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Offline Chemistry4me

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« on: 23/11/2008 02:04:23 »
I've got a STACK of chemistry questions :o, some I can do others I can't ???. I'm sure that there are plenty of talented chemists out there who are just crying out for some problems [:-'(]. I've got the answers to some but some I don't. Here are a few to kick us off.

1)A 0.200 g sample containing only CaCO3 and MgCO3 is titrated with 0.200 mol L-1 HCl. 20.75 mL is required to reach the methyl orange end-point, when all the carbonate has been converted to H2CO3. Calculate the mass of CaCO3 in the sample.

2)At 298 K, unequal amounts of BCl3(g) and BF3(g) are mixed in a container. The gases reacted in the two ways shown below. When equilibrium is finally reached, the four gases were present in these relative molar amounts:BCl3 (90), BF3 (470), BClF2 (200) BFCl2 (45)
2BCl3 + BF3 ⇄ 3BFCl2
BCl3 + 2BF3 ⇄ 3BClF2
What were the original molar amounts of BF3 and BCl3 present based on these results?

Hmm... I've asked a couple of chemistry teachers this question and so far no-one has been able to answer it. I haven't got the answer to this question and I don't know how to solve it either, but if someone gets an answer it'll be easy to check if it is right.


 

Offline Bored chemist

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« Reply #1 on: 23/11/2008 10:02:21 »
The second question is a bit more unusual than most.
If you add up all the clorines in the product and add up all the fluorines as well then you get the ratio of the F to Cl that must have been resent at the start.

How far have you got  with the first question?
 

Offline Chemistry4me

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« Reply #2 on: 23/11/2008 21:43:12 »
I've got
n(HCl)= 0.2 x 0.02075
      = 0.00415
n(CO32-)=0.002075
So there are 0.002075 moles of carbonate ions in the sample. So there must be 0.002075 moles of Calcium carbonate and Magnesium carbonate altogether, right?
I don't know if this is right but then I went: M(CaCO3) x n(CaCO3) + M(MgCO3) x n(MgCO3)=0.2   The two n's will add up to 0.002075
I'm sort of stuck there ???
Anyone got any ideas?
« Last Edit: 23/11/2008 22:07:35 by Chemistry4me »
 

Offline Bored chemist

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« Reply #3 on: 24/11/2008 19:25:17 »
You can put numbers in for M(CaCO3) and M(MgCO3) if you look up the molar masses of the two carbonates.

You have x grams of CaCO3 and y grams of MgCO3.
x+y= 0.2

Also,  you have x/(about 100) moles of CaCO3 and y/(about 84) moles of MgCO3

x/(about 100)+y/(about84) = some number you can calculate from the titration for the number of moles, (but don't forget 1 mole of HCl isn't equivalent to 1 mole of carbonate)
That should give you a pair of simultaneous equations to solve.

Another way to do it (if you don't know about simulatneous equations- or just don't like them) is to plot a graph. Calculate the volume of HCl you would need for different mixtures of the carbonates and see where the mixture gives closest to 20.75 ml
This is a messy time consuming way to do it, but it will work.
 

Offline Chemistry4me

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« Reply #4 on: 25/11/2008 00:09:52 »
Sweet, I've got it :o :o,
84.32(0.002075-x) + 100.09(x)=0.2
0.0174964 - 84.32x + 100.09x=0.2
-0.025036=-15.77x
x=0.00158757 moles
So there are 0.00158757 moles of CaCO3 and 0.00048743 moles of MgCO3 in the sample. If you substitute these back in you get about 0.1999999 so that must be right :)
Cheers Bored Chemist
Hmmm... just got to get the other one now... :-\
 

Offline Chemistry4me

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« Reply #5 on: 25/11/2008 00:14:58 »
In the mean time here are a couple simpler ones for people to try.

1)A compound consists of 14.29% carbon, 57.14% oxygen, 1.190% hydrogen and an element having the same number of moles as there are moles of carbon. Determine the formula of the compound.

