What is the force needed to cause a 80kg object to accelerate at 1.6m/s??
1 answer:
You might be interested in
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
Answer:
a) galvanic cell
b)electrolytic cell
c) i) K=6.27x10'34
ΔG°=198790 J
ii) K=3.58x10'-34
ΔG°= 191070 J
d) E°=0.278 v
ΔG°= -26827 J
Explanation:
a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".
The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.
b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.
c) i) look image attached
ii) k = look image attached
ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)
ΔG° =-191070
d) E°= 0.0592 v/n x lg K
E°= 0.0592V / 1 X log 5.0X10'4
E°= 0.278 v
ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v
ΔG° = -26827 J
