VCE Chemistry Question Thread

[sweetcheeks]

What is the oxidation number for H in NaH?
According to my book, the oxidation of H in NaH is -1. However, this online calculator (https://www.periodni.com/oxidation_numbers_calculator.php) suggests two different solutions. With regards to the VCE course, is one answer more correct than the other or are they both perfectly acceptable?

Thanks 

Follow the rule that when hydrogen is with a metal (e.g. potassium hydride (KH), magnesium hydride (MgH2)) it is in the -1 oxidation state. I don't know why the calculator shows two ways to do it, I've never seen a case where anyone would assign Na as -1 and H and +1.

The best way to remember/follow that rule for hydrides is to consider that another oxidation number 'rule' is that all group 1 metals have a +1 oxidation state in their non-atomic forms and metals generally (only in circumstances outside of VCE chemistry) don't have negative oxidation states/numbers.

[lzxnl]

What is the oxidation number for H in NaH?
According to my book, the oxidation of H in NaH is -1. However, this online calculator (https://www.periodni.com/oxidation_numbers_calculator.php) suggests two different solutions. With regards to the VCE course, is one answer more correct than the other or are they both perfectly acceptable?

Thanks 

An oxidation number is essentially a measure of how many electrons the atom has gained or lost, assuming all bonds are ionic. In molecular oxygen, the oxidation number is zero as both oxygen atoms share the electrons in the covalent bond equally. However, in carbon dioxide, the oxygen atoms are much more electronegative than the carbon atom, so for the purposes of calculating the oxidation number, the oxygen takes both electrons and has oxidation number -2. Meanwhile, the carbon has just lost all of the four bonding electrons (by this calculation scheme) so has oxidation number -2.

Now consider NaH. Sodium hydride is an ionic material; you need to know this. Sodium ions can only be positive to empty the outer shell. Hydrogen atoms can either accept or donate an electron to either fill the inner shell or empty it. As sodium wants to lose electron, hydrogen must accept one, and thus its oxidation number is -1.

[persistent_insomniac]

Thermochemistry qs:
for e.g. 2C4H10 + 13O2 ----> 8CO2 + 10H2O    ∆H = -5772kJ/mol
What is the energy released if 116g of butane is burnt?
I know how to do the calculation = 5772. Do I need to include the + or - sign or is it already assumed to be - as it is asking the energy released. Also what are the units - is it just kJ or kJ/mol?
Thanks!

[Lear]

Thermochemistry qs:
for e.g. 2C4H10 + 13O2 ----> 8CO2 + 10H2O    ∆H = -5772kJ/mol
What is the energy released if 116g of butane is burnt?
I know how to do the calculation = 5772. Do I need to include the + or - sign or is it already assumed to be - as it is asking the energy released. Also what are the units - is it just kJ or kJ/mol?
Thanks!

Energy released is positive and units are only kJ.

[Quinapalus]

Thermochemistry qs:
for e.g. 2C4H10 + 13O2 ----> 8CO2 + 10H2O    ∆H = -5772kJ/mol
What is the energy released if 116g of butane is burnt?
I know how to do the calculation = 5772. Do I need to include the + or - sign or is it already assumed to be - as it is asking the energy released. Also what are the units - is it just kJ or kJ/mol?
Thanks!

It is looking for total energy released, so this will take the unit kJ.

[dream chaser]

Hi Guys,

Just have a question in regards to this question:

A commercial concrete cleaner contains hydrochloric acid. A 25.00mL volume of cleaner was diluted to 250.0mL in a volumetric flask. A 20.00mL aliquot of 0.4480M sodium carbonate solution was placed in a conical flask. Methyl orange indicator was added and the solution was titrated with the diluted cleaner. The indicator changed permanently from yellow to pink when 19.84mL of the diluted cleaner has been added.

The equation for the reaction is:
Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + CO2(g) + 2H2O(l)

Calculate the concentration of hydrochloric acid in the concrete cleaner.

I know how to calculate and get the answer. My question is why is the concentration of HCL in the burette(v=19.84mL) the same as the concentration of the diluted solution(v=250.0mL). Also, even though it does not say in the question, does HCL undergo an aliquot like the Na2CO3 solution does? Also, is the HCL burette the same as the HCL aliquot? And is the concentration of HCL that takes part in the reaction the same as the HCL that is in the burette?

[Yertle the Turtle]

Hi Guys,

Just have a question in regards to this question:

Do you understand the process that is going on? The aliquot is the measurement in the pipette, while the titre is the measurement in the burette. First you dilute the cleaner, and then you put the diluted solution into the burette. Therefore the concentration will be the same, though the volume is different, as the number of moles of HCl is only a portion of the HCl in the 250mL. The aliquot of the Na2CO3 is the measurement of 20mL in the pipette, while the HCl is in a titre of 19.84mL, in the burette. The HCl solution in the burette is what undergoes the reaction, and therefore it will have the same concentration as the diluted solution. I hope this helps, but before continuing make sure that you learn the process of a titration.

[dream chaser]

Thank you for the help  . Much appreciated. I understand the process of titrations now. I was just confused whether an aliquot of the diluted HCL solution was put into the burette. But your comment seems to clarify my misunderstanding. So in principal, if it does not say that the solution has undertaken an aliquot, we just assume that all of it is in the burette(which in my question, the diluted 250mL HCL solution) right?

[Yertle the Turtle]

Thank you for the help  . Much appreciated. I understand the process of titrations now. I was just confused whether an aliquot of the diluted HCL solution was put into the burette. But your comment seems to clarify my misunderstanding. So in principal, if it does not say that the solution has undertaken an aliquot, we just assume that all of it is in the burette(which in my question, the diluted 250mL HCL solution) right?

