HSC Chemistry Question Thread

[anotherworld2b]

Hi I was wondering could I get help in calculating the molar mass and composition of sodium bicarbonate?
Could I get a reply as soon as possible? I have to hand in my assignment tomorrow 
We used 4g of NaHCO3 and got 3.1g of NaCl

NaHCO3(s) +  HCl(aq) → NaCl (s) + H2O(l) + CO2(g)

Info I have: ( not entirely sure its correct)
1. Mass of tube + NaHCO3
51.6g
2. Mass of tube
47.6g
3. Mass of NaHCO3
4g
4. Moles of NaHCO3
0.0459 gmol -1
5. Mass of tube + NaCl
50.7g
6. Mass of NaCl
3.1g
7. Moles of NaCl
0.0530 gmol -1
Mass of heating basin + water
71.5g
Mass of water
20.8g
Mass of carbon dioxide
23.9g
Ratio: Moles NaCl /Moles NaHCO3
0.0530 : 0.0459

[amandali]

can you check my response thanks

In this option you studied one natural product that was not a fossil fuel. Describe the issues associated with shrinking world supplies of this natural product, and evaluate progress being made to solve the problems identified. 7 marks
A natural product is one that is used with little or no modification. An example include raw rubber which is a naturally occurring addition polymer formed from the polymerisation of 2-methyl-1,3-butadiene (isoprene) and obtained from sap of rubber trees. Many issues arise as its supply starts to shrink thus a replacement is required. In the early 20th century, the demand for rubber outstripped supply as rubber trees can only produce a certain amount of rubber each year. Rubber was needed for tyres for military vehicles thus its limited supply adversely affected countries’ performance at war. In ww2, Japan had control of rubber producing areas which threatened the supply of natural rubber in other countries. Moreover, automobile industry was growing and car manufacturers needed rubber. Due to the rising in demand for rubber, a first replacement for rubber is produced which is Styrene Butadiene rubber (SBR) formed from monomers of 1,3-butadiene and styrene. It is vulcanised with short Sulfur chains forming cross-links between polymer chains. It is more favourable than raw rubber as it has the improved properties ie. more durable, more resistant to chemical attack, and stronger. The progress made solved the problems as it increased the supply of rubber such that its demand is being met and cost is able to be maintained low. 

[jakesilove]

Hi I was wondering could I get help in calculating the molar mass and composition of sodium bicarbonate?
Could I get a reply as soon as possible? I have to hand in my assignment tomorrow 
We used 4g of NaHCO3 and got 3.1g of NaCl

NaHCO3(s) +  HCl(aq) → NaCl (s) + H2O(l) + CO2(g)

Info I have: ( not entirely sure its correct)
1. Mass of tube + NaHCO3
51.6g
2. Mass of tube
47.6g
3. Mass of NaHCO3
4g
4. Moles of NaHCO3
0.0459 gmol -1
5. Mass of tube + NaCl
50.7g
6. Mass of NaCl
3.1g
7. Moles of NaCl
0.0530 gmol -1
Mass of heating basin + water
71.5g
Mass of water
20.8g
Mass of carbon dioxide
23.9g
Ratio: Moles NaCl /Moles NaHCO3
0.0530 : 0.0459

Hey,

Unfortunately, I can't do calculations for you, and I won't be able to get to this properly tonight. I definitely can't do your assignments for you. All I can recommend is having a proper think about what you're actually trying to find, and utilising the equations at your disposal to the best of your ability.

Good luck,

Jake

[jakesilove]

can you check my response thanks

In this option you studied one natural product that was not a fossil fuel. Describe the issues associated with shrinking world supplies of this natural product, and evaluate progress being made to solve the problems identified. 7 marks
A natural product is one that is used with little or no modification. An example include raw rubber which is a naturally occurring addition polymer formed from the polymerisation of 2-methyl-1,3-butadiene (isoprene) and obtained from sap of rubber trees. Many issues arise as its supply starts to shrink thus a replacement is required. In the early 20th century, the demand for rubber outstripped supply as rubber trees can only produce a certain amount of rubber each year. Rubber was needed for tyres for military vehicles thus its limited supply adversely affected countries’ performance at war. In ww2, Japan had control of rubber producing areas which threatened the supply of natural rubber in other countries. Moreover, automobile industry was growing and car manufacturers needed rubber. Due to the rising in demand for rubber, a first replacement for rubber is produced which is Styrene Butadiene rubber (SBR) formed from monomers of 1,3-butadiene and styrene. It is vulcanised with short Sulfur chains forming cross-links between polymer chains. It is more favourable than raw rubber as it has the improved properties ie. more durable, more resistant to chemical attack, and stronger. The progress made solved the problems as it increased the supply of rubber such that its demand is being met and cost is able to be maintained low.

