Answer:
A)
,
, 
A = 1.5×
, A = 1.9×
, A=1.5×
B) 4.469
Explanation:
From Arrhenius equation

where; K = Rate of constant
A = Pre exponetial factor
= Activation Energy
R = Universal constant
T = Temperature in Kelvin
Given parameters:




taking logarithm on both sides of the equation we have;

since we have the rate of two different temperature the equation can be derived as:


= 19846.04×7.544×
= 1.497
=
= 4.469
3rd one:
it is very reactive because it does not have a full Valence shell.
this is because it's in group 1 so it has one electron in its outer shell, and it wants to have a full outer shell ( which it can gain by losing the electron in a reaction).
Hope this helps :)
Answer:
we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
Explanation:
when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate