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October 01, 2025, 11:01:31 pm

Author Topic: VCE Chemistry Question Thread  (Read 2911861 times)  Share 

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Kel9901

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Re: VCE Chemistry Question Thread
« Reply #2985 on: March 10, 2015, 09:23:03 am »
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Can someone help with this question

during acid/base titration of Na2CO3 and HCl, calculated value of HCl is found to be lower than true value. this could be due to:

a) student rinsed pipette with water instead of primary standard Na2CO3 solution
b) student filled pipette with Na2CO3 solution so the top of the meniscus was level with calibration mark
c) Concial flask was rinsed with water
d) Student overshoots the end point

So the answers a OR b (shady). And the other shady thing is idk whats being titrated against what. Assuming its the HCl thats in the titration flask as implied by a), why would the answer be a as wouldn't rinsing the pipette with water instead of Na2CO3 mean  the concentration of Na2CO3 would be lower and therefore the concentration of HCl would be percieved to be higher, not lower than true value?

CoNfUsEd :o

The pipette dispenses a 'fixed' volume of solution regardless of concentration (while the burette is more like a 'tap'). The rinsing with water reduces the concentration of Na2CO3 (water dilutes the solution), and a fixed total amount of solution is entering the flask, so less Na2CO3 enters the conical flask. This means less HCl is needed to dilute the lower amount of Na2CO3, and hence calculated HCl concentration is lower.
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cosine

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Re: VCE Chemistry Question Thread
« Reply #2986 on: March 10, 2015, 07:49:43 pm »
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When we work out the percentage composition of a compound of each element, why do we work out the mole? Can we not just work out the mass of each element, and divide it by the mass of the whole compound and work from there? Thank you
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #2987 on: March 10, 2015, 10:36:23 pm »
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When we work out the percentage composition of a compound of each element, why do we work out the mole? Can we not just work out the mass of each element, and divide it by the mass of the whole compound and work from there? Thank you

There's two types of composition for a solid: mass composition and mole composition. The question is probably asking for the latter, but you're right. What you've given is a perfectly valid way of defining elemental composition.
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cosine

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Re: VCE Chemistry Question Thread
« Reply #2988 on: March 10, 2015, 10:39:35 pm »
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There's two types of composition for a solid: mass composition and mole composition. The question is probably asking for the latter, but you're right. What you've given is a perfectly valid way of defining elemental composition.
Mole composition is the empirical formula, right? Why exactly do we find number of mole for the empirical formula? Thanks
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chocolate.cake.1

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Re: VCE Chemistry Question Thread
« Reply #2989 on: March 10, 2015, 10:42:01 pm »
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Hello  :)

My textbook says that the following steps are involved in preparing a standard solution:
1. Place the weighed sample in volumetric flask
2. Half fill with water. Shake to dissolve sample.
2. Add water to the calibration line. Shake again.

So my question is: why do we have to half fill the volumetric flask with water first and then shake it? (in other words, why aren't we allowed to fill the volumetric flask up to the calibration line all in one go?)

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #2990 on: March 10, 2015, 10:59:08 pm »
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Mole composition is the empirical formula, right? Why exactly do we find number of mole for the empirical formula? Thanks

So if you have 10 moles of carbon, 10 moles of hydrogen and 16 moles of oxygen in your compound and you're sure it's pure, the empirical formula is just C5H5O8, the lowest common factor of the three coefficients.

Hello  :)

My textbook says that the following steps are involved in preparing a standard solution:
1. Place the weighed sample in volumetric flask
2. Half fill with water. Shake to dissolve sample.
2. Add water to the calibration line. Shake again.

So my question is: why do we have to half fill the volumetric flask with water first and then shake it? (in other words, why aren't we allowed to fill the volumetric flask up to the calibration line all in one go?)

It's harder to shake the flask when it's full. I've been guilty of this myself many times in the lab and it does make a difference.
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IntelxD

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Re: VCE Chemistry Question Thread
« Reply #2991 on: March 10, 2015, 11:09:37 pm »
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The pipette dispenses a 'fixed' volume of solution regardless of concentration (while the burette is more like a 'tap'). The rinsing with water reduces the concentration of Na2CO3 (water dilutes the solution), and a fixed total amount of solution is entering the flask, so less Na2CO3 enters the conical flask. This means less HCl is needed to dilute the lower amount of Na2CO3, and hence calculated HCl concentration is lower.

