin, then you need to be adding 273 to both the initial and the final temperatures (to convert both into kelvin) as shown here for example:

you cannot just add 273 to the final change as you have done, you must convert both the initial and final temps before finding the change in temperature. Note that if you simplify the above equation as such:




which shows that when calculating the change in temperature, you do not need to convert anything into kelvin
Hopefully by taking on these tips you can fix up your answers for question 2

Question 3 is a similar question, however is not quite the same thing. Question 2 asks you to calculate the change in energy of the total reaction, however question 3 wants you to calculate the change in energy for just 1 mol of reaction. For reaction A this would be calculating the energy change for 1 mol of HCl to react with 1M of NaOH.
Hopefully this helps :)
« Last Edit: March 08, 2020, 05:22:36 pm by Erutepa »
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Einstein_Reborn_97

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Re: Chemistry Question Thread
« Reply #3819 on: March 16, 2020, 07:20:53 pm »
0
HF(aq) + CF3COO-(aq) ⇌ F(aq) + CF3COOH(aq)
Keq = 3.80 x 10-4

Identify the strongest acid and the strongest base in this system.

I know HF is a weak acid and thus it's conjugate base will be a strong base. I also know that the equilibrium constant indicates that there are more reactants than products. I'm just not sure where to go from there.
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BlackFrost

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Re: Chemistry Question Thread
« Reply #3820 on: April 16, 2020, 08:24:55 pm »
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Hey guys,
I was just wondering if anyone has any resources on Preliminary Chemistry past papers (2015 onwards) and any topic tests on Module 1 and 2.

Thanks

Einstein_Reborn_97

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Re: Chemistry Question Thread
« Reply #3821 on: April 17, 2020, 06:34:43 pm »
+3
Hey guys,
I was just wondering if anyone has any resources on Preliminary Chemistry past papers (2015 onwards) and any topic tests on Module 1 and 2.

Thanks

The NESA website has tons of previous past papers (exams and marking guidelines) for pretty much every subject, which you can download and print. Here's the link: https://educationstandards.nsw.edu.au/wps/portal/nesa/11-12/resources/hsc-exam-papers
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stels

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Re: HSC Chemistry Question Thread
« Reply #3822 on: May 27, 2020, 05:14:40 pm »
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You add 25 milliliters of 2.0 mM HCl solution to 75 milliliters of 1.0 mM Ca (OH) 2 solution. What is the pH of the resulting solution?

Einstein_Reborn_97

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Re: HSC Chemistry Question Thread
« Reply #3823 on: May 27, 2020, 08:16:36 pm »
+5
You add 25 milliliters of 2.0 mM HCl solution to 75 milliliters of 1.0 mM Ca (OH) 2 solution. What is the pH of the resulting solution?
Hi stels,

Step 1: Balanced chemical equation ALWAYS:
\(2HCL_{(aq)}+Ca(OH)_{2(aq)}→CaCl_{2 (aq)}+2H_2O_{(l)}\)

Step 2: Find the limiting reagent and the solution in excess (and how much)
\(n(HCl)=cV=(2.0\times10^{-3})(25\times10^{-3})=50\times10^{-6} mol\)
\(n(Ca(OH)_2)=cV=(1.0\times10^{-3})(75\times10^{-3})=75\times10^{-6} mol\)

Reaction stoichiometry = 2:1
∴ HCl is the limiting reagent (all of it will be neutralised) and \(Ca(OH)_2\) is in excess.

\(Ca(OH)_2\) required (used up) = \(50\times10^{-6} \div 2 = 25\times10^{-6} mol\)
Excess \(Ca(OH)_2\) = \(75\times10^{-6} - 25\times10^{-6} = 50\times10^{-6} mol\)

Step 3: Work out the concentration of the solution in excess, then the concentration of \(H^+\) or \(OH^-\) ions and finally the pH
Total volume of resulting solution = 25 + 75 = 100 mL = 0.10 L
c(\(Ca(OH)_2\)) = \(\frac{n}{V}\) = \(\frac{50\times10^{-6}}{0.10}\) = 0.0005 mol/L
\([OH^-]\) = 2\(\times\)0.0005 = 0.001 mol/L
\(pOH = -log_{10}(0.001) = 3\)
pH + pOH = 14
∴ pH = 14 - 3 = 11

Hope that helps! ;)
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006896

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Re: HSC Chemistry Question Thread
« Reply #3824 on: June 28, 2020, 12:37:51 pm »
0
Hello!
Could someone help me with this question about redox reactions and galvanic cells?
Thanks!