CHEM AND BIO HELP AND TIPS HERE

[The Golden Girl =D]

Usually, small chained hydrocarbons are gases, and the state changes from gas to liquid to solid as the chain length increases.

Eg: methane--> gas

hexane--> liquid

thx

[Ivo]

people there is smthn i wanna kno ........when r hydrocarbons solids , and when r they liquid and when r they gases? ....

Boiling point increases with chain length.  So for example, for alkanes:

Methane: -164^OC
Ethane: -87^OC
Propane: -42^OC
Butane: -0.5^OC

First four alkanes - gases at room temperature.
Next twelve are liquids.
The rest are solids.

The explanation for this is that as chain length increases, the attraction between the molecules increases - so it takes more energy to separate them.

[The Golden Girl =D]

Boiling point increases with chain length.  So for example, for alkanes:

Methane: -164^OC
Ethane: -87^OC
Propane: -42^OC
Butane: -0.5^OC

that's gd info , thx man 

[Dana]

OK right, here goes:

i)  Moles of butane: 0.01/24=0.000417
    From the equation, 1 mole of butane reacts with 6.5 moles of oxygen
    Therefore, 0.000417*6.5=0.00271
    So volume of oxygen reacted: 0.00271*24*1000=65cm³
    Therefore, volume of oxygen left: 100-65=35cm³

ii) From the equation, 1 mole of butane gives 4 moles of carbon dioxide
    Therefore, 0.000417*4=0.00167
    So volume of carbon dioxide formed: 0.00167*24*1000=40cm³

That was the technical way of doing it.  Here is a simply way, but only applies if calculations involve gases only:

Consider this formula:

                      Volume of gas (in dm3)
Moles of gas = --------------------------
                                      24

Because we know one mole of gas occupies 24dm³ at r.t.p., therefore we can go about it this way:

i)  From the equation, 1 mole of butane reacts with 6.5 moles of oxygen
    So, 10cm³ of butane reacts with 10*6.5=65cm³ of oxygen
    Therefore, volume of oxygen left: 100-65=35cm³
od tho
ii) From the equation, 1 mole of butane gives 4 moles of carbon dioxide
    So 10cm³ of butane gives 10*4=40cm³ of carbon dioxide
However, the way I've gone about it using ratios can only be applied to gas volumes calculations, as explained above.  For a lot of the others, you'll still need to apply the formula:

             Mass
Moles = ------
               Mr
If you still don't get it, let me know!  

thanks so much i also had this doubt. i dont get the second method tho :/ can u explain it further cuz it seems so simple

[Ivo]

thanks so much i also had this doubt. i dont get the second method tho :/ can u explain it further cuz it seems so simple

The first method is guaranteed to work.  However, the second way to do it, only for gases, I suppose is because 1 mole of gas occupies the same volume, at the same temperature and pressure.  At temperature and pressure, this volume is 24dm³.

So I suppose because the formula for finding out moles is: volume/24.  Because this 24 is a constant value when working out the number of moles for a gas, then you could just look at the volumes of gas used and the ratio in the equations, as the ratio would be the same. 

However, for things like working out concentration, the moles of the substance using the formula varies, and not at 24, so I guess that's why it can be so simply done for gas calculations.

I hope that clears your doubt somewhat.

[Dana]

The first method is guaranteed to work.  However, the second way to do it, only for gases, I suppose is because 1 mole of gas occupies the same volume, at the same temperature and pressure.  At temperature and pressure, this volume is 24dm³.

So I suppose because the formula for finding out moles is: volume/24.  Because this 24 is a constant value when working out the number of moles for a gas, then you could just look at the volumes of gas used and the ratio in the equations, as the ratio would be the same. 

However, for things like working out concentration, the moles of the substance using the formula varies, and not at 24, so I guess that's why it can be so simply done for gas calculations.

I hope that clears your doubt somewhat.

somewhat thanks anyway

[Dana]

@ Ivo i found this in the examiner's report

Question 7
(a) (i) Candidates did not seem to be familiar with this type of calculation and did not realise that the mole
ratio for gases is the same as the reacting volume ratio. With this knowledge, the calculation is
easy and involves some simple arithmetic.
butane oxygen
mole ratio 1 6.5
volume ratio 10 65
Therefore 65 cm3 of oxygen used. 100 – 65 = 35 cm3 left.
By the same reasoning, 10 cm3 of butane will form 40 cm3 of carbon dioxide.
you're right and i get it

[Ivo]

@ Ivo i found this in the examiner's report

Question 7
(a) (i) Candidates did not seem to be familiar with this type of calculation and did not realise that the mole
ratio for gases is the same as the reacting volume ratio. With this knowledge, the calculation is
easy and involves some simple arithmetic.
butane oxygen
mole ratio 1 6.5
volume ratio 10 65
Therefore 65 cm3 of oxygen used. 100 – 65 = 35 cm3 left.
By the same reasoning, 10 cm3 of butane will form 40 cm3 of carbon dioxide.
you're right and i get it

Gosh, that one sentance from the examiner's report explained 3 paragraphs of what I said!  I really need to work on explaining consisely!  ;P

By the way, which session/year was this question from?

