VCE Chemistry Question Thread

[lzxnl]

If a substance absorbs less strongly to the stationary phase in TLC does it move further or less from the solvent front?

Generally, it moves less away from the solvent front; as it'd be swept along more by the solvent, it would get  closer to the solvent front.

[Reus]

I'm so screwed for chem

[LiquidPaperz]

few questions here please

1) for V1, in the first attachment why have they added the values?
2) for V2, in the second attachment why have they added the values?  both 1) and 2) would you also be able to add when we know when to add it to v1 or to v2
3) how do i do question 24?

thanks

[ikiwi]

1) for V1, in the first attachment why have they added the values?

V1 is the total volume of the container when the valve is opened because when the valve is opened, the gases are able to take up the space of both containers

2) for V2, in the second attachment why have they added the values?  both 1) and 2) would you also be able to add when we know when to add it to v1 or to v2

As with the first question, V1 is the original volume of the container in which the gas is in. For O2, its originally in the 3L container. For He, its originally in the 5L container. V2 is the final volume, encompassing both containers because the valve has been opened. To get V2, you have to find the total volume, which is the two original volumes added together.

3) how do i do question 24?

Draw the diagram first. Use the working out they showed in the examples except with the different values.

[LiquidPaperz]

V1 is the total volume of the container when the valve is opened because when the valve is opened, the gases are able to take up the space of both containers
As with the first question, V1 is the original volume of the container in which the gas is in. For O2, its originally in the 3L container. For He, its originally in the 5L container. V2 is the final volume, encompassing both containers because the valve has been opened. To get V2, you have to find the total volume, which is the two original volumes added together.
Draw the diagram first. Use the working out they showed in the examples except with the different values.

Ok thanks for that. I thought V1 was the initial volume and V2 the final volume? why is V1 the total volume and V2 the final volume? im a bit confused on this aspect

thanks

[ikiwi]

I thought V1 was the initial volume and V2 the final volume? why is V1 the total volume and V2 the final volume?

You're right. It's just that in this case, the initial volume is equal to the total volume because the syringe is fully open at the start.

[grannysmith]

Determine the new pressure when a 2.0 L flask at 101 kPa is connected to a 1.0 L flask at 202 kPa. Assume constant temperature.

Edit: never mind, got it

[psyxwar]

VCAA Unit 4 2010, q1 b) ii).

Energy density may be defined as the amount of energy released per gram of fuel. Use molar enthalpy of
combustion data to calculate the energy density of hydrogen gas in kJ g–1.
Can someone explain why the answer is 143kJ/g, and not 71.5kJ/g?

Data book reports delta H as -286kJ/mol, balanced combustion equation is 2H2 + O2 -> 2H2O

Therefore 2 mol of H2 will combust to release 286kJ of energy, meaning 4g releases this much.

Energy density = 286/4 = 71.5kJ/g?

Unless the delta H reported is for H2 + 1/2 O2 -> H2O, but I was under the impression its for the balanced equation with integer coefficients? O__O

[ikiwi]

The molar heat of combustions in the data book is for the equation where the co-efficient of the fuel is 1, so its for the equation: H2 + 1/2 O2 -> H2O

[psyxwar]

The molar heat of combustions in the data book is for the equation where the co-efficient of the fuel is 1, so its for the equation: H2 + 1/2 O2 -> H2O

waiiit seriously? wtf ok thanks, that clears up a lot @__@.

[nhmn0301]

VCAA Unit 4 2010, q1 b) ii).

Energy density may be defined as the amount of energy released per gram of fuel. Use molar enthalpy of
combustion data to calculate the energy density of hydrogen gas in kJ g–1.
Can someone explain why the answer is 143kJ/g, and not 71.5kJ/g?

Data book reports delta H as -286kJ/mol, balanced combustion equation is 2H2 + O2 -> 2H2O

Therefore 2 mol of H2 will combust to release 286kJ of energy, meaning 4g releases this much.

Energy density = 286/4 = 71.5kJ/g?

Unless the delta H reported is for H2 + 1/2 O2 -> H2O, but I was under the impression its for the balanced equation with integer coefficients? O__O

I think you might misunderstand the meaning of kJ/mol. The "mol" here does not necessarily refer to the mole of any particular substance in the equation, it refers to the sum of all the substances, aka mole of the whole reaction. Hence, in your integer balanced equation, we have 2 x 286 kJ of energy released from H2 and 1 x 286 kJ of energy released from O2. Same for fraction balance equation, we have 1 mole of H2 release 286 kJ and O2 release 143 kJ. Should give you the same answer no matter how you balance it.

[psyxwar]

I think you might misunderstand the meaning of kJ/mol. The "mol" here does not necessarily refer to the mole of any particular substance in the equation, it refers to the sum of all the substances, aka mole of the whole reaction. Hence, in your integer balanced equation, we have 2 x 286 kJ of energy released from H2 and 1 x 286 kJ of energy released from O2. Same for fraction balance equation, we have 1 mole of H2 release 286 kJ and O2 release 143 kJ. Should give you the same answer no matter how you balance it.

Yeah I am aware. I thought the 286kJ/mol was for the thermochemical equation 2H2 + O2 -> 2H2O instead of the equation H2 + 1/2O2 -> H2O

[nhmn0301]

Yeah I am aware. I thought the 286kJ/mol was for the thermochemical equation 2H2 + O2 -> 2H2O instead of the equation H2 + 1/2O2 -> H2O

sorry I misinterpret !

[Yacoubb]

VCAA Unit 4 2010, q1 b) ii).

Energy density may be defined as the amount of energy released per gram of fuel. Use molar enthalpy of
combustion data to calculate the energy density of hydrogen gas in kJ g–1.
Can someone explain why the answer is 143kJ/g, and not 71.5kJ/g?

Data book reports delta H as -286kJ/mol, balanced combustion equation is 2H2 + O2 -> 2H2O

Therefore 2 mol of H2 will combust to release 286kJ of energy, meaning 4g releases this much.

Energy density = 286/4 = 71.5kJ/g?

Unless the delta H reported is for H2 + 1/2 O2 -> H2O, but I was under the impression its for the balanced equation with integer coefficients? O__O

I usually just divide the molar heat of combustion value by the molar mass of the species under analysis and this gives you the energy density. So I would have gone 286kJ/mol * 1mol/2g (recipricol of 2g/mol) = 143J/g. Would that have impacted anything? Because I've never had this problem, and I presume it's because I do it ^ this way.

[0KxBaNa2]

Help with b ii) please

I understand that removal of products would have caused the back reaction to drop but why would the forward reaction rate decrease? Wouldn't it increase as the system shifts right to increase [products]?