2)Calculate the pH of the solution that results from mixing at 25 C

a) 10 mL of 0.15 mol L-1 NaOH with 25 mL of 0.10 mol L-1 HCl

b) 20 mL of 0.15 mol L-1 NaOH with 25 mL of 0.10 mol L-1 HCl

KW = 1 x 10-14
pH= -log [H3O+]
 

Offline Chemistry4me

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« Reply #6 on: 25/11/2008 05:42:30 »
This might help   :)

1 H 1.008     23 V 50.94    45 Rh 102.9   67 Ho 164.9 
2 He 4.003    24 Cr 52.00   46 Pd 106.4   68 Er 167.3 
3 Li 6.941    25 Mn 54.94   47 Ag 107.9   69 Tm 168.9 
4 Be 9.012    26 Fe 55.85   48 Cd 112.4   70 Yb 173.0 
5 B 10.81     27 Co 58.93   49 In 114.8   71 Lu 175.0 
6 C 12.01     28 Ni 58.69   50 Sn 118.7   72 Hf 178.5 
7 N 14.01     29 Cu 63.55   51 Sb 121.8   73 Ta 180.9 
8 O 16.00     30 Zn 65.38   52 Te 127.6   74 W 183.9   
9 F 19.00     31 Ga 69.72   53 I 126.9    75 Re 186.2 
10 Ne 20.18   32 Ge 72.59   54 Xe 131.3   76 Os 190.2 
11 Na 22.99   33 As 74.92   55 Cs 132.9   77 Ir 192.2 
12 Mg 24.31   34 Se 78.96   56 Ba 137.3   78 Pt 195.1 
13 Al 26.98   35 Br 79.90   57 La 138.9   79 Au 197.0 
14 Si 28.09   36 Kr 83.80   58 Ce 140.1   80 Hg 200.6 
15 P 30.97    37 Rb 85.47   59 Pr 140.9   81 Tl 204.4 
16 S 32.07    38 Sr 87.62   60 Nd 144.2   82 Pb 207.2 
17 Cl 35.45   39 Y 88.91    61 Pm (145)   83 Bi 209.0 
18 Ar 39.95   40 Zr 91.22   62 Sm 150.4   84 Po (209) 
19 K 39.10    41 Nb 92.91   63 Eu 152.0   85 At (210) 
20 Ca 40.08   42 Mo 95.94   64 Gd 157.3   86 Rn (222) 
21 Sc 44.96   43 Tc (98)   65 Tb 158.9   87 Fr (223) 
22 Ti 47.88   44 Ru 101.1   66 Dy 162.5   88 Ra 226.0
« Last Edit: 25/11/2008 05:45:06 by Chemistry4me »
 

Offline Chemistry4me

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« Reply #7 on: 02/01/2009 09:28:20 »
Speaking of homework...(lightarrow) ;) ;). If ever there was a question in which physics and chemistry went hand in hand, this would be the one (well atually, I'm not that sure [:I]). I saw this question in the International Chemistry Olympiad paper (and they give this kind of stuff to secondary school students :o :o :o) The question goes as follows:

The periodic system of the elements in our world is based on the four quantum numbers n = 1,2,3...; l = 0,1...(n-1); m1 = 1,2,...,l; and ms = 1/2

Let us move to Flatlandia. It is a 2-dimensional world where the periodic system of the elements is based on three electron quantum numbers: n = 1,2,3...; m = 0,1,2,...(n-1); and ms = 1/2. m plays the combined role of l and m1 of the three dimensional worlds (for example s,p,d...levels are related to m). The following tasks and the basic principles relate to this  Flatlandia where the chemical and physical experience obtained from our common 3-D world is applicable.

a) Draw the first 4 periods of the Flatlandia periodic table of the elements. Number the elements according to their nuclear charge. Use the atomic numbers (Z) as the symbols of the elements. Write the electron configuration of each element.

b) Draw the hybrid orbital of the elements with n = 2. Which element is the basis for the organic chemistry in Flatlandia (use the atomic number as a symbol)? Find the Flatlandia analogues for ethane, ethene and cyclohexane. What kind of aromatic ring compounds are possible in Flatlandia?

c) Which rules in Flatlandia correspond to the octet and 18-electron rules in the 3-D world?

d) Predict graphically the trends in the first ionisation energies of the Flatlandian elements with n = 2. Show graphically how the electronegativities of the elements increase in the Flatlandian periodic table.

e) Draw the molecular orbital energy diagrams of the neutral homonuclear diatomic molecules of the elements with n = 2. Which of these molecules are stable in Flatlandia?

f) Consider simple binary compounds of the elements (n = 2) with the lightest element (Z=1). Draw their Lewis structures, predict geometries and propose analogues for them in the 3-D world.

g) Consider elements with n ≤ 3. Propose an analog and write the chemical symbol from our world for each Flatlandian element. On the basis of this chemical and physical analogy predict which elements are solid, liquid or gas at the normal pressure and temperature.

