Ans to (02 Nov) Qns

Posted by Kyo on November 02, 2007 at 08:19:22:

In Reply to: (02 Nov) Qns posted by Kyo on November 01, 2007 at 16:36:03:

Ans to Q23
First, determine formula of cream solid : AgBrO3 (s).
Next, find no. of mol of AgBrO3 (0.005 mol) and AgCl (0.03 mol).
Then, find mol ratio of BrO3- to Cl- (1:6).
Hence, 3Cl2 + NaBr + 6NaOH --> NaBrO3 + 6NaCl + 3H2O


Ans to Q24
First, find no. of mol of I2 (0.002 mol).
Next, determine mole ratio between X and I2 (1:2).
Then, by comparing stoichiometric coefficients of 1:[(n+1)/2] with 1:2 ratio (step 2 above), calculate that n=3.
Hence, equation is I2 + 3Cl2 --> 2ICl3


Ans to Q25



Ans to Q26
a) Total no. of mol of gas at equilibrium = (0.55)(1)N2 + (0.55)(3)H2 + (0.45)(2)NH3 = 3.1 mol.
Let the total pressure at equilibriuim be x.
Substituting x and partial pressure values into Kp expression, solve for x.
x = total pressure at equilibrium = 1.2 x 10^3 atm
b) Total no. of mol of gas at equilibrium = (0.9)(1)N2 + (0.9)(3)H2 + (0.1)(2)NH3 = 3.8 mol.
Notice that 400 deg C = 673K.
Remember that Kp (or Kc) values for forward reaction and reverse reaction, are simply the reciprocal values of each other. Hence, Kp for dissociation of NH3, ie. backward reaction, is 1/(2.18 x 10^-6).
Let the total pressure at equilibriuim be x.
Substituting x and partial pressure values into Kp expression, solve for x.
x = total pressure at equilibrium = 122 atm

Ans to Q27
Since mol of H+ ions in 10cm3 of FA1 is 2.1 x 10-3 mol, no. of mol of H2YO4 in 10cm3 of FA1 is 0.5 x 2.1 x 10-3 = 1.05 x 10-3 mol.
Molarity = (concentration in g/dm3)/(molar mass in g)
Molar mass of H2YO4 = 98.1g, hence molar mass of Y = 32.1g.
Y is sulphur.
Sulphurous acid H2SO3 is weaker than sulphuric acid H2SO4;
because the capacity of sulphur to draw away electrons from the OH group (thus making the O-H bond more polar and weakening it), and/or the capacity of sulphur to draw away electrons from the O- conjugate base, thus stabilizing the negative charge by induction, increases with the no. of electronegative / electron withdrawing O atoms attached to sulphur.


Ans to Q28
Notice that the oxidation state (OS) of sulphur in sulphate ion is +6, in sulphur dioxide is +4, in hydrogen sulphide is -2. The reducing power of the halide ions increases down the group (ie. iodide is a more powerful reducing agent compared to bromide), because as we go down the group, the no. of electron shells increases, and not only is the distance between nucleus and valence shell increased, but in addition the valence electrons are also progressively better shielded from the positive nuclear charge. Consequently, the iodide ion loses its valence electrons more readily (and is hence a more powerful reducing agent) compared to the bromide ion.

Follow Ups:

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