HSC Chemistry Question Thread

[jakesilove]

Hi, how do i calculate the percentage by mass of a mixture? And also finding the empirical formulas? I'm having trouble understanding this concept

Could you please provide a specific question? That way, we can figure out what you're struggling with.

[Kekemato_BAP]

Could you please provide a specific question? That way, we can figure out what you're struggling with.

What is the percentage by mass of oxygen (O) in sodium hydroxide (NaOH)?
Suppose 3.2g of sulfur reacts with oxygen to produce 6.4g of sulfur oxide. What is the formula of the oxide?
Thank you so much

[jakesilove]

What is the percentage by mass of oxygen (O) in sodium hydroxide (NaOH)?
Suppose 3.2g of sulfur reacts with oxygen to produce 6.4g of sulfur oxide. What is the formula of the oxide?
Thank you so much

Hey! To find the percentage mass of Oxygen in Sodium hydroxide, we just find the mass of Oxygen, and divide by the total mass of Sodium hydroxide. So,



Now, the next one is a bit tougher. We have some reaction,



Where A and x are some constants. Now, the molar ratio is 1:1, so there are



of the Sulfur oxide. If 0.1 moles of sulfur oxide weighs 6.4 grams (in the question), then the molar mass of the sulfur oxide must be 64g!

So, the molar mass of sulfur, plus the molar mass of how ever many oxygen atoms we have, must be 64.



Clearly, x=2, and so the oxide is SO2(g)

[Kekemato_BAP]

Hey! To find the percentage mass of Oxygen in Sodium hydroxide, we just find the mass of Oxygen, and divide by the total mass of Sodium hydroxide. So,



Now, the next one is a bit tougher. We have some reaction,



Where A and x are some constants. Now, the molar ratio is 1:1, so there are



of the Sulfur oxide. If 0.1 moles of sulfur oxide weighs 6.4 grams (in the question), then the molar mass of the sulfur oxide must be 64g!

So, the molar mass of sulfur, plus the molar mass of how ever many oxygen atoms we have, must be 64.



Clearly, x=2, and so the oxide is SO2(g)

Thank you that helps a lot

[Kle123]

Why do we need a secondary solution? Is its concentration supposed to be more accurate than the primary solution? How can measure something to be more accurate than the solution it was used to create it?

These are thoughts that pop up when i learned titration. Teacher never explained it. Could someone please help me? THANKS

My question kinda got lost in the vast ocean of posts. I'm just going to re submit it. If anyone can help it'd be much appreciated thank you!

[bsdfjnlkasn]

Hey there,

I was wondering if anyone could help me out with clarifying some things about Le Chatellier's principe - I wrote these notes but am unsure of how true they are

o   Pressure refers to the pressure acting on the entire system
•   Adding more of one thing just increases its concentration (if aqueous)
-   This then increases the systems volume and actually decreases pressure

•   If pressure is increased, the concentration of all the products and reactants increases also

EDIT:
Why do dilute acids have higher pH readings (I know it's kind of intuitive but could I get an explanation through Le Chatelier's principle)

THANK YOU!

[jakesilove]

My question kinda got lost in the vast ocean of posts. I'm just going to re submit it. If anyone can help it'd be much appreciated thank you!

Hey! I have to say, I never actually learned about the 'secondary' solution. I've got questions about it before, but the fact that I don't know anything about it means that you don't need to know it for the HSC. However, if your teacher would expect you to use it in a method, then you probably should. Maybe just clarify with the rest of the class (or maybe someone on the forum can help!). Unfortunately, I'm not your man here

[jakesilove]

Hey there,

I was wondering if anyone could help me out with clarifying some things about Le Chatellier's principe - I wrote these notes but am unsure of how true they are

o   Pressure refers to the pressure acting on the entire system
•   Adding more of one thing just increases its concentration (if aqueous)
-   This then increases the systems volume and actually decreases pressure

•   If pressure is increased, the concentration of all the products and reactants increases also

EDIT:
Why do dilute acids have higher pH readings (I know it's kind of intuitive but could I get an explanation through Le Chatelier's principle)

THANK YOU!

