-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
chem paper1 2007 help needed
- Thread starter 1992
- Start date
- Messages
- 677
- Reaction score
- 190
- Points
- 53
q2: find number of moles of the oxide by 2 diff means....
n = 0.23/(14x + 15y)
n= 120/24000
equate them...you'll get an equation...14x + 16y = 46...now a hit n trial method.....from options A to D, put x n y values until u get 46 :-D
q5: CN- has a lone pair on N atom and triple covalent bond b/w C and N. carbon has a lone pair too as it has one of its own electron in its outer shell and one gained giving a negative charge to CN..so bonding
pair of electrons are 3, lone pairs on C n N are 1...
q8: delta H of a reaction = H of products - H of reactants....calculate it using this formula....all elements have zero enthalphy changes of formation...
q9: nw Kc for 2 shuld be 1/2...but there r molar differences...if u multiply eq one by 0.5 or 1/2...u'll get eq 2.....so its (1/2)^0.5 = 1/underroot 2
q16: down the group, enthalphy changes of formation of hydrogen halides become more endothermic...b/c bond energies of hydrogen halides decrease...which means bond formation is less exothermic...
q26: the alcohol must be a tertiary one since it is nt oxidised....and a chiral one too....so on a carbon atom, one side -OH group will be attached...and on the rest three different alkyl groups...but to give minimum no of Carbon atoms to the compound...so R-groups are CH3, C2H5, and C3H7..in total..7 carbons must be present!
n = 0.23/(14x + 15y)
n= 120/24000
equate them...you'll get an equation...14x + 16y = 46...now a hit n trial method.....from options A to D, put x n y values until u get 46 :-D
q5: CN- has a lone pair on N atom and triple covalent bond b/w C and N. carbon has a lone pair too as it has one of its own electron in its outer shell and one gained giving a negative charge to CN..so bonding
pair of electrons are 3, lone pairs on C n N are 1...
q8: delta H of a reaction = H of products - H of reactants....calculate it using this formula....all elements have zero enthalphy changes of formation...
q9: nw Kc for 2 shuld be 1/2...but there r molar differences...if u multiply eq one by 0.5 or 1/2...u'll get eq 2.....so its (1/2)^0.5 = 1/underroot 2
q16: down the group, enthalphy changes of formation of hydrogen halides become more endothermic...b/c bond energies of hydrogen halides decrease...which means bond formation is less exothermic...
q26: the alcohol must be a tertiary one since it is nt oxidised....and a chiral one too....so on a carbon atom, one side -OH group will be attached...and on the rest three different alkyl groups...but to give minimum no of Carbon atoms to the compound...so R-groups are CH3, C2H5, and C3H7..in total..7 carbons must be present!
- Messages
- 677
- Reaction score
- 190
- Points
- 53
Nov07:
q1:that is quite simple....for 150cm^3 , 0.216g of Ag was coated...so for 1cm^3 , mass quoted will be 0.216/150 = 1.44 x 10^-3 ....so moles of silver per 1cm^-3 = 1.44 x 10^-3/ 107 = 1.35 x 10^-5...........
No of atoms = 1.35 x 10^-5 x 6.02 x 10^23 = 8 x 10^18................................................................
q9: ive answerd tht before...check out the correct topic...
q10: 1 mole of oxygen requires 2 moles of NO2....0.8 moles of oxygen require 0.8 x 2 =1.6 moles of NO2...
so moles of NO2 left at equlibrium = 4-1.6= 2.4......similarly find the no of moles of NO produced from 1.6 moles of NO2...which are 1.6moles...
so nw u'll find it easy to arrange an expression for Kc.......
q37: hey just c organic chemistry 1nce!!!!!!!
q1:that is quite simple....for 150cm^3 , 0.216g of Ag was coated...so for 1cm^3 , mass quoted will be 0.216/150 = 1.44 x 10^-3 ....so moles of silver per 1cm^-3 = 1.44 x 10^-3/ 107 = 1.35 x 10^-5...........
No of atoms = 1.35 x 10^-5 x 6.02 x 10^23 = 8 x 10^18................................................................
q9: ive answerd tht before...check out the correct topic...
q10: 1 mole of oxygen requires 2 moles of NO2....0.8 moles of oxygen require 0.8 x 2 =1.6 moles of NO2...
so moles of NO2 left at equlibrium = 4-1.6= 2.4......similarly find the no of moles of NO produced from 1.6 moles of NO2...which are 1.6moles...
so nw u'll find it easy to arrange an expression for Kc.......
q37: hey just c organic chemistry 1nce!!!!!!!