Hey! I didn't do this area of Chemistry, but I think your answer is spot on! The only thing I would mention is that the last part of the question asks you to 'evaluate'. When you talk about SBR, you've mentioned its benefits, but are there any detrimental aspects? Impact on the environment, cost effectiveness, timeline etc? It is good to mention something like this, even if you decide that, overall, the process is beneficial.

Jake

[zoeh]

Hi,
I just have a few questions about electrolysis because it wasn't really explained well in class…
When the electrodes used are inert how does this impact on the reaction? Is the anode and cathode still the 2 electrodes? And where does the oxidation occur? For example if graphite electrodes were used in copper sulfate electrolyte would copper reduce at the cathode and what would be oxidised?
thanks so much

[jakesilove]

Hi,
I just have a few questions about electrolysis because it wasn't really explained well in class…
When the electrodes used are inert how does this impact on the reaction? Is the anode and cathode still the 2 electrodes? And where does the oxidation occur? For example if graphite electrodes were used in copper sulfate electrolyte would copper reduce at the cathode and what would be oxidised?
thanks so much

Hey! Basically, if an electrode is inert, it is non-reactive (bit of a tautology there, sorry). It won't oxidise or reduce at any point in the reaction. Generally, Graphite or Platinum electrodes are used to signify that the electrode is to be inert. However, something else needs to be happening for this cell to move forward.
An inert electrode acts only as a location for electrons to move from one place to another, and thus for some kind of oxidation/reduction to occur. For instance, if you had a half-cell with a platinum electrode in water, the platinum doesn't actually do anything; it is the water that is oxidised or reduced! This goes for any solution in which an inert electrode is placed (the standard one used in the HSC, I believe, is bubbling Hydrogen gas through a liquid. The gas attaches to the Platinum electrode, is oxidised or reduced by the inflow/outflow of electrons, and changes states accordingly).

I hope that this clears things up for you! If you see an inert electrode, you need to look for something else to move the reaction forward. Then, compare this other substance to the other half cell, using your table of reduction potentials, in order to determine which side oxidises and which side reduces (as with any standard galvanic cell).

Let me know if I can clear any of this/anything else up!

Jake

[zoeh]

Hey! Basically, if an electrode is inert, it is non-reactive (bit of a tautology there, sorry). It won't oxidise or reduce at any point in the reaction. Generally, Graphite or Platinum electrodes are used to signify that the electrode is to be inert. However, something else needs to be happening for this cell to move forward.
An inert electrode acts only as a location for electrons to move from one place to another, and thus for some kind of oxidation/reduction to occur. For instance, if you had a half-cell with a platinum electrode in water, the platinum doesn't actually do anything; it is the water that is oxidised or reduced! This goes for any solution in which an inert electrode is placed (the standard one used in the HSC, I believe, is bubbling Hydrogen gas through a liquid. The gas attaches to the Platinum electrode, is oxidised or reduced by the inflow/outflow of electrons, and changes states accordingly).

I hope that this clears things up for you! If you see an inert electrode, you need to look for something else to move the reaction forward. Then, compare this other substance to the other half cell, using your table of reduction potentials, in order to determine which side oxidises and which side reduces (as with any standard galvanic cell).

Let me know if I can clear any of this/anything else up!

Jake

Thanks so much that clears it up!

[conic curve]

What are the most important things in prelim chem (other than chemical equations and moles) which are important in HSC chem (and will reappear in HSC chem)?

Thanks

[anotherworld2b]

I'm sorry if my question seemed like I was taking advantage of this thread.
I was unsure how to calculate molar mass, moles and ect and what formula I needed to use.
I am still quite confused about molar mass and moles calculations in general.
Could I have help in understanding when to use what formula and where to use it?

[jakesilove]

What are the most important things in prelim chem (other than chemical equations and moles) which are important in HSC chem (and will reappear in HSC chem)?

Thanks

There is very little actual crossover between the courses. In fact, I'd hesitate to say that there is absolutely none, whatsoever. What is important is a good understanding of chemical reactions (ie. Neutralization reactions, that kind of thing), a working knowledge of chemical terms (solutions, mixtures, moles, gas etc.) and not much else. Don't worry too much about the content you are learning now, and its relation to the HSC course. Just learn it for your exams, and then move on to the HSC curriculum later this year!

Jake

[jakesilove]

I'm sorry if my question seemed like I was taking advantage of this thread.
I was unsure how to calculate molar mass, moles and ect and what formula I needed to use.
I am still quite confused about molar mass and moles calculations in general.
Could I have help in understanding when to use what formula and where to use it?

That's not a problem

Okay, so let's do a brief refresher of moles, molar mass and mass, and how to calculate each from the other.

A mole is a collection of atoms (in fact, it is an exact number of atoms, Avagadro's number. This isn't important in the HSC, but knowing this makes understanding the concepts a lot easier). 2 moles is two times Avagadro's number of atoms, 1/2 a mole is 50% Avagadro's number etc. As you will imagine, each element's atom will have a different mass. So, if a single atom of Nitrogen was heavier than a single atom of Helium, you would expect a mole of Nitrogen to weigh more than a single mole of Helium (despite them having the same number of atoms!). The molar mass of a substance is the mass of one mole of the element.

