Homework questions thread

[soccerboi]

Can someone refresh my memory of polar and non polar? I seem to have forgotten the difference.

[Hamdog17]

Polar=uneven distribution of electrical charge in a molecule. This uneven charge results from molecules being assymetical in a particular plane of reference. Water, ethanol, methanol, etc. are polar. Polar solvents dissolve polar substances (eg. Water dissolves sucrose).

Non-polar=even distribution of charge within a molecule. Molecules are symetrical in all planes of reference (generally, exc. fats). Octane, heptane, fats, etc. are non-polar. Non-polar solvents dissolve non-polar substances (eg. Petrol (Octane) dissolves oil)

Generally non-polar solvents will not dissolve polar substances and vice versa.

[mihir94]

Hi, can someone help me with this question?

A mixture of butane gas C4H10, and excess oxygen, which occupied 170ml at SLC, was sparked. When the rection ceased, the mixture was returned to SLC. The final volume of the gas mixture was 100ml. When this gas mixture was bubbled through concentrated NaOH solution, all of the carbon dioxide was removed. The volume of gas remaining was 20ml. The equation for the combustion is:

2C4H10 (g) + 13O2 (g) --> 8CO2 (g) + 10H2O (g)

What was the volume of butane in the original mixture?

[Shenz0r]

Hi, can someone help me with this question?

A mixture of butane gas C4H10, and excess oxygen, which occupied 170ml at SLC, was sparked. When the rection ceased, the mixture was returned to SLC. The final volume of the gas mixture was 100ml. When this gas mixture was bubbled through concentrated NaOH solution, all of the carbon dioxide was removed. The volume of gas remaining was 20ml. The equation for the combustion is:

2C4H10 (g) + 13O2 (g) --> 8CO2 (g) + 10H2O (g)

What was the volume of butane in the original mixture?

Here's some food for thought, I'm not too sure that I'm right though, dunno what to do with the excess oxygen occupying 170 mL at SLC. But rule of thumb: work out the mol for everything regardless, it might have something to do with it lol. Although I'm pretty sure that it doesn't because it is in excess, and therefore C4H10 must be the limiting reagant.

But to be safe, n(excess O2) = volume/molar volume = 0.17/24.5 = 0.0006939 mol

First step, find out the volume of the CO2 removed.
1. V(CO2) = 100 - 20 = 80mL
Next, find the mol of CO2 using PV = nRT.
2. n(CO2) = PV/RT = 101.3 x 0.08 / 8.31 x 298 = 0.003273 mol
Now find the number of mol of C4H10 using the molar ratios of the equation.
3. n(C4H10) = 2/8 x 0.003273 = 0.000818 mol
Now use PV = nRT to find the volume of butane.
4. V(C4H10) = nRT/P = 0.000818 x 8.31 x 298 / 101.3 = 0.02 L (correct to 3 sig figs - is that even right?) = 20mL.

[Hamdog17]

Just a tip to save time and minimise errors during exams/SACs use the equation: . The data booklet gives the values for STP and SLC btw.

[depressedchild]

please help me with this 

cream of tarta when mixed with an excess of bicarb soda forms baking powder. when dissolved in water, baking powder produces carbon dioxides which makes cake rise. equation is

KHC4H406 (aq) + NaHCO3 (aq) --> KNaC4H4O6 (aq) + h20 (l) + CO2 (g)

at slc, the volume of carbon dioxide produced when 0.25 of cream tartar is used is what?

[Hamdog17]

Assuming you meant 0.25g of cream of tartar? (The states aren't right in that case then) then you go,
m(KHC4H4O6)=0.25g
n=m/M    n(KHC4H4O6)=0.25/(4*12+5+6*16+39.1)=0.25/188.1=0.0013291mol
n(KHC4H4O6)reacted=n(CO2)formed
:. n(CO2)formed=0.0013291mol
Vm=V/n
:. V=Vm*n    V=24.5*0.0013291=0.03256L
V(CO2)evolved=33mL (2 sig figs)

[depressedchild]

thanks, i realised that was a very simple question sighh. One more please

the emprical formula of the hydrated ionic compund with composition barium 56.2% and chlorine 29.1% by mass is

a. BaCl2.H20
b. BaCl2.2H2o
c. BaCl2.3H20
d. Ba2cl.H20

[soccerboi]

When finding the mol of oxygen, do we find the mol of O or O2?(i dont know when to divide the mass by 16 and when by 32)

For example, A student burned 1.234g of Aluminium in pure oxygen and obtained 2.322g of a white oxide. From these results, determine the empirical formula of the oxide of Aluminium.

In this question the solutions divided by 16 when calculating the mol of oxygen but why? I thought that because its Aluminum being burnt in oxygen then it must be O2 gas?

[Hamdog17]

Ok so we can from the outset determine the water content,
%(H2O)=100-56.2-29.1=14.7%
:. Per 100g of the compound 14.7g is water
n=m/M    n(H2O)=14.7/18=0.81667mol

n(Ba2+)=56.2/137.3=0.409mol
n(Cl-)=29.1/35.5=0.8197mol
n(Ba2+):n(Cl-):n(H2O)
0.409:0.8197:0.81667   (roughly)   1:2:2
:. The compound is BaCl2.2H2O
:. Option B.

[soccerboi]

thanks, i realised that was a very simple question sighh. One more please

the emprical formula of the hydrated ionic compund with composition barium 56.2% and chlorine 29.1% by mass is

a. BaCl2.H20
b. BaCl2.2H2o
c. BaCl2.3H20
d. Ba2cl.H20

Ba                      :  Cl                      : H20
56.2/137.3          29.1/35.5              (100-56.2-29.1)/18
=0.4093mol        =0.8197mol            =0.8167mol
1                      :        2                   :          2
:.  BaCl2.2H20
:. its option B

^Beat me to it

[Kaille]

thanks for making this thread

this is probably a seriously easy question...

Q. In a 2M CuCl2 solution, what is the concentration of Cl ions?

[Hamdog17]

[CuCl2]=[Cu2+]=2M
So for every mole of CuCl2, there are 2 moles of Cl- ions.
:. [Cl-]=2*[Cu2+]
    [Cl-]=2*2=4M
   

[mihir94]

Thanks Shenz0r that's what i thought. The excess oxygen tripped me up. 

[Kaille]

[CuCl2]=[Cu2+]=2M
So for every mole of CuCl2, there are 2 moles of Cl- ions.
:. [Cl-]=2*[Cu2+]
    [Cl-]=2*2=4M
   

oh.my.gosh i didn't know it was that simple.

thanks!