Homework questions thread

[thushan]

I need help with this, i have no idea where to start

Q) When solid lead dioxide,PbO2, is heated, it forms solid lead monoxide, PbO, and oxygen. Heating solid barium peroxide, BaO2, yields solid barium monoxide, BaO, and oxygen. A mixture of lead dioxide and barium peroxide was heated until both decompositions were complete. If the initial mass of the mixture was 15g and the final mass was 13.8g, what mass of lead dioxide was present in the original mixture? Ans is 7.873g

Just if you were interested, I had another way of doing this. Very similar to Panicmode's way, but involving only one variable.

Let m(BaO2) = x, => m(PbO2) = 15 - x

Now, the mass lost is effectively oxygen, so we can derive the following eqn:

x.((137.3+16)/(137.3+32)) + (15-x)((207.2 + 16)/(207.3+32)) = 13.8 --> solve for x => 15-x is the answer you are looking for

(137.3+16)/(137.3+32) = M(BaO2)/M(BaO) and similarly for the lead compounds.

[ggxoxo]

I was just wondering, how much in-depth should we study the biomolecules for chem? I did bio 3/4 last year and so would that knowledge be sufficient, say for like the enzymes and DNA component of the chem course?  And what about for the other biomolecules?

[Panicmode]

I was just wondering, how much in-depth should we study the biomolecules for chem? I did bio 3/4 last year and so would that knowledge be sufficient, say for like the enzymes and DNA component of the chem course?  And what about for the other biomolecules?

Yeah it should be sufficient. Just note that in chemistry the focus is more on what makes up the biomacromolecules rather than their function and use.

I did both 3/4 chem and 3/4 bio last year so the two kinda blended into each other. This was the section I did the best on (given my love of bio), so especially if you have a bio-background, I wouldn't worry too much about it.

[Truck]

Hey all, one question from me here...

A 2.203g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32g of water and the carbon was oxidized to 3.23g of carbon dioxide.
A) find the empirical formula of the compound.

[Hamdog17]

Ok so,
m(H2O)=1.32g
n(H2O)=m/M=1.32/18=0.073333mol
n(H)=2*n(H2O)=2*0.073333=0.146666mol

m(CO2)=3.23g
n(CO2)=3.23/(12+2*16)=0.073409mol
n(C)=n(CO2)=0.073409mol

~We aren't told whether it contains oxygen or not so assume it does. If it doesn't, the value of m(O) will be zero grams.
m(O)=2.203-m(H)-m(C)=2.203-(12*0.073409)-(0.146666)=2.203-1.02757=1.17543g
n(O)=1.17543/16=0.073464mol

:. n(C):n(H):n(O)      0.073409:0.146666:0.073464
:. (roughly)  1:2:1
:. The empirical formula of the compound is CH2O

[Kaille]

hi guys

simple question, but i have no idea what to do.

Q. How much water must be added to 1.0 L of a solution of a strong base of pH 13.0 in order to decrease the pH to 12.0?

thanks

[thushan]

ok, pH decrease by 1 from 13 to 12 means a 10fold dilution (logarithmic scale)

So you want to turn 1 L of solution into 10 L. Hence add 9 L of water.

[Aurelian]

hi guys

simple question, but i have no idea what to do.

Q. How much water must be added to 1.0 L of a solution of a strong base of pH 13.0 in order to decrease the pH to 12.0?

thanks

ok, pH decrease by 1 from 13 to 12 means a 10fold dilution (logarithmic scale)

So you want to turn 1 L of solution into 10 L. Hence add 9 L of water.

Just in case you want to augment this with hard calculations to help understand, you can view it like so;

Q. How much water must be added to 1.0 L of a solution of a strong base of pH 13.0 in order to decrease the pH to 12.0?

pH 13.0 = pOH 1.0
Therefore, [OH-] = 10^-1M
Therefore, n(OH-) = 1.0L * 10^-1M = 0.10mol

pH 12.0 = pOH 2.0
Therefore, [OH-] = 10^-2M

That is, we're trying to have a solution of concentration 10^-2M of OH- ions, and we know the amount of OH- we have is 0.10mol. Since c = n/v, we can work out the volume of such a solution;

v = n/c = 0.10mol/0.010M = 10L

We already had 1L, so add 9L more.

That either overcomplicated things for you, or helped clarify them - I hope it was the latter!

[dinosaur93]

Q1.
The Hydrogen Peroxide in the following reaction is acting as a REDUCTANT:

5H₂O₂(aq) + 2MnO₄^-(aq) + 6H⁺(aq) －－＞5O₂(g) + 8H₂O(l) + 2Mn²⁺(aq)

Explain...

[paulsterio]

The oxidation number of Mn when it's going from MnO₄^- to Mn²⁺ is a decrease, it goes from +7 to +2. Hence, it has undergone reduction, meaning that it is the oxidant. The only other species present is the Hydrogen Peroxide, meaning that this has to be the reducing agent, it causes the reduction in the Mn, hence, it is a reductant.

[dinosaur93]

tnx! 

[Bismuth]

The oxidation number of Mn when it's going from MnO₄^- to Mn²⁺ is a decrease, it goes from +6 to +2. Hence, it has undergone reduction, meaning that it is the oxidant. The only other species present is the Hydrogen Peroxide, meaning that this has to be the reducing agent, it causes the reduction in the Mn, hence, it is a reductant.

Not strictly part of the question, but I think you meant the oxidation state for Mn in MnO₄^-(aq) is +7, not +6.

H₂O₂ is also one of those 'exceptions' to oxidation number rules, as oxygen usually takes -2 in compounds, but in peroxides it is -1.

[Kaille]

another question...

q. 30 ml of 10 M NaOH is mixed with 20 mL of 0.1 M HCL. The pH of the resulting solution is?

thanks

[Hamdog17]

Here you go,

OH-(aq) + H+(aq) ---> H2O(l)
n(NaOH)=n(OH-)=cV=(30/1000)*10=0.3mol
n(HCl)=n(H+)=cV=(20/1000)*0.1=0.002mol
:. OH-(aq) is in excess
n(OH-)excess=n(OH-)original-n(OH-)reacted
n(OH-)reacted=n(H+)reacted=0.002mol
:. n(OH-)excess=0.3-0.002=0.298mol
[OH-]resultant=0.298/((20+30)/1000)=5.96M
pOH=-log[OH-]=-log(5.96)=-0.77525
pH=14-pOH=14+0.77525=14.78
One very basic solution!

[paulsterio]

Not strictly part of the question, but I think you meant the oxidation state for Mn in MnO₄^-(aq) is +7, not +6.

H₂O₂ is also one of those 'exceptions' to oxidation number rules, as oxygen usually takes -2 in compounds, but in peroxides it is -1.

Sorry, you're right, it's +7, I think I took the charge as +2 when I evaluated it, my bad, I'll fix it now