HSC Chemistry Question Thread

[Kekemato_BAP]

just out of curiosity ( i don't take chem but this is somewhat related to bio test and i don't have a chem background)
what would happen if you heated hydrogen peroxide to like 100°C?

Cool question! Hydrogen peroxide is an unstable compound of H2O2 which slowly decomposes into H2 gas and O2 gas at room temperature. The drastic increase in heat rapidly speeds up the decomposition rate. Even though the boiling point is 150'C, it would likely have already decomposed into the explosive mixture of hydrogen and oxygen gas.

[Kekemato_BAP]

Hi I was having a bit of trouble with the following 2 questions, I was hoping you could explain them to me \
1. A compound has a percent composition of 38.1% C, 6.3% H and 55.6% Cl and a molar mass of approximately 125 g/mol. What is the empirical and molecular formula?
2. What is the concentration of a sodium carbonate solution prepared by diluting 10.0mL of a 5.0 M Na2CO3 solution to 250mL?
Thanks!

Hey! The second question has already been answered, but not the first one. Here's my answer:
First put the percentages as grams.
38.1g C, 6.3g H, 55.6g Cl
Now you can convert these into moles.
38.1/12.01= ~3.17
6.3/1.008= ~5.95
55.6/35.45= ~1.57
It can be seen that the moles of C, H, and Cl are about a 3:6:1.5 ratio if you round them up. But you can't have 1.5 of a Cl, so it would be doubled to 6:12:3 which simplifies into 2:4:1.
Therefore, the empirical formula is C2H4Cl.
The molar mass of C2H4Cl is ~63.5g/mol and the molar mass is supposed to be 125g/mol. Thus, the molecular formula is C4H8Cl2 because 63.5 is about half of 125.

Hope this makes sense!!

[bananna]

Hi
Please help!! 
If during combustion an alcohol lost 30% of heat to the environment, would I multiply the MOLES by 70/100 or is it totally different

Thank youuuuuu
Much appreciated

[Kekemato_BAP]

Hi
Please help!! 
If during combustion an alcohol lost 30% of heat to the environment, would I multiply the MOLES by 70/100 or is it totally different

Thank youuuuuu
Much appreciated

Hi! If the combustion loses 30% of its heat, this means that the recorded value of heat is 70% of the actual value of heat. Divide the recorded value by 7 then multiply by 10. That should work in giving you the actual heat of combustion. Divide this value by the number of moles of fuel consumed in the combustion to get your kJ/mol.
Hope this helps

[bananna]

Hi! If the combustion loses 30% of its heat, this means that the recorded value of heat is 70% of the actual value of heat. Divide the recorded value by 7 then multiply by 10. That should work in giving you the actual heat of combustion. Divide this value by the number of moles of fuel consumed in the combustion to get your kJ/mol.
Hope this helps

is there any way you could put it in terms of multiplying ?
 like value multiplied by 30/100 or 70/100 ?
my teacher explained it once like that and I got it
but now i forgot
thanks

[Kekemato_BAP]

is there any way you could put it in terms of multiplying ?
 like value multiplied by 30/100 or 70/100 ?
my teacher explained it once like that and I got it
but now i forgot
thanks

The "theoretical" heat of combustion is the heat released without any loss. It is measured in kJ/mol or J/mol such as ethanol 1360kJ/mol.
I don't think there is a formula for it, but I made this little formula for calculating heat of combustion:

Finding actual heat (heat with loss):
Theoretical x [(100-%)/100]
%=how much was lost

Finding theoretical heat (heat without loss):
(Actual) / [(100-%)/100]
%=how much was lost

For example, you combust 1 mole of ethanol with 30% of heat lost to the environment. Ethanol=1360kJ/mol.
To calculate the actual heat, this is 1360kJ x [(100-30%)/100]
=1360kJ x 70/100
=952kJ
Therefore, the heat absorbed by the thing being heated is 952kJ.
Like you said, you multiply it by 70/100. (100-%/100)

Hope this makes sense!!

[Bubbly_bluey]

Hey guys I need help with these questions especially Q14.
(I'm having trouble uploading the picture vv)
13)The gases sulfur dioxide and hydrogen sulfide reacted when mixed, to form sulfur and water. SO₂(g) + 2H₂S (g) --> 3S (s) + 2H₂O (l)
What is the role of hydrogen sulfide in the above reaction?
a) Bronsted Lowry acid
b) Arrhenius acid
c)oxidant
d) reactant

14) What mass of sulfur is formed when 2L of sulfur dioxide is mixed with 2L of hydrogen sulfide, under 273K and100kPa?
a) 1.5g
b)2g
c)4g
d) 8g
Thanks

[thenerdygangster]

Hey guys I need help with these questions especially Q14.

Not too sure about this question but I'll give it a crack. For 13, you'll need to keep the basic definitions of both Arrhenius acid and Bronsted Lowry acid in your head. An Arrhenius acid is any species that can increase the concentration of H⁺ in an aqueous solution. A Bronsted Lowry acid can be used even without an aqueous solution present. Oxidants are electron acceptors and oxygen donators so you can cross (c) out . I would say since the two reactants are gases, H₂S would be a Bronsted Lowry acid.

For 14, you'll have to figure out which reactant is the limiting reagent. In this case, you have 1 mole of SO₂ reacting with 2 moles of H₂S. This means that for every one mole of SO₂ used, 2 moles of H₂S would be used as well. So finding the number of moles of each reactant (in this particular reaction), you can figure out the limiting reactant (think Ideal Gas Law; PV = nRT). From there, use stoichiometry to figure out the mole of S formed and consequently, the mass.

