Q1
(1) Write a balance equation for the reaction between potassium chlorate and glucose that produces potassium chloride, carbon dioxide and water.
(15 marks) (read the section on “Naming Compounds” to find the chemical formula)
4 KClO3 + C6H12O6 = 4 KCl + 6 CO2 + 6 H2O <-from #7
(2) Write down the electronic configuration for atomic oxygen and indicate the number of unpaired electrons. (15 marks)
oxygen :1s2 2s2 2p4 the number of unpaired electrons : 2
(3) Define electronegativity and give an example for each of the following:
Electronegativity : a chemical property that describes the tendency of an atom or a functional group to attract electrons (or electron density) towards itself and thus the tendency to form negative ions <--copy from wiki
An ionic compound (5 marks)
NaCl
A non-polar diatomic molecule (5 marks)
O2
A polar diatomic molecule with a non-zero dipole moment indicated. (5 marks)
HCl :H一Cl : direction of non-zero dipole moment: → ( that is H→ Cl)
(4) Using VSEPR theory, predict the structure of the compound HClO4. What sort of hybridization will the central chlorine atom adopt? (15 marks)
hybridization state of Cl : sp3
shpae of HClO4 : tetrahedral shape since there are 4 bonds (O) attached with Cl
(5) Use molecular orbital theory to draw molecular energy level diagrams for H2, H2+, and HHe+. (15 marks)
do it by yourself .
作者: 36661124 時間: 11-9-6 09:55 PM
Question 2 (25 marks)
(1) One mole of chlorine corresponds to 2 × (6.023 × 1023) particles of chlorine atoms. By use of electrolysis, calculate the number of electrons required to generate 0.1 liter of chlorine gas at 25°C, 0.98 atm of external pressure? (Assume ideal gas behavior and the vapor pressure of water is negligible.) (10 marks)
Cl2 + 2e - -> 2Cl-
PV=NRT
(0.98x101325)(0.1x1000x10^-6)=n(8.314)(25+273)
n=4x10^-3 mole gas
so electrons required : 2x(6.023 × 10^23) x4x10^-3=4.82x10^21 e-
(2) A 1.268 g sample of metal carbonate MCO3 was treated with 100.00 mL of 0.1083 M H2SO4, yielding CO2 gas and a soluble metal sulfate solution. The solution was then boiled to remove all soluble carbon dioxide and titrated with 0.1241 M NaOH. A volume of 71.02 mL NaOH was consumed to neutralize the excess H2SO4.
a. What is the identity of metal M? (10 marks)
M is a group 2 metal because of M2+ .M is not Ca since CaSO4 is insoluble .so M maybe Mg2+(Mg metal) which MgSO4 is a soluble metal sulfate solution
b. If the density of carbon dioxide at room temperature is 1.799 g/L, calculate the volume of CO2 generated in the reaction with H2SO4. (5 marks)
2NaOH + H2SO4 ->Na2SO4 + 2H2O
mole of naoh used : (0.07102)x( 0.1241 M)=8.81x10^-3 mole
mole of excess h2so4 :8.81x10^-3/2=4.41x10^-3 mole
MCO3 + H2SO4 -> MSO4 + CO2 + H2O
total mole of h2so4 : (0.1x0.1083)=0.01083 MOLE
MOLE OF H2SO4 USED TO REACT TO MCO3 : 0.01083-4.41x10^-3=6.42X10^-3 MOLE
MOLE OF CO2: 6.42X10^-3 MOLE
pv=nrt : (101325)(V)= 6.42X10^-3 x8.314 x(25+273)
V=1.57x10^-4M³作者: 36661124 時間: 11-9-6 10:00 PM