Wouldn't the calculated concentration of HCl be higher as your titre volume will be lower?
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grannysmith

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Re: VCE Chemistry Question Thread
« Reply #2992 on: March 11, 2015, 08:08:43 am »
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Wouldn't the calculated concentration of HCl be higher as your titre volume will be lower?
Except we're standardising the HCl solution if I'm not mistaken. So in this case, less titre volume is required. Using mole ratios, this means that a lower amount of Na2CO3 translates into a lower amount of HCl. Ultimately, you're getting a lower concentration of HCl.

Conversely however, if we were standardising the Na2CO3 solution, then the HCl concentration would be known prior to the titration. If the same thing happened with the Na2CO3 (i.e. diluted) in this case, then this would mean less HCl titre volume is needed for a given volume of diluted Na2CO3.  But it would be assumed that, using mole ratios, the amount of Na2CO3 is the same. Hence, the number of moles of HCl is 'constant' in all cases. If you're dividing a constant value by a lower titre volume, you're getting a higher [HCl].

Not sure if this is correct or I'm just confusing myself ahah

Kel9901

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Re: VCE Chemistry Question Thread
« Reply #2993 on: March 11, 2015, 11:25:14 am »
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Wouldn't the calculated concentration of HCl be higher as your titre volume will be lower?
Basically what happens is this:
Lower c (Na2CO3) (and equal V)
--> Lower n (Na2CO3)
--> Lower n (HCl) needed
--> Lower V (HCl)
--> Higher calculated c (HCl)

So basically n (HCl) appears to be the same (but is actually lower), and V (HCl) is lower, so c=n/V will be higher
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IntelxD

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Re: VCE Chemistry Question Thread
« Reply #2994 on: March 11, 2015, 07:34:56 pm »
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Basically what happens is this:
Lower c (Na2CO3) (and equal V)
--> Lower n (Na2CO3)
--> Lower n (HCl) needed
--> Lower V (HCl)
--> Higher calculated c (HCl)

So basically n (HCl) appears to be the same (but is actually lower), and V (HCl) is lower, so c=n/V will be higher

In your previous post you stated that the calculated HCl concentration would be lower. That's why I asked the question.

Except we're standardising the HCl solution if I'm not mistaken. So in this case, less titre volume is required. Using mole ratios, this means that a lower amount of Na2CO3 translates into a lower amount of HCl. Ultimately, you're getting a lower concentration of HCl.

Conversely however, if we were standardising the Na2CO3 solution, then the HCl concentration would be known prior to the titration. If the same thing happened with the Na2CO3 (i.e. diluted) in this case, then this would mean less HCl titre volume is needed for a given volume of diluted Na2CO3.  But it would be assumed that, using mole ratios, the amount of Na2CO3 is the same. Hence, the number of moles of HCl is 'constant' in all cases. If you're dividing a constant value by a lower titre volume, you're getting a higher [HCl].

Not sure if this is correct or I'm just confusing myself ahah

The problem with this is that the proposed amount of Na2CO3 doesn't change in our working. Thus our calculations will not reflect the actual decrease in the amount of Na2CO3 (due to inadvertent dilution). Consequently, we will obtain a lower titre volume (less HCl is needed to neutralise the smaller amount of Na2CO3) and thus a greater perceived concentration of the HCl solution.
« Last Edit: March 11, 2015, 07:43:19 pm by IntelxD »
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Re: VCE Chemistry Question Thread
« Reply #2995 on: March 11, 2015, 07:39:46 pm »
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Can someone please explain mole ratios to me, thanks
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knightrider

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Re: VCE Chemistry Question Thread
« Reply #2996 on: March 11, 2015, 08:14:29 pm »
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How do the electronic configurations of the different transition metals explain their location in the periodic table?

Redoxify

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Re: VCE Chemistry Question Thread
« Reply #2997 on: March 11, 2015, 08:25:41 pm »
+1
How do the electronic configurations of the different transition metals explain their location in the periodic table?

When it comes to each transitional metals, live every other element they have different molar masses, this means that they have different numbers of electrons, following the sub shell configuration, 1s,2s etc they are placed in order
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Redoxify

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Re: VCE Chemistry Question Thread
« Reply #2998 on: March 11, 2015, 08:30:31 pm »
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Can someone please explain mole ratios to me, thanks

Hi cosine, the mole ratio in reference to compounds is the amounts of moles of any two compounds involved in a chemical reaction, for example, 2H2(g) + O2(g) → 2 H2O(g) where every 1 mole of O2 used, 2 moles of H2Ore formed. Therefore the mole ratio between O2 and H2O is 1:2.
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Re: VCE Chemistry Question Thread
« Reply #2999 on: March 11, 2015, 08:58:23 pm »
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When it comes to each transitional metals, live every other element they have different molar masses, this means that they have different numbers of electrons, following the sub shell configuration, 1s,2s etc they are placed in order

Thanks Redoxify :)