ssNake's Chemistry [u2] Q's

[luken93]

So the volume this HCl is in is the volume of the cleaner you added (as it is what is in the cleaner). So divide the mol of HCL by the volume (in L) so  0.01792mol/0.01948L to give the concentration in M. This will be the concentration of the dilute cleaner. So times it back by the dilution factor, 250/20 which will give the conc of the cleaner.

N(HCl)=0.01792mol
so c=n/v=0.01792/0.1948
=0.09199M
that is in the diluted sample
so in the inital sample the concentration will be 250/20 *0.09199
=1.150M

EDIT: Fixed it

I think you forgot a step
n(HCl) in 19.84=0.01792mol
n(HCl) in 250 ml = 0.2258 mol
c(HCl) of diluted = 0.2258/0.25 = 0.903M

In undiluted; C1V1 = C2V2
0.903 x 0.25 = C2 x 0.025
C2 = 9.03M

Hopefully that's right, I miss titration Qs in unit 4 haha

[b^3]

Yeh I did, I left the mol of HCL in 250ml and 19.48ml the same, where as it is the mol of HCL in 250ml and 20ml that is the same. I haven't done titrations for a while, I'm a little rusty. I'm making so many mistakes lately on here .

[luken93]

Yeh I did, I left the mol of HCL in 250ml and 19.48ml the same, where as it is the mol of HCL in 250ml and 20ml that is the same. I haven't done titrations for a while, I'm a little rusty. I'm making so many mistakes lately on here .

Hahaha yeah I know the feeling!

[REBORN]

Argh I'm not getting vol Q's!

Q: A household ammonia solution was analysed to determine its ammonia content. 9.97g of the household ammonia was placed in a volumetric flask and made up to a 250mL of solution. 20mL of the resulting solution required 24.62mL of 0.11M Hcl to achieve a methyl orange end point. Calculate the mass percentage of ammonia in in the initial household ammonia.

So I got:

n(NH₃) = 9.97/17 = 0.58647mol.
c(NH₃ in vol flask) = 0.58647/0.25 = 2.34588M

We write the eq as NH₃ + HCl ----> NH₄Cl

n(HCl) = 0.11 x 0.02462 = 2.7082 x 10^-3 mol.

And now?

Also can someone please explain what's happening when they work it out.

[b^3]

We are going to work out the concentration (and from that mol and mass) of the ammonia solution from the know volume and concentration of HCL.
So n(HCl)=0.11*0.2462=0.0027082mol
We know that this reacts with NH3 in a 1:1 ratio
So the mol of NH3 reacted is nreacted(NH3)=n(HCL)=0.0027082 mol
This is the amount in the 20ml that you took (i.e. the alliquot) since this is the amount reacting with the HCL. So the amount in the 250ml solution should be in proportion as these two are the same concentration (since we are just taking it out of one and putting it in the other, no diluting here).
So the amount in the 250ml volumetric flask is 250/20*0.0027082=0.0338525mol.
The original amount has been diluted to get this. So the amount in the original is the same, but the concentration is different (Volume has changed, amount in mol hasn't. Thus the amount in the original is 0.0338525mol.
Now we can work out the original mass of the ammonia in the original solution. so moriginal(NH3)=0.0338525*17=0.5754925g
The original sample (not ammoina, it is the sample i.e. ammonia mixed in other stuff, water, impurities, other ingredients) has mass 9.97g.
So the percentage by mass of ammonia will be 0.5754925/9.97*100=5.77% (w/w)

So basically the steps here are:
1. Determine the amount of mole of the known solution (the HCL)
2. From this work out the amount of the unknown solution reacted (using mol ratios)
3. Next work out how much the original sample will have by using ratios of volume originally over volume taken out
4. Find mass of original (NH3)
5. Divide mass of (NH3) by total mass of the original sample *100
6. Bingo.

[REBORN]

Which group would be more likely to be oxidised or reduced? Provide explanation.

Group A: sodium, calcium, magnesium
Group B: sulfur, oxygen, chlorine

[b^3]

Group A will all want to lose electrons, since they have one or two outer shell electrons and so are more likely to be oxidised.
Group B will want to gain electrons to complete their outer shell and so will more likely be reduced.

Remember: AN OIL RIG CAT
ANode - Oxidation Is Loss || Reduction Is Gain - CAThode

[REBORN]

Cd + NiO₂ + 2H₂O ----> Cd(OH)₂ + Ni(OH)₂

When assigning oxidation numbers to the (OH)₂ is it just -1 under the OH? And this means both O and H are -1?

[b^3]

O.N of O is -2, O.N of H is +1
-2+1= -1
O.N of OH^- is -1
so (OH^-)₂ in total will be -2 making Ni²⁺ +2

[REBORN]

Assign oxidation numbers for photosynthesis.

I don't get it.

Is glucose 0 as a whole? :S

and the oxygen by product is a free element so 0 O.N ?

[b^3]

Assign oxidation numbers for photosynthesis.
I don't get it.
Is glucose 0 as a whole? :S
and the oxygen by product is a free element so 0 O.N ?

Yes glucose is 0 as a whole
Yes the O2 is 0
O on it's own will be 0
O inside others will be -2 (except things like H₂O₂)
H will be +1
C₆H₁₂O₆(aq)+ 6O_2(g) --> 6CO₂(g) +6H₂O(g) (respiration sorry, reverse it)
  +1(12)             0                  -2(2)      +1(2)
           -2(6)                                              -2(1)
so in C6H12O6
H will be +1
O will be -2
C will be 6x+12-2(12)=0 so C will be 0
In O2
O is 0
In CO2
O is -2
so C is x+2(-2)=0
x=+4, so C will be +4
In H2O
O is -2 and H is +1

EDIT: sorry thats respiration, reverse the equation for photosynthesis, my bad

[REBORN]

Using electrochemical series - will a spont. reaction occur b/w solutions of potassium chloride and copper chloride?

so which elements do I look at in the series? how do I know?

Also b/w iodine and aluminium. If so, write an ionic eq.

ty

[|ll|lll|]

Using electrochemical series - will a spont. reaction occur b/w solutions of potassium chloride and copper chloride?

so which elements do I look at in the series? how do I know?

Also b/w iodine and aluminium. If so, write an ionic eq.

ty

Look at Potassium and Copper.
Since Potassium is more reactive than copper, this reaction will take place:
K + CuCl2 --> KCl + Cu
since potassium is more reactive than copper and will have a higher tendency to form octet bond with chlorine, thus displacing copper. Same applies for iodine and aluminium

[REBORN]

The answer was 'no reaction'.

[illuminati]

The answer is no reaction because they are both oxidants.
We know that redox will only occur if reduction and oxidation occur simultaneously, and since they are both oxidants that cause reduction, there is no reaction.