Pheeew... That took a while :-X
It a b****y difficult question I know ::) ::)
Anyone want to try it? Or even have the slightest clue where to begin? ;D
« Last Edit: 02/01/2009 10:35:48 by Chemistry4me »
 

Offline Chemistry4me

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« Reply #8 on: 03/01/2009 03:22:37 »
Perhaps here is one more suited for physics?
According to Henry's law, the mole fraction of a solute gas in an ideal liquid solution is directly proportional to the partial pressure of that gas above the solution. The solubility of nitrogen gas in water at 20C and 500 kPa is 87 mg/L. Under the same conditions, the solubility of oxygen gas is 205 mg/L. Assume the solutions to be ideal.
a) Calculate the mole fractions of N2 and O2 in water at 20C if the partial pressure of each gas is 500 kPa.
b) What is the mass fraction of nitrogen and oxygen in water kept in open air at 20о C for a long time?
 

UG

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« Reply #9 on: 08/02/2009 03:12:45 »
A compound consists of 14.29% carbon, 57.14% oxygen, 1.190% hydrogen and an element having the same number of moles as there are moles of carbon. Determine the formula of the compound.
CHO3Na?
 

Offline Chemistry4me

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« Reply #10 on: 08/02/2009 03:13:51 »
Haha, I didn't think anyone was going to answer! Yep, CHO3Na is correct :)
 

UG

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« Reply #11 on: 08/02/2009 03:34:37 »
How do u do this?

0.075g of impure sodium bicarbonate is dissolved in water and titrated with 0.032 mol L sulfuric aicd using methyl orange indicator. 12.8 mL of sulfuric acid was required.
Calc the % of sodium bicarbonate in the solid sample.
 

Offline Chemistry4me

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« Reply #12 on: 08/02/2009 03:46:35 »
Balanced equation: 2NaHCO3 + H2SO4 → Na2SO4 + 2CO2 + 2H2O

n(NaHCO3)/2 = n(H2SO4)/1
n(NaHCO3) = 2 x n(H2SO4)

2 x c(H2SO4) x V(H2SO4) = m(NaHCO3)/M(NaHCO3)

M(NaHCO3)=84 g mol-1

2 x 0.032 x 0.0128 = m(NaHCO3)/84
m = (2 x 0.032 x 0.0128) mol x 84
  = 0.0688g

This is the mass of pure NaHCO3 in 0.075 g of impure solid.
%NaHCO3= 91.7 %  :)
 

UG

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« Reply #13 on: 08/02/2009 03:57:37 »
Cool, thanks.
 

Offline Chemistry4me

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« Reply #14 on: 08/02/2009 03:58:07 »
No worries :)
 

UG

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« Reply #15 on: 09/02/2009 04:09:29 »
Are u good with organic chemistry? cause how do you synthesis m-chloronitrobenzene staritng from bromine?
 

Offline Chemistry4me

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« Reply #16 on: 09/02/2009 04:10:17 »
Bromine? I think you meant benzene... :):)
 

UG

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« Reply #17 on: 09/02/2009 04:11:39 »
lol, ha, yea sorry, benzene!! [:I]
 

Offline Chemistry4me

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« Reply #18 on: 09/02/2009 04:12:52 »
Well Mr UG, I'll have a go :).
 

Offline Chemistry4me

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« Reply #19 on: 09/02/2009 04:26:59 »
Working backwards to find the immediate precursor of m-chloronitrobenzene.



There are two substituents on the ring, a -Cl group, which is ortho- and para-directing, and an -NO2 group, which is meta-directing. We cannot nitrate chlorobenzene because the wrong isomers (o- and p-chloronitobenzenes) would form, but chlorination of nitrobenzene should give the desired product.



Then we find the immediate precursor of nitrobenzene, which is benzene, which can be nitrated. :)



The problem is solved. :)
 

UG

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« Reply #20 on: 09/02/2009 04:29:06 »
dat's cool man! did you work that out in yor head?
 

Offline Chemistry4me

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« Reply #21 on: 09/02/2009 04:30:48 »
did you work that out in yor head?
Nah, nah, of course I didn't :) I had to go over an old book to remind me of the stuff :) Got any more while we're at it :D?
 

UG

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« Reply #22 on: 09/02/2009 04:35:40 »
Bro, yor other questions are to hard! But i wish I coud do them.
 

Offline Chemistry4me

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« Reply #23 on: 09/02/2009 04:40:06 »
Okay, how about this one?
Nitrogen combines with bromine to form a molecule. What is the shape of the molecule?
 

UG

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« Reply #24 on: 09/02/2009 04:44:30 »
Is dat NBr3, nitrogen bromide?
 

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