Random Question Thread.

[cns1511]

so that 'two water molecules per nucleotide' accounts for the base/phosphate too yeap

Yeah, sorry should have worded that a bit better.

[d0minicz]

Hey what are the potential errors associated with calorimetry?

thanks

[monokekie]

energy losses will always occur despite the insulation of the calorimeter. a solution to prevent the errors associated is to calculate the efficiency of the calorimeter. (which is called calibration)

this is the only one i could think of at the moment.

[d0minicz]

Predict the products at each electrode during electrolysis (using unreactive solutions) of 1M lead(II) nitrate.
alright what i dont get is ... which unreactive electrode do we use ?

thanks ...

[d0minicz]

oh wait its unreactive so we dont include it right >_>

[d0minicz]

100mL of water is being heated using 1.7g of butane. The temperature of the water rose by 60 degrees celcius. If only 30% of the chemical energy from butane is converted to thermal energy in the water, find the heat of combustion of butane in kJ/g.

thanks...

[monokekie]

100mL of water is being heated using 1.7g of butane. The temperature of the water rose by 60 degrees celcius. If only 30% of the chemical energy from butane is converted to thermal energy in the water, find the heat of combustion of butane in kJ/g.

i would find the total heat absorbed by water first, so (assuming density is 1 g per ml) 100g x 4.184(specific heat capacity of water)x delta T ( change in temperature) =418.4 x 60 j = then you needa convert this into Kj


now we have the total energy absorbed by water. and from the question we are given the info that it is only 30% of the chemical energy from 1.7 g of butane.
let the heat of combustion per gram of butane be X

(0.3)(1.70)X=4.184x60
then just workout X.

is X the correct answer?

[chem-nerd]

E used to heat water = 100g x  4.18 Jg^-1C^-1 x 60C = 25.08 kJ

This is only 30% of the energy available. Thus 100% is 25.08 kJ/0.3 = 83.6 kJ

83.6kJ/1.7g = 49 kJ/g

[d0minicz]

thx guys
sorry i read the question wrong, and times it by 0.3 instead of divide lol

[rajah21]

Hey, just wondering what points I could use to answer this question:

"Suggest some practical improvements that could be made to the calorimeter you have calibrated."

Many thanks!

[d0minicz]

Do we have to memorise the exact composition of cells such as the membrane ... Downs cell ... etc?

and the equations?

[dekoyl]

Do we have to memorise the exact composition of cells such as the membrane ... Downs cell ... etc?

and the equations?

Nope.

However, it'll do no harm to know what they look like. Just you won't need to remember the specific equations.

[kendraaaaa]

Hey, just wondering what points I could use to answer this question:

"Suggest some practical improvements that could be made to the calorimeter you have calibrated."

Many thanks!

Insulation is one point you could discuss.

[d0minicz]

A dilute aqueous solution known to be either potassium chloride or potassium bromide was electrolysed. Bubbles of colourless gas formed at the positive electrode of the cell during electrolysis.

a) Which solution was being electrolysed? Explain your choice.

thanks =]

[ed_saifa]

Potassium bromide.
Looking at the electrochemical series Bromide ions are a stronger reductant than water and chloride ions.
Water is the strongest oxidant. Looking at the series, we see Hydrogen gas is produced.
In electrolytic cells, anode is negative and cathode is positive.

Key note: strongest oxidant and strongest reductant
oxidation at the anode, reduction at the cathode
same in both galvanic and electrolytic cells.