Hope that helps 

[Kekemato_BAP]

Hey guys I need help with these questions especially Q14.
(I'm having trouble uploading the picture vv)
13)The gases sulfur dioxide and hydrogen sulfide reacted when mixed, to form sulfur and water. SO₂(g) + 2H₂S (g) --> 3S (s) + 2H₂O (l)
What is the role of hydrogen sulfide in the above reaction?
a) Bronsted Lowry acid
b) Arrhenius acid
c)oxidant
d) reactant

14) What mass of sulfur is formed when 2L of sulfur dioxide is mixed with 2L of hydrogen sulfide, under 273K and100kPa?
a) 1.5g
b)2g
c)4g
d) 8g
Thanks

Great question!
13) I'm not that great at the topic, but the hydrogen sulfide acts as a base because sulfur dioxide is an acidic oxide, and the solid sulfur is the salt equivalent in the neutralisation (?)
14) This one I like
SO2 + 2H2S -> 3S + 2H2O
(1:2:3:2 stoichiometric mole ratio)
(n=moles)
v(SO2)= 2L
n(SO2)= 2L / 22.71L/mol (273K=0'C)
=0.0881mol
1:3 mole ratio, n(S)= 0.2642mol
m(S)= n*MM = 0.2642mol*32.07g/mol
=8.4g (round to 8g)
Therefore, the answer is D!!

Hope this helps sorry I couldn't answer the first one properly!!

[Bubbly_bluey]

Not too sure about this question but I'll give it a crack. For 13, you'll need to keep the basic definitions of both Arrhenius acid and Bronsted Lowry acid in your head. An Arrhenius acid is any species that can increase the concentration of H⁺ in an aqueous solution. A Bronsted Lowry acid can be used even without an aqueous solution present. Oxidants are electron acceptors and oxygen donators so you can cross (c) out . I would say since the two reactants are gases, H₂S would be a Bronsted Lowry acid.

Hope that helps 

Yea i agree both Arrhenius acid and oxidants are incorrect. I was stuck between reductant and bronsted lowry. Maybe I missed apoint but could you explain why its bronsted lowrys? Because I was thinking since I dont know which reactant is acting as an acid, so I looked on to how sulfur dioxide has been reduce??
Thanks for the help tho! much appreciated

Great question!

14) This one I like
SO2 + 2H2S -> 3S + 2H2O
(1:2:3:2 stoichiometric mole ratio)
(n=moles)
v(SO2)= 2L
n(SO2)= 2L / 22.71L/mol (273K=0'C)
=0.0881mol
1:3 mole ratio, n(S)= 0.2642mol
m(S)= n*MM = 0.2642mol*32.07g/mol
=8.4g (round to 8g)
Therefore, the answer is D!!

Hope this helps sorry I couldn't answer the first one properly!!

YES! thats what i did at first(taking the stoichiometric ratios of SO₂: 3S) but when you take the ratios of 2H₂S: 3S its a different answer (which is apparently the correct one C). But I don't understand why you have the only consider a certain ratio   

[kiwiberry]

YES! thats what i did at first(taking the stoichiometric ratios of SO₂: 3S) but when you take the ratios of 2H₂S: 3S its a different answer (which is apparently the correct one C). But I don't understand why you have the only consider a certain ratio   

We're given the volumes of both reactants, so we have to find the limiting reagent first. In this case, n(H2S) = n(SO2) = 0.0881.. mol. However, we have to consider the 1:2 stoichiometric ratio- 2 moles of H2S are required to react with 1 mole of SO2. Hence SO2 will be in excess and H2S will be the limiting reagent, meaning we must use the 2H₂S:3S ratio when calculating

[Kekemato_BAP]

We're given the volumes of both reactants, so we have to find the limiting reagent first. In this case, n(H2S) = n(SO2) = 0.0881.. mol. However, we have to consider the 1:2 stoichiometric ratio- 2 moles of H2S are required to react with 1 mole of SO2. Hence SO2 will be in excess and H2S will be the limiting reagent, meaning we must use the 2H₂S:3S ratio when calculating

Yea i agree both Arrhenius acid and oxidants are incorrect. I was stuck between reductant and bronsted lowry. Maybe I missed apoint but could you explain why its bronsted lowrys? Because I was thinking since I dont know which reactant is acting as an acid, so I looked on to how sulfur dioxide has been reduce??
Thanks for the help tho! much appreciated
YES! thats what i did at first(taking the stoichiometric ratios of SO₂: 3S) but when you take the ratios of 2H₂S: 3S its a different answer (which is apparently the correct one C). But I don't understand why you have the only consider a certain ratio   

Yes!!! It actually is C!! I didn't read the full question sorry!! I didn't see that there was 2L of H2S which would have been a limiting reagent. My calculations were correct, but completely ignored the 2L of H2S sorry!!

[Alalamc]

Now you can convert these into moles.
38.1/12.01= ~3.17
6.3/1.008= ~5.95
55.6/35.45= ~1.57

Thanks so I am actually doing a chemistry online course because I haven't done it in yr 11/12, so do you have to round up the moles to the nearest .5 before you start to take them into consideration for the subscripts?

[Kekemato_BAP]

Thanks so I am actually doing a chemistry online course because I haven't done it in yr 11/12, so do you have to round up the moles to the nearest .5 before you start to take them into consideration for the subscripts?

It depends.. sometimes the values are exact whole numbers like 2, 4, 3. Sometimes, they're like 1.98, 4.2, 2.99; and you just have to round them to their nearest half-number like 1, 2, 1.34. You have round to nearest .5 and then make them all whole numbers, eg. 1, 2, 1.5 then change to 2, 4, 3 because you can't have half an atom.
Hope this makes sense

[ellipse]

Just to confirm, is the water and ethanol gas or liquid during hydration and dehydration process?