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Google tells us this is a verse from Proverbs 4:7 (NIV)
Perhaps http://www.sgforums.com/forums/1381 would help you.
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Wow I like the long paragraphs though I don't get everything.
I'll type out my solution to Q1.
1. Find pH.
30cm^3 of 0.1mol/dm^3 CH3COOH mix with
10cm^3 of 0.1mol/dm^3 NaOH.
[ Ka of CH3COOH = 1.7 * 10^-5 ]
No. of moles of NaOH = 1*10^-3
No. of moles of CH3COOH = 3*10^-3
NaOH is the limiting reactant. 2*10^-3 moles of CH3COOH would be left.
1*10^-3 moles of CH3COONa would be formed.
A weak acid with it's sodium salt forms a buffer system?
CH3COOH <--> CH3COO- + H+
CH3COONa -> CH3COO- + Na+ (Ionic salt dissociates completely)
CH3COO- + H2O -> CH3COOH + OH-
Kb = [CH3COOH][OH-] / [CH3COO-]
[CH3COO-] = (2*10^-3) * (1000/40) = 0.05 (I think this is wrong)
(1*10^-14) / (1.7*10^-5) = [OH-]^2 / 0.05 (Assume at 298K)
[OH-] = 5.42326*10^-6
pH = 14 - (-log[OH-]) = 8.73 (But answer is 4.5)
Edited by secretliker 13 Jul `08, 2:52AM
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1. Find pH.
30cm^3 of 0.1mol/dm^3 CH3COOH mix with
10cm^3 of 0.1mol/dm^3 NaOH.
[ Ka of CH3COOH = 1.7 * 10^-5 ]
2. Calculate25cm^3 of 0.050mol/dm^3 ethanoic acid titrated with
0.100mol/dm^3 NaOH
(i) initial pH
(ii) final pH
(iii) pH at equivalence point
(iv) after 10.00cm^3 NaOH was added
(v) after 15cm^3 NaOH was added.
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Originally posted by bonkysleuth:
and there;s this question. if is of a car moving on a horizontal section until it reaches a ramp and moves downwards. length of the ramp is 0.9 m .
(ii) the car has 0.30 J of kinetic energy at the topof the ramp and loses 0.50 of potential energy as it moves to the bottom of the ramp. calculate kinetic energy of car at bottom of the ramp.
i tried to equate using principle of conservation of energy.
At top of ramp, car has KE (0.30J) and GPE (this is equal to 50J since the car loses 50J of GPE when it travels to the bottom).
Assuming no energy is lost as thermal energy, all GPE is converted to KE.
Thus KE at bottom of ramp = 80J?
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Just part (a), I think my answer key is wrong.
In the manufacture of glass panels, small bubbles occur at random at an average of 1 bubble in every glass panel. The number of bubbles detected in a panel is denoted by X and follows a Poisson distribution.
(a) find the probability that, in a randomly chosen glass panel, there are at least three bubbles. [ans=0.0144?]
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When barium metal is burned in air, an oxide is formed with 81.1% by mass of barium. Calculate the formula of the compound. [Mr of barium=137, oxygen=16]
The answer is simple (BaO2), but I'm looking for a proper working instead of guess & check.
I tried doing this but I failed:
Let compound be Ba(x)O(y)
137x / (137x + 16y) = 0.811
...
Edited by secretliker 02 Jul `08, 12:53AM
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Q1) If a^(2x-1) = b^(1-3y) and a^(3x-1) = b^(2y-2), show that 13xy = 7x + 5y -3.
(2x-1)(2y-2) = (1-3y)(3x-1)
4xy-2y-4x+2 = 3x-9xy-1+3y
13xy = 7x + 5y - 3
Q2) Given that log5 (x) = 4 logx (5), calculate the possible values of x.
log5(x) / log5(5) = 4 log5(5) / log5(x)
[ log5(x) ] ^2 = 4 [ log5(5) ] ^2
[ log5(x) ] = ±(2)
x=5^(-2) or 5^2
x=1/25 or 25
Edited by secretliker 29 Jun `08, 2:38AM
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It's explained in the equations already. Because both AgCl and AgOH are white ppt, simply adding AgNO3 cannot be a complete test for presence of chloride ions.
It's because if there are XOH and XCl present in the solution, then the HNO3 added would undergo acid-base reaction with the HNO3 to form XNO3 and H2O. XCl would react with HNO3 to form XNO3 and HCl.
Now, after adding AgNO3, you can be sure that the white ppt is AgCl.
[I hope I'm right.]
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fail what