- Messages
- 40
- Reaction score
- 44
- Points
- 28
Thanks a lotQ1. the mass that is given is the mass of Ag so use it to find out the moles of Ag
n of Ag = 0.216/108 = 2 x 10^-3
now multiply this with avagadro's constant to find the no. of atoms in this much mass (or moles) of Ag
no. of atoms of Ag present = [2 x 10^-3] X [6.02 x 10^23] = 1.204 x 10^21
now they asked us no.of atoms per unit area (on 1 cm2) and have already given us the area 150 cm2
divide the total no. of atoms by the given area and you will have atoms per unit area
no. of atoms per unit area = [1.204 x 10^21]/150 = 8.03 x 10^18, thus A is your answer
Q10. moles of O2 is given use this to find out the moles of NO and moles of NO2 is also given
use cross multiplication by seeing the ratio of moles from the equation provided
2 : 1
x : 0.8
x = 2 x 0.8 = 1.6
so we have moles of each as
O2 = 0.8
NO = 1.6
NO2 = 4 - 1.6 = 2.4 (because 1.6 moles of NO2 have been converted to NO)
Kc = ([NO]^2 x [O2])/[NO2]^2
Kc = ([1.6]^2 x [0.8])/ [2.4]^2
thus our answer is D
Q19. I was stuck at it so i opened the Examiner's Report
and this was the answer in it
In Question 19 candidates were asked to recognise any chiral centre in three compunds for which only the
molecular formula was given. Many candidates failed to notice that C3H6I2 could have the structure
CH3CHICH2I.
so our answer is B
Q37. no idea sorry
