In Reply to: (01 Nov) Qns posted by Kyo
on October 31, 2007 at 16:16:03:
Ans to Q17
3 isomers in total.
1) 1-Bromo-2-Fluoro-ethane
AND
2) Dextrorotatory-1-bromo-1-fluoro-ethane optical isomer, and 3) Levorotatory-1-bromo-1-fluoro-ethane
optical isomer
OR
2) Rectus-1-bromo-1-fluoro-ethane optical isomer, and 3) Sinister-1-bromo-1-fluoro-ethane
optical isomer
OR
2) (+)-1-bromo-1-fluoro-ethane optical isomer, and 3) (-)-1-bromo-1-fluoro-ethane
optical isomer
Ans to Q18
Silicon is in period 3, carbon is in period 2.
Bond length between silicon and oxygen is too long to form pi bond
in double bond, compared to smaller bond length between carbon and silicon
so easier to form pi bond in double bond; there is decreased ease of overlap
of silicon 3p orbitals to form pi bond in double bond (with oxygen), compared
to greater ease of overlap of carbon 2p orbitals to form pi bond in double
bond (with oxgen). Consequently, it is thermodynamically favourable for
silicon and oxygen to form single bonds, and for carbon and oxygen to form
double bonds.
Ans to Q19
a) Enthalpy of reaction that produces liquid water will be more exothermic
than enthalpy of reaction that produces gaseous water; because some of
the heat energy is used up to convert liquid water to gaseous water.
b) (i) -282.5 kJ/mol (ii) +342 kJ/mol
Ans to Q20
a) 52.8% b) 1.59 atm-1
2NO + O2 <--> 2NO2
Initial (atm) | (2/3)(5) | (1/3)(5) | 0
Change (atm) | -1.76 | -0.88 | +1.76
Equilibrium (atm) | 1.573 | 0.787 | 1.76
Ans to Q21
Formula is C3H6
#1 - CxHy + (x + y/4) O2 --> x CO2 + y/2 H2O
Note : don’t memorize the above equation, derive it yourself as follows
:
Let coefficient of O2 on LHS be k; hence 2k (oxygen atoms) = 2x + y/2
(oxygen atoms) ; k = (x + y/4)
#2 Soda lime = sodium hydroxide (soda) + calcium hydroxide (lime).
Carbon dioxide is an acidic oxide (forms carbonic acid), which is 'removed'
from the air by neutralization with alkaline sodium hydroxide.
Ans to Q22
a) 0.075 mol
Apply stoichiometry to balanced equation 4KClO3 --> 3KClO4 + KCl
b) In sodium hypochlorite / chlorite / chlorate / perchlorate (Na+ and
ClO- / ClO2- / ClO3- / ClO4-), because oxygen is more electronegative than
chlorine, the oxidation state of O in ClO- is -2, which means the oxidation
state of Cl in ClO- / ClO2- / ClO3- / ClO4- must be +1, +3, +5, +7 respectively.
Since 10 mole of thiosulphate ions (S2O4)2- reduces 5 moles of iodine
molecules, it means 10 moles of iodide ions must be have oxidized by the
sodium hyplochlorite.
Since each mole of iodie ions being oxidized must donate 1 mole of
electrons to 2 moles of the oxidizing agent, this means 2 ClOn- receives
10 electrons.
From the information given in the question, as well as from observing
the oxidation states of chlorine versus oxygen, the student can surmise
that the 'positively-charged' chlorine atoms are the ones who receive the
electrons, so we leave the oxygen out of the calculations.
2 moles of hypochlorite ions (ClO-) can only receive 2 electrons (since
oxidation state of chlorine here is only +1)
2 moles of chlorite ions (ClO2-) can only receive 6 electrons (since
oxidation state of chlorine here is only +3)
2 moles of chlorate ions (ClO3-) can only receive 10 electrons (since
oxidation state of chlorine here is only +5)
2 moles of perchlorate ions (ClO-) can only receive 14 electrons (since
oxidation state of chlorine here is only +7)
Therefore, the answer n = 3.