You can pick it up and move it
Those are both correct! great job, keep up the good work (-:
Answer:
The answer is "
"
Explanation:
Please find the complete question in the attached file.
Equation:
at 
at equilibrium 
![p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\](https://tex.z-dn.net/?f=p%3D%200.47%20%5C%20%5C%20atm%5C%5C%5C%5CSO_2%3D3.3-0.47%20%3D%202.83%20%5C%20%5C%20atm%5C%5C%5C%5CO_2%3D%200.74%20-%5Cfrac%7B0.47%7D%7B2%7D%3D0.74-0.235%3D0.555%20%5C%20atm%5C%5C%5C%5CK_P%3D%5Cfrac%7B%5BPSO_3%5D%5E2%7D%7B%5BPSO_2%5D%5E2%5BPO_2%5D%7D%5C%5C%5C%5C)
Increase at the temperature
Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
