Posted by Kyo on November 03, 2007 at 08:26:38:
In Reply to: (03 Nov) Qns posted by Kyo on November 02, 2007 at 09:40:49:
Ans to Q29
Let no. of mol of the two hydrated salts be x and y.
Since no. of mol of water evaporated away = 0.111 mol,
5x + 7y = 0.111 --- (eqn 1)
Since 3g of salts are left over after evaporating the water,
159.6x + 120.4y = 3.0 --- (eqn 2)
Solve simultaneous eqns for x; x = 0.0148 mol.
Hence % by mass of CuSO4.5H2O in mixture = 73.86%
Ans to Q30
From the Data Booklet,
MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O ~ ~ ~ E = +1.52 V
Fe3+ + e- --> Fe2+ ~ ~ ~ E = +0.77 V
At X (equivalent to the point of half-neutralization in an acid-base reaction), [Fe2+] = [Fe3+], E = +0.77 V
At Y (equivalent to the point of half-neutralization in an acid-base reaction), [MnO4-] = [Mn2+], E = +1.52 V
By stoichiometry and from graph, mol of Fe2+ oxidized by MnO4 = 5 x (mol of MnO4-) = 5 x (0.02 x 16.5/1000) = 1.65 x 10-3 mol.
Ans to Q31
From mass of bubble, determine (using weighted average of O2 and CO2) the number of moles of gas. Just before it bursts, pressure = 1 atm and temperature = 25 deg C = 298 K. Apply PV = nRT; volume can be found, hence diameter can be found. Ans to (i) is 3.0 cm.
Body temperature of cold blooded reptile = surrounding water temperature at 6.4 atm with number of moles of gas known and volume of bubble known. Ans to (ii) is 8 deg C.
Apply PV = nRT for "bubble in sun", where number of mol is same as (i) and (ii). Ans to (iii) is 0.0159 cm.
Students should be aware that there are two values for gas constant R :
EITHER use
Gas constant R = 8.31
Pressure = pascals (Pa not kPa or atm)
Volume = m3 (not dm3 or cm3)
OR
Gas constant R = 0.0821
Pressure = atm (not Pa or kPa)
Volume = dm3 (not m3 or cm3)
If you want to stick to using only one of the above gas-constant-R values all the time, then you must know how to convert Pa to atm and vice-versa : 1 atm = 1.01325 x 10^5 pascals.
Ans to Q32
a) Carbocation intermediates are stabilized by electron donating ‘R’ groups, which is the underlying reason why Markonikov’s rule works in most cases. In 3,3,3 trichloroprop-1-ene however, the 3 electron withdrawing Cl groups, further destabilize the carbocation, causing the Markonikov product to be the minor product.
b) Under aqueous conditions, the OH- ion acts as a nucleophile, attaching itself to an electron-poor atom. Under alcoholic (eg. ethanolic) conditions, the OH- ion acts as a base, plucking off a proton. Zaitsev’s rule works in most cases because (alkenes with) double bonds are stabilized by (more) ‘R’ groups. Thus, but-2-ene is more stable (and thus is the major product) compared to but-1-ene.
