Answer:
Look on the picture.
Explanation:
He could find only 2 isomers of n-hexane alkenes for this reaction. Other two could be marked from other direction.
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?
n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)
c₂v₂=2c₁v₁
c₂=2c₁v₁/v₂
c₂=2*2.00*0.25/2.00=0.5 mol/L
0.5 M NaOH
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
Amines are derivatives of
Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;
Primary Amines: In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.
Secondary Amines: In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.
Tertiary Amines: In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.
Below are three isomers of tertiary amines with molecular formula
C₅H₁₃N.
<span>1 mole of benzene (78g) requires 30.8 kJ/ of heat, so 11.5g will need ..... (it's a proportion calculation.) Temperature does not change at BPt and is not relevant if the temp of the liquid is already at the BPt
ne definition of entropy is qrev/T, where qrev is the heat added in reversible operation (for complicated reasons pertaining to heat as a path function) and T is the temperature at which this is done.
Phase changes are particularly good examples for calculations of changes in entropy, since temperature will not change will the bonds of a state are being broken.
The calculations required boils down to:
1) finding the moles of benzene given from molar mass.
2) multiplying that moles by the heat of vaporization.
3)diving the heat energy required by the temperature of boiling point.</span>