[ambitious94]

Heyy guys I need some help in the acids bases chapter.
Describe and explain the importance of controlling acidity in soil.
Is it just that different plants require different acidity for growth. Can anyone please explain a little more?

[Ivo]

Heyy guys I need some help in the acids bases chapter.
Describe and explain the importance of controlling acidity in soil.
Is it just that different plants require different acidity for growth. Can anyone please explain a little more?

Soil may become acidic after a number of years.  This can be due to acid rain, bacteria and fungi rotting the vegetation so that it releases acids or use of fertilisers containing ammonium salts.

Many crop plants such as onions, cabbage and beans grow better if the soil is neutral.  If soil acidity drow below pH 5.5 many plants will not grow well.

We can remove excess acidity from the soil by adding crushed limestone - calcium carbonate.  This neutralises the acid.  The calcium carbonate and the products are neutral.

CaCO₃ + 2H⁺ -> Ca²⁺ + CO₂ + H₂O

Farmers often add lime (calcium oxide) to the soil.  This also neutralises excess acid.

CaO + 2H⁺ -> Ca²⁺ + H₂O

However, care must be taken to ensure not too much lime is added to the soil.  This is because lime is strongly alkaline when it dissolves in water.  Most plants do not survive alkaline conditions.  Therefore, if the soil does become too alkaline, farmers spray the soil with manure or even with very dilute sulphuric acid.

I hope this has cleared your doubt. 

[Helium]

OK right, here goes:

i)  Moles of butane: 0.01/24=0.000417
    From the equation, 1 mole of butane reacts with 6.5 moles of oxygen
    Therefore, 0.000417*6.5=0.00271
    So volume of oxygen reacted: 0.00271*24*1000=65cm³
    Therefore, volume of oxygen left: 100-65=35cm³

ii) From the equation, 1 mole of butane gives 4 moles of carbon dioxide
    Therefore, 0.000417*4=0.00167
    So volume of carbon dioxide formed: 0.00167*24*1000=40cm³

That was the technical way of doing it.  Here is a simply way, but only applies if calculations involve gases only:

Consider this formula:

                      Volume of gas (in dm3)
Moles of gas = --------------------------
                                      24

Because we know one mole of gas occupies 24dm³ at r.t.p., therefore we can go about it this way:

i)  From the equation, 1 mole of butane reacts with 6.5 moles of oxygen
    So, 10cm³ of butane reacts with 10*6.5=65cm³ of oxygen
    Therefore, volume of oxygen left: 100-65=35cm³

ii) From the equation, 1 mole of butane gives 4 moles of carbon dioxide
    So 10cm³ of butane gives 10*4=40cm³ of carbon dioxide

However, the way I've gone about it using ratios can only be applied to gas volumes calculations, as explained above.  For a lot of the others, you'll still need to apply the formula:

             Mass
Moles = ------
               Mr

If you still don't get it, let me know! 

Thank you so much
I fully understand now thanks to you

[Helium]

Gosh, that one sentance from the examiner's report explained 3 paragraphs of what I said!  I really need to work on explaining consisely!  ;P

By the way, which session/year was this question from?

2008 O/N first variant,

Althought i prefer the first method u explained, there might be enough place for

many calculations

[Ivo]

2008 O/N first variant,

Althought i prefer the first method u explained, there might be enough place for

many calculations

In which case the 2nd method is probably more straight-forward and less time consuming.  I just included the 1st method so that you guys can understand it completely!  

You could always use the blank pages for rough working.

I guess the key idea for the 2nd method, nicely provided by Dana is:

The mole ratio for gases is the same as the reacting volume ratio.

[CatAly$t]

can somebody answer this and explain to me pls?

[Ivo]

can somebody answer this and explain to me pls?

OK, Q24, we know that it is a sensory neurone so it must be D.  We know it's the sensory neurone because it has a long dendron and short axon.  Dendron bring information to the cell body, and axon bring information away from the cell body.  Also, the cell body is located outside the spinal cord, so we know it is a sensory neurone.

Q28, I think it's C, because it's developing because there is a vacuole.  Also, it is growing because cell number is increasing.

Q38, B, think of the log phase, exponential, because there are no limiting factors, so everything is in abundance (e.g. unlimited space, food etc)

Did I get them right?