Chemistry 3/4 2013 Thread

[lzxnl]

So if we're adding three stearic acid (C17H35COOH) molecules to glycerol
I'm sure we could write it as HOCH(OOC(CH2)16CH3)CH(OOC(CH2)16CH3)CH(OOC(CH2)16CH3)OH
although admittedly it looks ugly.

[Homer]

VCAA 2008 Exam 1, Short answers Question 2 bii

why do they multiply 35 with 100/25

why isn't it 35 x 25/100?

Thanks

[Edward21]

VCAA 2008 Exam 1, Short answers Question 2 bii

why do they multiply 35 with 100/25

why isn't it 35 x 25/100?

Thanks

Hmmm look, the way I look at it is if the concentration is 35mg/L, multiply that by 0.1L (100mL) to get 3.5mg of MnO4- in the 25mL! So the ratio is want/have ie. 1000mL (the whole initial solutions)/25mL (have) times the 3.5mg = 140mg present in total, which is the right answer. But that's just my way. If you want to work by the VCAA way (remember there's an endless way to do these things yet still get the same answer!)

How they did it is, they worked backwards via DILUTION FACTORS. Basically you calculate what the solution was made up to divided by what it was and times by the concentration you found to work backwards to the original concentration of the solution. Ie. 35(mg/L) is what the concentrations is X (100mL what is was made up to)/(25mL the original volume of the solution) therefore 35mg/L times 100/25 = 140mg/L is the original concentration, and we have 1L of this solution (oh how very neat VCAA....) therefore the amount in 1L is literally just the 140mg like I said before, as it's 140mg per litre, and we want to find how much is in 1L ie. multiply by 1  did that make sense?

[Edward21]

Hmmm look, the way I look at it is if the concentration is 35mg/L, multiply that by 0.1L (100mL) to get 3.5mg of MnO4- in the 25mL! So the ratio is want/have ie. 1000mL (the whole initial solutions)/25mL (have) times the 3.5mg = 140mg present in total, which is the right answer. But that's just my way. If you want to work by the VCAA way (remember there's an endless way to do these things yet still get the same answer!)

How they did it is, they worked backwards via DILUTION FACTORS. Basically you calculate what the solution was made up to divided by what it was and times by the concentration you found to work backwards to the original concentration of the solution. Ie. 35(mg/L) is what the concentrations is X (100mL what is was made up to)/(25mL the original volume of the solution) therefore 35mg/L times 100/25 = 140mg/L is the original concentration, and we have 1L of this solution (oh how very neat VCAA....) therefore the amount in 1L is literally just the 140mg like I said before, as it's 140mg per litre, and we want to find how much is in 1L ie. multiply by 1  did that make sense?

Dilution factor of 4! (100/25) sorry got a step ahead of myself. Remember dilution factor is V(made up to)/V(initial) then multiply this by the concentrated of the diluted solution, to find the concentration of the original solution before it was diluted. C1V1=C2V2 simplifies this, but essentially you are calculating a dilution factor

[random_person]

VCAA 2008 Exam 1, Short answers Question 2 bii

why do they multiply 35 with 100/25

why isn't it 35 x 25/100?

Thanks

Therefore, there is of in sample (as the sample is made up to 1.00 L

in =

  of 

to two sig figs

[Jeggz]

What would the equation be to prove that NaOH must be standardised? 

[Edward21]

What would the equation be to prove that NaOH must be standardised? 

2NaOH(s)+CO2(g) --> Na2CO3(s) + H2O(l)

The sodium hydroxide can't be used as a primary standard, it can react with CO2(g) in the air, thus it's composition cannot be entirely known unless recently standardised because of the above reaction

[Jeggz]

You're a life saver- thanks buddy 

[Edward21]

You're a life saver- thanks buddy 

No probs, remember I used (s) to represent this if it occurred whilst the NaOH in its solid powder form, reaction with CO2(g).
You could have both the NaOH and Na2CO3 with (aq) states if you left a solution of NaOH(aq) and CO2(g) reacts with it 

[LOLs99]

What is the state for I2 in half equation for the reaction of iodine. Is It in aq or liquid

[lzxnl]

Iodine is naturally found in solid form. However reactions with solids aren't as fast as reactions with the same solids in solution as the rate of collision is greatly maximized. Therefore generally we have dissolved iodine.

[LOLs99]

So aqueous or liquid form is acceptable?

[lzxnl]

Liquid form implies undissolved liquid iodine or pure liquid iodine, which generally we don't have, so aqueous, not liquid.

[LOLs99]

Okay thanks

[Edward21]

Has anyone here tried out the Heinamann textbook's exam paper 1? I'd like to know how you guys found it in terms of difficulty