1- we know that 4-tert-butylcyclohexanol is more polar than 4-tert-butylcyclohexanone (where the alcohols in general are more polar than ketons due to the hydrogen bond)
2- during separation via chromatography (in this case) the more polar solute will dissolve easily in polar solvents, where like dissolves like.
3- So, 4-tert-butylcyclohexanol will dissolve in ethyl acetete (which is polar) more than 4-tert-butylcyclohexanone, i.e, will have much higher Rf.
4- And also 4-tert-butylcyclohexanone will dissolve in dichloromethane (which lower in polarity than ethyl acetate) more than 4-tert-butylhexanol, i.e, will have much higher Rf
Answer:
39.1-32.5 and you will find your answer it always like that, you subtract your starting point from your ending point
Explanation:
<span><u>PA to PB 100 pm to the left of the nucleus, along the -x axis.</u>
<u>100 pm below the nucleus along the -z axis.</u>
PAPB 100 pm in front of the nucleus, along the -y axis. 100 pm behind the nucleus, along the +y axis.
PAPB 100 pm to the right of the nucleus, along the +x axis. 100 pm above the nucleus, along the +z axis. </span>
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
The temperature will change from 100K to 173.87 K
calculation
by use of law that is V1/T1=V2/T2
V1=3.75 L
T1=100k
V2=6.53 L
T2=?
make T2 the subject of the formula
T2=(V2 xT1)V1
=6.52 x100/3.75=173.87K