So, let's put this all together into some calculations. The formula for relating all of this is



So, if we had 3g of Carbon, which has a molar mass of approximately 12g, then we have 1/4 moles of Carbon!

Easy.

Now, there's some more complicated maths that I think your assignment required you to do. Say you had a reaction between two substances. You can balance the equation easily, by adding numbers out the front of each reactant and product to make sure the same number of elements/atoms are on each side. A 2 signifies 2 moles, a 1 signifies 1 mole etc. Now, say you had a reaction that looks something like this:



We know that for any two moles of A, we need 1 mole of B to react, which will produce 1 mole of C. Therefore, if we know we have 2g of A, we can use the above formula to figure out how many moles that is (using the molar mass). We then know we need HALF as many moles of B, and can figure out the relevant mass of B (using it's molar mass). This is stuff that's really important to the HSC syllabus; I'm not sure what level you're at but hopefully this sounds familiar.

That's the basics of the whole thing. I could go into much more depth, but that really is a bulk of the skills in this course. If what I've written above doesn't make sense, or you're having trouble with anything, please feel free to post specific questions because at the moment it's hard to gauge where you're having problems. If none of this is making any sense, I would talk to your teacher about having a few lessons with them. It's a big topic area, and it's important you understand what's going on.

Hope that this helped, at least a little.

Jake

[MysteryMarker]

Hey guys, just got a past trial paper question that i don't really understand. I don;t know what the answer is but i've been told it is A. Can anyone explain to me how this is so, and if possible, maybe through the aid of a lewis electron dot diagram? Thanks Guys.

Which Reaction involved the formation of a coordinate covalent bond?

a)   H+ + OH- → H2O
b)   Mg + 2H+ →Mg2+ + H2
c)   C2H4 + H2 → C2H6
d)   CH4 + Br2 → CH3Br + HBr

I originally thought the answer was b as the magnesium atom technically 'provides' both electrons to form the bonding pair for the hydrogen atom.

Cheers broskies.

[RuiAce]

Hey guys, just got a past trial paper question that i don't really understand. I don;t know what the answer is but i've been told it is A. Can anyone explain to me how this is so, and if possible, maybe through the aid of a lewis electron dot diagram? Thanks Guys.

Which Reaction involved the formation of a coordinate covalent bond?

a)   H+ + OH- → H2O
b)   Mg + 2H+ →Mg2+ + H2
c)   C2H4 + H2 → C2H6
d)   CH4 + Br2 → CH3Br + HBr

I originally thought the answer was b as the magnesium atom technically 'provides' both electrons to form the bonding pair for the hydrogen atom.

Cheers broskies.

B is wrong because if you react an acid with a metal you get a salt plus the hydrogen gas. E.g.
Mg_(s) +2 HCl_(aq) → MgCl_2(aq) + H_2(g)

C is obviously wrong because alkene → alkane cannot possibly become a coordinate bond

D is wrong mainly because it's just the formation of a CFC (or rather, BCFC here).


The answer is A. It should have been established clearly that the species H₃O⁺, that is, the hydronium ion features a coordinate bond between one of the hydrogens and the oxygen. This is because the acid H⁺ isn't actually present. What's actually present is indeed, the hydronium ion. The more accurate way equation A is written is this:

H₃O⁺ + OH^- ⇌ 2 H₂O_(l)

[trixqwe]

Hi there! I'm a little stuck on how to answer this question and I would appreciate any help 

Compare an 'amphiprotic' substance and an 'amphoteric' substance. Use examples and equations to support your answer.

Thanks in advance!!

[RuiAce]

Hi there! I'm a little stuck on how to answer this question and I would appreciate any help 

Compare an 'amphiprotic' substance and an 'amphoteric' substance. Use examples and equations to support your answer.

Thanks in advance!!

Technically amphiprotic substances are all amphoteric, but not vice versa. Just like how all squares are rectangles.

Amphoteric just means it exhibits both acidic and basic properties. You can take Al₂O₃ and compare two reactions. When aluminium oxide reacts with an acid, it's exhibiting basic properties. When it reacts with a base, it's exhibiting acidic properties. (Don't remember the equations right now, but either Al₂O₃ or ZnO is included in the Jacaranda textbook.)

Amphiprotic specifically relates to B-L theory. That is, an acid is a proton donor, whereas a base is a proton acceptor. Amphiprotic substances are both. An example is your HCO₃^- ion (or if that's too inadequate, just consider NaHCO₃).
It's easy to show that if you react sodium hydrogen carbonate with hydrochloric acid, you get water, carbon dioxide and sodium carbonate. Whereas if you react it with sodium hydroxide, you get sodium carbonate and water.