The aliquot is always in the pipette, the titre in the burette. A solution does not "undertake" an aliquot or a titre, but your idea is right. If not an aliquot, it is in the burette.

[Bri MT]

Thank you for the help  . Much appreciated. I understand the process of titrations now. I was just confused whether an aliquot of the diluted HCL solution was put into the burette. But your comment seems to clarify my misunderstanding. So in principal, if it does not say that the solution has undertaken an aliquot, we just assume that all of it is in the burette(which in my question, the diluted 250mL HCL solution) right?

cleaner (HCl): volumetric flask ---> diluted in volumetric flask (n stays the same, c & v change c1v1=c2v2)  ----> burette   ------>  titre added to conical flask

conical flask:  sodium carbonate pipetted out of source container (aliquot taken) / aliquot is added to conical flask   -----> indicator added to conical flask  ----> titre (HCl from burette) added to conical flask

recap on titrations

The burette contains a known concentration of a substance (this is why you rinse the burette with that substance). You check how much solution is in the burette. You let the solution go from the burette into a conical flask until you reach the end point (permanent colour change).  You check how much solution is now in the burette.     You calculate the volume of solution it took to reach the end point - this is your titre.   Because you now know the volume delivered in the titre, and the concentration, you can find the number of mols to took to reach the end point.

In the flask (placed below the burette) you have a known volume of a solution (this is why you rinse with water).

You use the formula for the reaction between the two solutions to figure out how many mols were in the flask. Because you now have number of mols & volume for the solution you put in the flask, you can figure put the concentration

When you put the solution in the flask in the first place, you might have used a pipette. When you add solution using a pipette you're delivering an aliquot. You wouldn't pipette solution into the burette - it would take forever & you don't need to put a set volume in there anyway (you just need to know the volume that leaves the burette / the titre)

[dream chaser]

Really appreciate the explanations Yertle the Turtle and miniturtle. Thanks 

[dream chaser]

Hi Guys,

Quick Question. Does one wavelength of light(for instance 400nm) have only one colour or a range of colours(ROYGBV)?

The reason why I am asking this is in regards to UV Visible Spectroscopy. I don't fully understand what happens after the light passes through the monochromator. All I know is that the role of a monochromator is to select a single wavelength of light.

So for example, if I observed a sample cell through uv visible spectroscopy to be red at a particular wavelength, does this mean that only green goes past the monochromator(because of complementary colours and colour observed vs colour absorbed) and that gets absorbed by the sample cell so that red transmits or do all colours of that wavelength pass through the wavelength and red transmits. Also, if it is the latter, then what happens to the other colours(orange, yellow, blue and violet) in this process?

Please could someone explain this to me properly.

All responses would be greatly appreciated
Thanks 

[studyingg]

Hi Guys,

Quick Question. Does one wavelength of light(for instance 400nm) have only one colour or a range of colours(ROYGBV)?

The reason why I am asking this is in regards to UV Visible Spectroscopy. I don't fully understand what happens after the light passes through the monochromator. All I know is that the role of a monochromator is to select a single wavelength of light.

So for example, if I observed a sample cell through uv visible spectroscopy to be red at a particular wavelength, does this mean that only green goes past the monochromator(because of complementary colours and colour observed vs colour absorbed) and that gets absorbed by the sample cell so that red transmits or do all colours of that wavelength pass through the wavelength and red transmits. Also, if it is the latter, then what happens to the other colours(orange, yellow, blue and violet) in this process?

Please could someone explain this to me properly.

All responses would be greatly appreciated
Thanks 

I'm pretty sure that a single colour of visible light corresponds to certain amount of energy associated with a wavelength of distinct frequency, so one wavelength (in the visible light spectrum) corresponds to particular colour. However, I think that with UV visible spectroscopy, a range of wavelengths is emitted from the light source. The image I attached is a pretty good visualisation of what happens. As you can see, the monochromator is comprised of an entrance slit, dispersion device and exit slit. Visible light enters the entrance slit from the light source (and it is comprised of wavelengths of the range ROYGBV), the dispersion device then separates these wavelengths so they disperse at separate angles, so that when they pass through the exit slit only a single wavelength of a certain frequency (and therefore colour) passes through, the amount of energy possessed by it depends on the sample used. I guess the other wavelengths are reflected using some type of filter.   

[dream chaser]

I'm pretty sure that a single colour of visible light corresponds to certain amount of energy associated with a wavelength of distinct frequency, so one wavelength (in the visible light spectrum) corresponds to particular colour. However, I think that with UV visible spectroscopy, a range of wavelengths is emitted from the light source. The image I attached is a pretty good visualisation of what happens. As you can see, the monochromator is comprised of an entrance slit, dispersion device and exit slit. Visible light enters the entrance slit from the light source (and it is comprised of wavelengths of the range ROYGBV), the dispersion device then separates these wavelengths so they disperse at separate angles, so that when they pass through the exit slit only a single wavelength of a certain frequency (and therefore colour) passes through, the amount of energy possessed by it depends on the sample used. I guess the other wavelengths are reflected using some type of filter.

Thank you studyingg for the help  . So in the picture, you gave, since green is what is getting absored, the sample will appear red right? Also, would the observed sample colour depend on the wavelength of colour that passes through the exit slit?

[turtlebanana]

Hey guys! A bit confused on this question about thermochemical equations, enthalpy etc. (Question 3 from Chapter 2.2 Heinemann Chemistry 2)

The combustion of octane to form carbon dioxide and liquid water can be
written as:
C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(l) ΔH = –5450 kJ mol–1
The combustion of octane to form carbon dioxide and steam can be
written as:
C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(g)


How would the energy released by the combustion of 1 mole of octane to
form steam compare with the energy released by 1 mole of octane to form
liquid water?