Hey! Just a quick summary of Le Chatelier's Principle, which states that a system in equilibrium, when introduced to a change, will shift in equilibrium to minimise that change.

Concentration

If an equation is in equilibrium, and you add more of the reactants, then the system will want to 'use up' the reactants. Thus, the system will move towards the products.

If an equation is in equilibrium, and you add more products, then the system will want to 'use up' the products. Thus, the system will move towards the products.

Temperature

If the process is exothermic, then it RELEASES heat when it moves forward. Thus, if you increase the temperature, moving forward will only increase the temperature more. So, the system will move BACKWARDS (ie towards the reactants), to 'use up' the heat.

If the process is endothermic, then it ABSORBS heat when it moves forward. Thus, if you increase the temperature, moving forward will 'use up' the excess heat. So, the system will move FORWARD (ie towards the products), to 'use up' the heat.

Pressure

If a system of GASES are at equilibrium, and you increase the pressure, then the system will move towards the side with fewer moles of gas. If a system of gases are at equilibrium, and you decrease the pressure, then the system will move towards the side with greater moles of gas.


With regards to your Acid question; when you dilute an acid, then by Le Chatelier's principles the equilibrium will shift towards the side with Hydrogen ions (ie. more of the acid will ionise, to increase the concentration of ions). However, despite the increase in ion numbers, the concentration will actually DECREASE overall as you have increased the volume of the solution. ie. you may have gone from 100 ions to 1000 ions, but the solution has gone from 10mL to 1000mL, and so the concentration of ions has actually decreased. Thus, the pH will increase, as it will be LESS acidic.

[kiwiberry]

My question kinda got lost in the vast ocean of posts. I'm just going to re submit it. If anyone can help it'd be much appreciated thank you!

Is this about how you have to standardise NaOH for example by titrating it against oxalic acid first before you can use it in a titration as a secondary standard? I believe it's because NaOH reacts with CO2 in the air, which neutralises part of the solution and so affects its concentration. NaOH is also hydroscopic (it absorbs water from the atmosphere) so it's very hard to measure out solid NaOH accurately. Basically it's an unsuitable primary standard solution because it's concentration is always changing, so you have to titrate it first with a suitable primary standard (like oxalic acid) to find its actual concentration at the time

[bsdfjnlkasn]

Hey Jake - THANK YOU SO MUCH for that awesome response.

I have a few more questions 

Why do alkenes have lower boiling points than alkanes?
Wouldn't a double bond be more stable as it's stronger than any of the others?
Or should we be looking at the bonding they exhibit? There are more dispersion forces associated with alkanes because they have 2 extra hydrogens than the corresponding alkenes - right? Realistically, the actual single bonds themselves are weaker but because there are so many of them this factor isn't really considered as it does little in determining the BP. When talking about boiling points we're discussing the energy required to break intermolecular bonds right? Because that's the only way I can understand us ignoring the intramolecular double bond...

What is it about the double bond that makes it so reactive and susceptible to addition reactions? One more general question - do we need to know all of the reactions associated with alkanes and alkenes?

Could someone please clarify if i've justified the differences in boiling points appropriately/address the associated queries? I know it's a lot so any help would be appreciated, thank you again!!


EDIT: Is an alkyl what replaces the hydrogen in an alkane or is it the name for the entire new molecule formed?

Is Ethylbenzene and phenylethene the same thing (i.e. styrene)?

[Kle123]

It can be shown that HCl is a B-L acid: HCl+H2O -> Cl^- +H3O^+

How is it shown that NaOH + H20 is a B-L base?

Thank YOU.

[jakesilove]

Hey Jake - THANK YOU SO MUCH for that awesome response.

