VCE Chemistry Question Thread

[blacksanta62]

Thanks zsteve

[jyce]

Thanks jyce!

For concentration(ppm), my notes say that the solution has to be in kgs, so why don't we need to do anything to the denominator when going from g/L -> ppm?

This is why I said you should work on your understanding of these units, rather than trying to memorise them! ppm can be expressed in many different ways. You just need to ensure that the unit for the solute (i.e., the first unit) is a million times smaller than the unit for the solvent (i.e., the second unit). For example, mg L^-1, mg kg^-1 and mL kL^-1 are all expressions of ppm (because mg is a million times smaller than L, mg is a million times smaller than kg, and mL is a million times smaller than kL). To answer your question, specifically, you could change the denominator, or you could change the numerator.

Attached is an image of a slide from one of my presentations for Chemistry tutoring. It might help you get a conceptual understanding of ppm. Basically, if you arrange all different units as I have on this slide, any two units that are separated by only one unit in between will work as ppm, because all of these units are separated from one another by a factor of 1,000 and 1,000 x 1,000 = a million.

Oh and I'm a complete fool when it comes to technology, so I don't even know if my attachment will work 

[bedigursimran]

Hey guys. I have a quick question.

In an equilibrium equation, what happens if volume of reaction is decreased? Thanks

[bananabreadbelle]

Hey guys. I have a quick question.

In an equilibrium equation, what happens if volume of reaction is decreased? Thanks

Hi! :-)

When volume is decreased, the reaction will shift in the direction of the side with LESS particles.

[bedigursimran]

Hi! :-)

When volume is decreased, the reaction will shift in the direction of the side with LESS particles.

Does it shift in the direction of side with less particles even if it's aqueous?

Thanks xd

[sweetiepi]

Does it shift in the direction of side with less particles even if it's aqueous?

Thanks xd

Yes.*

*At it is in solution, it will most certainly move in the direction with the least number of moles/concentration

[kimmytaaa]

Help with question 2
S2O3^-2(Aq) + 2H+ -> s(s) + so2 (g) + h20 (l)
2. Explain why an increase in temperature leads to a change in the rate of reaction.

[Sine]

Help with question 2
S2O3^-2(Aq) + 2H+ -> s(s) + so2 (g) + h20 (l)
2. Explain why an increase in temperature leads to a change in the rate of reaction.

The increase of temperature increase the average kinetic energy of the molecules therefore they are a greater proportion of molecules with sufficient energy for a successful collision, increasing the frequency of successful collisions and thus increasing the rate of the reaction. Additionally the increase of temperature increase the speed of particles therefore they collide more frequently increasing frequency of successful collisions and increasing the rate of the reaction.

Note :  temperature increases the proportion of successful collisions AND rate of collisions.

[lzxnl]

It IS possible for a reaction rate to decrease with temperature if it's multi-step, and the slowest step involves a reactant that is formed from an exothermic reaction that occurs before.

Suppose the reaction A + B -> E proceeds as:
A + B <-> C + D -> E
If the first reaction is very fast and exothermic such that it is essentially instantaneous, but reversible, increasing the temperature will reduce the equilibrium yield of C and D. Suppose now that the increase in rate in reaction 2 from temperature is not compensated for by the drop in concentrations of C and D. Then, the rate of formation of E will decrease in response to a temperature increase.

Just something fun I thought I'd share. In VCE, temperature increases always increase reaction rates. Both forwards and backwards (but to different extents).

[kimmytaaa]

How did changing the concentration of sodium thiosulfate affect the rate of the reaction?

[kimmytaaa]

Explain why the rate of a reaction depends upon the concentration of the reactants?

[Elizawei]

Explain why the rate of a reaction depends upon the concentration of the reactants?

An increase in concentration of aq solutions (or pressure in gaseous reactants) means that there are more particles present per unit of volume hence the frequency of collisions is increased, therefore leading to a greater number of successful collisions. This results in a faster reaction rate.

[mary1911997]

Hi there,
need help with some chemistry questions any assistance is appreciated
so i have done a prac and a question states
the actual values for the heat of formation of magnesium nitrate, zinc nitrate, lead nitrate are given below
magnesium nitrate = -790 kJ/mol
zinc nitrate = -564 KJ/mol
lead nitrate = -447 KJ/mol

how do your values compare and give two reasons which could account for any difference
knowing that my results were
magnesium nitrate = -434kJ/mol
zinc nitrate = -67KJ/mol
lead nitrate = -211KJ/mol



question 2
the chromium  cathode in an electrolytic cell increases in mass by 1.37g in 25.5 minutes at a current of 5 amps . what is the charge on the chromium ion in solution?

thanks

[sweetcheeks]

question 2
the chromium  cathode in an electrolytic cell increases in mass by 1.37g in 25.5 minutes at a current of 5 amps . what is the charge on the chromium ion in solution?

thanks

What we are looking for is how many electrons are required to reduce chromium from being a positive ion back to its atomic state.
The mass of the cathode has increased by 1.37g. We can assume that it is all chromium that has been reduced. We now need to find the ratio of moles of chromium deposited to moles of electrons supplied. What we are looking for is how many electrons are required to reduce chromium from being a positive ion back to its atomic state.

To find the moles of chromium it is simply n.moles = mass/molecular weight
1.37g/52.0gmol = 0.0263 mole of chromium 'reacted'

To find the mole of electrons we have to use this equation (Amps X seconds)/96500 (96500 is found in the data booklet.)
(5 amps x (25.5*60)seconds)/96500=0.0792 mole of electrons.

We now must ratio by going electrons/chromium =0.0792/0.0263 =3

So for every atom of chromium deposited we needed 3 electrons. For every mole of chromium we needed 3 mole of electrons. Therefore the charge must be +3.

[mary1911997]

What we are looking for is how many electrons are required to reduce chromium from being a positive ion back to its atomic state.
The mass of the cathode has increased by 1.37g. We can assume that it is all chromium that has been reduced. We now need to find the ratio of moles of chromium deposited to moles of electrons supplied. What we are looking for is how many electrons are required to reduce chromium from being a positive ion back to its atomic state.

To find the moles of chromium it is simply n.moles = mass/molecular weight
1.37g/52.0gmol = 0.0263 mole of chromium 'reacted'

To find the mole of electrons we have to use this equation (Amps X seconds)/96500 (96500 is found in the data booklet.)
(5 amps x (25.5*60)seconds)/96500=0.0792 mole of electrons.

We now must ratio by going electrons/chromium =0.0792/0.0263 =3

So for every atom of chromium deposited we needed 3 electrons. For every mole of chromium we needed 3 mole of electrons. Therefore the charge must be +3.

thanks it makes more sense now