ssNake's Chemistry [u2] Q's

[Lasercookie]

I just left it in g/L and mL whereas you guys went to n/L for some reason?

It generally doesn't matter what unit of concentration you use, as long as you're consistent throughout while working out.

[b^3]

/[u2] chem checkpoints fail!

Not as much as methods checkpoints.

[b^3]

I just left it in g/L and mL whereas you guys went to n/L for some reason?

It generally doesn't matter what unit of concentration you use, as long as you're consistent throughout while working out.

Thankyou, I thought I'd forgotten basics of yr 11 chem for a minute.

[REBORN]

A chem student wants to determine the conc of HCl. She makes up 250mL of a standard solution of sodium carbonate in which the solution contains 1.358g of sodium carbonate. She then takes 20mL aliquots of this standard solution and titrates it with acid. The average titre is 20.24mL to read the end point.

a) Balanced eq? 2HCl + Na2CO3 ---> 2NaCl + H20 + CO2

b) calc the conc. of sodium carbonate solution? ---> I got this to be 0.05125M

c) conc of HCl? --> NO CLUE

[b^3]

a) 2HCl_(aq) + Na₂CO_3(aq) ---> 2NaCl_(aq) + H₂O_(l) + CO_2(g)
b) n(Na₂CO₃=1.358/((2*22.99)+12+48)
=0.0128mol
c(Na₂CO₃)=0.0128/0.25
=0.05125M
c) now this is a tritration, when you take the aliquot (the 20ml) of the Na₂CO₃ it will have the same concentration.
So you know the voulme of this is 20ml=0.02L
so the mol of ths sodium carbonate is n=cv
n_aliquot(Na₂CO₃)=0.05125*0.02
=1.025*10^-3
so from the equation you know that 1 mol of Na₂CO₃ will react with two mols of HCL
so n(HCL)=2*n(Na₂CO₃)
=2*1.025*10^-3
=2.050mol
now the volume of HCL used is 20.24ml=0.02024L
so c=n/v
c(HCL)=2.050/0.02024
=0.1013M

[REBORN]

Use the kinetic molecular theory of gases to explain why a given volume of a gaseous substance weighs less than the same volume of the substance in the liquid state.

[Graphite]

No idea what that theory is but I hope my explanation evolves around that theory. Generally for most substances they are most dense as it changes from a gas to a liquid to a solid with a few exceptions such as water when changing from a liquid to solid, it is less dense.
So that being said, the same volume would have a larger amount of substance due to being more dense.
Reason: gases are of a higher energy state so they are able to overcome the various intermolecular forces between molecules and so occupy a larger volume in the first place

[REBORN]

Boyle's law of gases: V = k/p

Okay...so PV = k.

But what is k? :S

I don't get it...

[Graphite]

P and V are inversely related
If P inc, V dec. If V inc, Pdec.

So PV=k
4x3=12
Let P dec 1 and V inc 1
3x4=12

So k=a constant

[REBORN]

So what does it do....like when would I ever use that in a calculation? :S

[Graphite]

Since any PV pair equates to a constant we know P1V1=P2V2 as P1V1=k and P2V2=k
So given a PV pair and another value P OR V, we can solve for the other unknown value

[REBORN]

Calc the molar volume of an ideal gas at -10 deg C and 90.0 kPa.

I don't get how they used P1V1/n1t1 = P2V2/n2t2 .......V stands for VOLUME not MOLAR volume?

so lost :/

[Lasercookie]

http://en.wikipedia.org/wiki/Molar_volume#Ideal_gases

"For ideal gases, the molar volume is given by the ideal gas equation: this is a good approximation for many common gases at standard temperature and pressure"

Seems that it's okay to use that ideal gas equation.

P1V1/n1t1 = P2V2/n2t2  is derived from/somehow related to that ideal gas equation isn't it?

[REBORN]

Right so I just somehow need to know that V = Vm. :S

[Lasercookie]

Right so I just somehow need to know that V = Vm. :S

V = Vm at standard lab conditions.

They set v1 to be a molar volume at SLC, so therefore v2 would also be a molar volume?

I can't really explain it that well, but it makes sense to me.