VCE Chemistry Question Thread

[knightrider]

What would 15.999 be to 1 significant  figure?

Would it be 20?

[jessica666]

What would 15.999 be to 1 significant  figure?

Would it be 20?

You would have to put it in scientific notation i think, 2 x 10^1

[Redoxify]

Example if you are trying to work out how much one atom of Argon weighs, why would you divide the molar mass of Argon by 6 x 10^-23

[izzywantsa97]

Example if you are trying to work out how much one atom of Argon weighs, why would you divide the molar mass of Argon by 6 x 10^-23

Are you asking why you'd divide it by 6 x 10^23? If so, it's just because a mole of anything is just 6.023 x 10^23 of that thing. So to find the mass of a single atom, you divide the molar mass by the number of atoms in that mole, which is 6 x 10^23

[Fusuy]

Hi Guys,
For UV-Visible Spectroscopy, does the sample solution have to be coloured? Or can solutions that absorb UV light also be analysised? Thanks

[Kel9901]

Hi Guys,
For UV-Visible Spectroscopy, does the sample solution have to be coloured? Or can solutions that absorb UV light also be analysised? Thanks

both coloured (absorbing visible light) and those that only absorb UV light can be analysed

[Fusuy]

both coloured (absorbing visible light) and those that only absorb UV light can be analysed

Thanks

[chocolate.cake.1]

Hello

How much water must be added to 100mL of a solution of a strong base of pH 13.0 in order to decrease the pH to 11.0?

I tried subbing in the H+ concentrations using the c1 x v1 = c2 x v2 formula:
10^(-13) x 0.1 = 10^(-11) x v2
v2 = 0.001 L
So yeah obviously my answer is wrong lol

Thankyou

[alchemy]

Hello

How much water must be added to 100mL of a solution of a strong base of pH 13.0 in order to decrease the pH to 11.0?

I tried subbing in the H+ concentrations using the c1 x v1 = c2 x v2 formula:
10^(-13) x 0.1 = 10^(-11) x v2
v2 = 0.001 L
So yeah obviously my answer is wrong lol

Thankyou

Alright, so your method is on the right track BUT since we're dealing with bases your concentration in the equation must be of [OH-] rather than [H+]. Likewise, if you're dealing with acids, the concentration in the equation should be kept as [H+] rather than [OH-]. Think about the reason for this mathematically...

[H+] in the original solution = 10^-13 so [OH-] = 10^-1
[H+] in the diluted solution = 10^-11 so [OH-] = 10^-3
c1v1=c2v2 since we're dealing with dilutions.
10^-1*0.1=10^-3*v2
v2=10
Since we have 100mL already, we need to add 10-0.1=9.9L

[lzxnl]

Hello

How much water must be added to 100mL of a solution of a strong base of pH 13.0 in order to decrease the pH to 11.0?

I tried subbing in the H+ concentrations using the c1 x v1 = c2 x v2 formula:
10^(-13) x 0.1 = 10^(-11) x v2
v2 = 0.001 L
So yeah obviously my answer is wrong lol

Thankyou

There is another reason why this working failed so abysmally.
Water is not pure. 'Pure' water is still a solution of 10^-7 M of hydroxide and hydrogen ions at 298 K. You're adding a 10^-7 M solution of H+ to a 10^-13 M solution of H+. You're not diluting the acid there.

For the base, however, you are diluting it because you're adding a 10^-7 M solution of base to a 10^-1 M solution, and the 10^-7 M is negligible.

[RazzMeTazz]

For an alkane and an alkene with the same number of carbon atoms, which would have the greater dispersion forces and why?

Also when an alkene isn't given without a number indicating where the double bond is located (for e.g. propene) is it safe to assume that it is referring to  prop-1-ene/1-propene?

Thanks

[mahler004]

For an alkane and an alkene with the same number of carbon atoms, which would have the greater dispersion forces and why?

Also when an alkene isn't given without a number indicating where the double bond is located (for e.g. propene) is it safe to assume that it is referring to  prop-1-ene/1-propene?

Thanks

For ethene/propene the double bone can only occur in one place (1-propene is the same as 2-propene). In other cases, you need to indicate where the double bond is.

[RazzMeTazz]

For ethene/propene the double bone can only occur in one place (1-propene is the same as 2-propene). In other cases, you need to indicate where the double bond is.

Thanks!

[lzxnl]

For an alkane and an alkene with the same number of carbon atoms, which would have the greater dispersion forces and why?

Also when an alkene isn't given without a number indicating where the double bond is located (for e.g. propene) is it safe to assume that it is referring to  prop-1-ene/1-propene?

Thanks

Erm. I'm assuming you mean the alkene and alkane are the same except the alkene has been dehydrogenated between two carbons once only.

Anyway, if that were the case, you'd expect the alkanes to pack together slightly better due to the lack of rigidity in the C=C bond. Also there are slightly more electrons there in an alkene available for dispersion forces to act upon. However, if you look at pentane and pent-1-ene's boiling points, the difference isn't huge.

[knightrider]

How would you do this question?

Calculate the amount (in mol) of  oxygen atoms in 1.5 mol of sodium sulfate .