HSC 2016 Chemistry Marathon (2 Viewers)

[Flop21]

Re: HSC Chemistry Marathon 2016

Calculate the number of oxygen atoms present in 0.843 mol of silver sulfate

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

Calculate the number of oxygen atoms present in 0.843 mol of silver sulfate

The formula for silver sulfate is Ag₂SO₄. Knowing this, this question becomes similar to the previous ones you posted here.

 

[Flop21]

Re: HSC Chemistry Marathon 2016

The formula for silver sulfate is Ag₂SO₄. Knowing this, this question becomes similar to the previous ones you posted here.

Can you help please, instead of 0.843 mol of silver sulfate, let's do 0.697 mol.

So is the calculation...

4(#of oxygen atoms per mole of silver sulfate) *217.33157 (g of silver sulfate we are looking at)
___________
311.81 (molar mass of silver sulfate)

= 2.788

what is this result in? do I use this to * by the mole number?

 

[leehuan]

Re: HSC Chemistry Marathon 2016

What.

4 moles of oxygen are in 1 mol of silver sulfate
Therefore 4*0.697 = 2.788mol of oxygen is present in 0.697mol of silver sulfate

So moles of oxygen = 2.788mol

Thus the number of atoms in oxygen is just 2.788 * 6.022*10^-23 because by definition

Once you have the moles and you want the number of atoms going back to the mass is pointless. Rather, if they gave you the mass then you'd want to use n=m/M to get the value for n

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

Can you help please, instead of 0.843 mol of silver sulfate, let's do 0.697 mol.

So is the calculation...

4(#of oxygen atoms per mole of silver sulfate) *217.33157 (g of silver sulfate we are looking at)
___________
311.81 (molar mass of silver sulfate)

= 2.788

what is this result in? do I use this to * by the mole number?

Had a closer look at the Q. The answer is simply 4*0.697*N_A, where N_A is Avogadro's number, which is on the HSC Chemistry formula sheet.

This is because in 0.697 mol of silver sulfate, there are 4*0.697 mol of oxygen atoms (from the silver sulfate formula).

Now that we know how many moles of oxygen atoms there are, we just multiply this by Avogadro's number to find the actual number of oxygen atoms.

Edit: Done above

 

[Speed6]

Re: HSC Chemistry Marathon 2016

 

[DatAtarLyfe]

HSC Chemistry Marathon 2016

Edit: i was meant to say the answer is D, 0.38kg

 

[Speed6]

Re: HSC Chemistry Marathon 2016

Edit: i was meant to say the answer is D, 0.38kg

Thank you!

 

[Flop21]

Re: HSC Chemistry Marathon 2016

Can someone help me with this?

I got 3.2421g, but not sure if it's correct.


And soz if I'm asking about chem a lot, but I get not answers to any tut problems and they only go through a tiny amount of problems in the tut. And they don't explain how to do these questions anywhere else.

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

Can someone help me with this?

I got 3.2421g, but not sure if it's correct.


And soz if I'm asking about chem a lot, but I get not answers to any tut problems and they only go through a tiny amount of problems in the tut. And they don't explain how to do these questions anywhere else.

Using the formula c = n/v, you rearrange it to get n = c x v, then using the molar mass given you just use the formula m = n x mw (this is obviously rearranged from n / m/mw)

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

Can someone help me with this?

I got 3.2421g, but not sure if it's correct.


And soz if I'm asking about chem a lot, but I get not answers to any tut problems and they only go through a tiny amount of problems in the tut. And they don't explain how to do these questions anywhere else.

The solution's concentration of sodium carbonate is 0.2296 mol/L, and there is 0.365 L of it, so the quantity of sodium carbonate used is 0.2296 mol/L * 0.365 L = 0.083804 mol.

Now that we know the quantity of sodium carbonate used to prepare Solution A, since we are given its molar mass, we can easily find the mass required using mass = (quantity)*(molar mass).

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

So the working out is 0.2296 x (0.365) as you have to have it in litres rather that milliliters, which = 0.08 (2.sig.), then 0.08 x 105.99 = 8.4792

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

The solution's concentration of sodium carbonate is 0.2296 mol/L, and there is 0.365 L of it, so the quantity of sodium carbonate used is 0.2296 mol/L * 0.365 L = 0.083804 mol.

Now that we know the quantity of sodium carbonate used to prepare Solution A, since we are given its molar mass, we can easily find the mass required using mass = (quantity)/(molar mass).

Sorry InteGrand, I know you helped me with alot of questions in the past but you're wrong there

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

Sorry InteGrand, I know you helped me with alot of questions in the past but you're wrong there

Oops careless me. I meant *, not / at the end. Thanks!

 

[Flop21]

Re: HSC Chemistry Marathon 2016

Thanks guys seems I was misunderstanding how to use the formula.

Now for part b:

So like we did before 0.2296*0.095 = 0.021812 mol of Na2CO3, then because there's 2 sodium ions for every mol of that - we * 2 to get 0.043624 mol.

Is this correct?

And for carbonate ions, it would be * 3 right?

 

[leehuan]

Re: HSC Chemistry Marathon 2016

Thanks guys seems I was misunderstanding how to use the formula.

Now for part b:

So like we did before 0.2296*0.095 = 0.021812 mol of Na2CO3, then because there's 2 sodium ions for every mol of that - we * 2 to get 0.043624 mol.

Is this correct?

And for carbonate ions, it would be * 3 right?

Yes to the sodium part.

No to the carbonate. There's only one carbonate ion present. If you meant oxygen though you would have to *3 because obviously each carbonate ion has 3 oxygen atoms within it.

 

[Flop21]

Re: HSC Chemistry Marathon 2016

Yes to the sodium part.

No to the carbonate. There's only one carbonate ion present. If you meant oxygen though you would have to *3 because obviously each carbonate ion has 3 oxygen atoms within it.

Oh yeah of course, thanks!

 

[zeefar]

Re: HSC Chemistry Marathon 2016

This is irrelevant but I do prelim chem and can't seem to grasp my head around redox reactions I understand oxidation is loss reduction is gain but my teacher didn't really describe it in depth anyone have any adequate resource to help this?

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

This is irrelevant but I do prelim chem and can't seem to grasp my head around redox reactions I understand oxidation is loss reduction is gain but my teacher didn't really describe it in depth anyone have any adequate resource to help this?

What are you having trouble with?

 

[leehuan]

Re: HSC Chemistry Marathon 2016

Here's an example

Ag₂SO_4(aq) + Ba_(s) -> BaSO_4(aq) + 2 Ag_(s)

We note that silver ions are being REDUCED to silver metal.
Ag⁺ + e^- -> Ag_(s)

This is because silver ions GAINED electrons to become silver metal.
(Note that the electron was offered by barium)

Conversely, we note that barium metal is being OXIDISED into barium ions.
Ba_(s) -> Ba²⁺ + 2e^-

This is because barium metal LOST electrons to become barium ions.
(Note that indeed the electrons were donated to silver)