VCE Chemistry Question Thread

[Syndicate]

Hi I don't understand Q10 of the 2015 VCAA exam at all. like how should i even start a feul cell questions like this?
Can someone please help me?  Thanks

Hi lovebiology,

Do you need help with all of it? or are there certain parts you don't understand?

Is there a way to write and balance the half equations in a fuel cell? Or will I always be given them?

I am not exactly sure what you really mean, however, in most cases you do have to work out the half equations given that you have enough information provided (ie. the electrolyte, type of fuel cell, reactants etc...).

[vcestressed]

Are all ionic compounds polar? I'm so confused 

[sweetiepi]

Are all ionic compounds polar? I'm so confused 

Ionic compounds are made of fully charged particles- i.e. NaCl: Na⁺ and Cl^-.
Polar compounds are made of covalent compounds that have a "dipole" (partial charge) on each atom that doesn't fully cancel out or an "ionic-dipole" compound- a fusion of a fully charged and partial charge. Therefore, I believe not all ionic compounds are polar.

(Excuse my super unclear response- I'm out and about lol)

[Syndicate]

Are all ionic compounds polar? I'm so confused 

Ionic compounds can neither be polar or non-polar. Polarity is related to covalent bonding (non-metal to non-metal), whereas ionic bonds are between atoms that metals and non-metals (ie. NaCl). (I don't know what really happens in advanced chemistry, but for high school chemistry, this is all you need to know).

To summarise, polar and ionic bonds are two different types of bonds.

[sweetiepi]

Ionic compounds can neither be polar or non-polar. Polarity is related to covalent bonding (non-metal to non-metal), whereas ionic bonds are between atoms that metals and non-metals (ie. NaCl). (I don't know what really happens in advanced chemistry, but for high school chemistry, this is all you need to know).

+1 to this, I did use some of bioorganic knowledge for my response.

[QueenSmarty]

Hello, may I please get some help with these questions?
1. The complete combustion of propane is represented by the thermochemical equation: C3H8+5O2 --> 3CO2+4H2O, (delta)H=-2220kJ per mol. Calculate the energy released per tonne of carbon dioxide produced when propane undergoes complete combustion.

2. An inexperienced hiker wishes to use his lighter, which contains 3g of butane, to heat the water in his mug. If his mug contains 150mL of water initially at 20 degrees Celsius, and 70% of the heat generated by the lighter is lost to surroundings, what will be the temp of the water when the fuel in the lighter is exhausted?

[Syndicate]

Hello, may I please get some help with these questions?
1. The complete combustion of propane is represented by the thermochemical equation: C3H8+5O2 --> 3CO2+4H2O, (delta)H=-2220kJ per mol. Calculate the energy released per tonne of carbon dioxide produced when propane undergoes complete combustion.

2. An inexperienced hiker wishes to use his lighter, which contains 3g of butane, to heat the water in his mug. If his mug contains 150mL of water initially at 20 degrees Celsius, and 70% of the heat generated by the lighter is lost to surroundings, what will be the temp of the water when the fuel in the lighter is exhausted?

1).
Step 1: Calculate the mol of CO2 by using the formula n = m/M

 n(CO2) = \( \frac{10^6}{44} \) =22727.27 mol

Step 2: Using stoichiometric ratios, calculate the mol of propane

\( \frac{n(propane)}{n(CO2} = \frac{1}{3} \implies n(propane) = \frac{n(CO2)}{3} \) = 7575.76 mol

Step 3: Use the formula H = n x Hc to calculate energy release.

H = 7575.76 x 2220 = 16.82 GJ

Therefore 16.82 Gj of energy is released per tonne of CO2 emitted.

2)
Step 1: write the thermochemical equation for butane (it isn't necessary but may help if there are questions that follow).

2C4H10 (l) + 13O2(g) -> 8CO2(g) + 10H2O(l) (low pressure so water is liquid) H = -5748 kJ mol^-1

Step 2: Calculate the mol of butane
n(butane) = 3/58 = 0.0517 mol

Step 3: Calculate energy release

H = 0.0517 x 2784 = 144 kJ

Step 4: Determine useful energy

144 x 0.3 = 43.2 kJ = 43200 J

St3p 5: Use Q =mcT

43200= 150 x 4.18 x (Tf - 20)

Tf - 20 = 68.9
Tf = 88.9 Celsius

[QueenSmarty]

1).
Step 1: Calculate the mol of CO2 by using the formula n = m/M

 n(CO2) = \( \frac{10^6}{44} \) =22727.27 mol

Step 2: Using stoichiometric ratios, calculate the mol of propane

\( \frac{n(propane)}{n(CO2} = \frac{1}{3} \implies n(propane) = \frac{n(CO2)}{3} \) = 7575.76 mol

Step 3: Use the formula H = n x Hc to calculate energy release.

H = 7575.76 x 2220 = 16.82 GJ

Therefore 16.82 Gj of energy is released per tonne of CO2 emitted.

2)
Step 1: write the thermochemical equation for butane (it isn't necessary but may help if there are questions that follow).

2C4H10 (l) + 13O2(g) -> 8CO2(g) + 10H2O(l) (low pressure so water is liquid) H = -5748 kJ mol^-1

Step 2: Calculate the mol of butane
n(butane) = 3/58 = 0.0517 mol

Step 3: Calculate energy release

H = 0.0517 x 2784 = 144 kJ

Step 4: Determine useful energy

144 x 0.3 = 43.2 kJ = 43200 J

St3p 5: Use Q =mcT

43200= 150 x 4.18 x (Tf - 20)

Tf - 20 = 68.9
Tf = 88.9 Celsius

Thank you!!

[Ahmad_A_1999]

Could I please get help with this question

Part D, i) and ii)

http://imgur.com/wmkWapT

[LPadlan]

100mL of ethane cas was mixed with 500 mL oxygen gas at 20C and 1 atmosphere pressure and sparked. The reaction shown by the following equation occurred: 2C2H6(g) + 7O2(g) ----> 4CO2(g) + 6H2O(l)
What would be the total volume of gas present when the reaction had cooled to the original temperature and pressure?

A. 200 mL B. 350mL C. 550mL D. 650mL
this is question 27, page 7 in checkpoints 2017

[cookiedream]

100mL of ethane cas was mixed with 500 mL oxygen gas at 20C and 1 atmosphere pressure and sparked. The reaction shown by the following equation occurred: 2C2H6(g) + 7O2(g) ----> 4CO2(g) + 6H2O(l)
What would be the total volume of gas present when the reaction had cooled to the original temperature and pressure?

A. 200 mL B. 350mL C. 550mL D. 650mL
this is question 27, page 7 in checkpoints 2017

Here's my working out. Excuse my poor handwriting!

[cookiedream]

Could I please get help with this question

Part D, i) and ii)

http://imgur.com/wmkWapT

Here's my working with explanations. Hope it helps!

[LPadlan]

Here's my working out. Excuse my poor handwriting!

Hey, can you please clarify on what you were doing on the top of the page? Thanks

[cookiedream]

Hey, can you please clarify on what you were doing on the top of the page? Thanks

Working out mole ratios, using the coefficients of the compounds in the reaction.
You then use these to work out excess and limiting reagents.

[Ahmad_A_1999]

Here's my working with explanations. Hope it helps!

Hey thanks for that! Had the SAC today on fuel and galvanic cells, went pretty good I think!