I have a few more questions 

Why do alkenes have lower boiling points than alkanes?
Wouldn't a double bond be more stable as it's stronger than any of the others?
Or should we be looking at the bonding they exhibit? There are more dispersion forces associated with alkanes because they have 2 extra hydrogens than the corresponding alkenes - right? Realistically, the actual single bonds themselves are weaker but because there are so many of them this factor isn't really considered as it does little in determining the BP. When talking about boiling points we're discussing the energy required to break intermolecular bonds right? Because that's the only way I can understand us ignoring the intramolecular double bond...

What is it about the double bond that makes it so reactive and susceptible to addition reactions? One more general question - do we need to know all of the reactions associated with alkanes and alkenes?

Could someone please clarify if i've justified the differences in boiling points appropriately/address the associated queries? I know it's a lot so any help would be appreciated, thank you again!!


EDIT: Is an alkyl what replaces the hydrogen in an alkane or is it the name for the entire new molecule formed?

Is Ethylbenzene and phenylethene the same thing (i.e. styrene)?

Alkenes and Alkanes actually have very similar boiling points. Remember, boiling point is to do with INTERmolecular bonds, not INTRAmolecular bonds. So, the double bond etc plays no role; you have to look at the molecule as a whole. You're absolutely right; this tends to come down to dispersion forces, which increase for bigger molecules.

A double bond ISN'T more stable; it would much prefer to be two single bonds. You don't need to know specific reactions, but you do need to understand what addition/condensation reactions are, and be able to show the 'splitting' of a double bonded alkene.

You've definitely justified it! Make your point a bit clearer (ie. 'The difference in boiling point is associated with the different in intermolecular dispersion forces' etc) and you'll be golden

Not sure about the Alkyl; I would just ignore that if I were you. Yes, they are the same

[jakesilove]

It can be shown that HCl is a B-L acid: HCl+H2O -> Cl^- +H3O^+

How is it shown that NaOH + H20 is a B-L base?

Thank YOU.

Hey! We just show it causing water to donate a Hydrogen ion; the easiest way is like this



The first Hydroxide ion has PULLED a hydrogen ion off the water, forming water itself. The important thing is that a component of the base acted as a proton acceptor; NaOH is a hard example, though, and one which I tend to avoid.

[bsdfjnlkasn]

Thanks again Jake for your comprehensive response!

I was just wondering if I could get an indication of how common the following topics are in exams:

1. The different types of cracking and their purposes (catalytic and thermal)
2. The LDPE process
3. Explanations of what the actual properties of polymers (i.e. chain branching and stiffening)

Also just a general question:

What additional information apart from crude oil/petroleum can we include in the following outcome: "1.1   Identify the industrial source of ethylene from the cracking of some of the fractions from the refining of petroleum:"

[RuiAce]

Thanks again Jake for your comprehensive response!

I was just wondering if I could get an indication of how common the following topics are in exams:

1. The different types of cracking and their purposes (catalytic and thermal)
2. The LDPE process
3. Explanations of what the actual properties of polymers (i.e. chain branching and stiffening)

Also just a general question:

What additional information apart from crude oil/petroleum can we include in the following outcome: "1.1   Identify the industrial source of ethylene from the cracking of some of the fractions from the refining of petroleum:"

Not really that often for the first one. It's nice to know both exist but it suffices to know just one of them, and you can include a description if you wish. Haven't seen them get examined explicitly because I think the description is not necessary for the course.

The second one was actually in the HSC exam of my year. That's worth considering. That being said, don't recall if it appeared before that.

With chain-branching vs stiffening the important thing is you know roughly what they look like first. LDPE vs HDPE is a pretty good example. That being said, you should know a few properties (e.g. flexible, transparent v.s. stiff, transluscent).
This has a somewhat high tendency of appearing in half yearly exams, but only an average tendency in HSC exams.

I'll leave Jake for that general question. I didn't really study that dot point in depth; I just knew that it was true and went with it.