The best method for separating two liquids in a mixture is distillation among the choices. Distillation separates liquids in terms of their volatility. Evaporation separates liquid from solid. This also applies for filtration which applies porous media for separation. Sorting is applied for two or more different kinds of solids. Answer is distillation.
Answer: In atoms, there are a total of four quantum numbers: the principal quantum number (n), the orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms).
Answer: remains constant.
Justification:
1) The phase changes are:
i) Boiling: pass from liquid to gas (absorbs heat energy)
ii) Condensation: pass from gas to liquid (release heat energy)
iii) Melting: pass from solid to liquid (absorb heat energy)
iv) Freezing: pass from liquid to solid (release heat energy)
v) Sublimation: pass from solid to gas (absorbs heat energy)
vii) Deposition: pass from gas to solid (release heat energy)
2) When a phase change occurs, whichever it is, the heat energy related with the process, either absortion or release, is used, to overcome the intermolecular forces (in the case of heat energy absortion) or to create stronger intermolecular forces (in the case of heat energy release).
Because of that, the heat energy exchange does not change the temperature of the substance.
Answer:
CH₄ - 162 ⁸C
CH₃CH₃ -88.5 ⁸C
(CH₃)₂ CHCH₂CH₃ 28 ⁸C
CH₃3(CH2)₃CH₃ 36 ⁸C
CH₃OH 64.5 ⁸C
CH₃CH₂OH 78.3 ⁸C
CH₃CHOHCH₃ 82.5 ⁸C
C₅H₉OH 140 ⁸C
C₆H₅CH₂OH 205 ⁸C
HOCH₂CHOHCH₂OH 290 ⁸C
Explanation:
To answer this question we need first to understand that for organic compounds:
a. Non polar compounds have lower boiling points than polar ones of similar structure and molecular weight.
b. Boiling points increase with molecular weight. In alkane compounds if we compare isomers, the straight chain isomer will have a higher boiling point than the branched one (s) because of London dispersion intermolecular forces.
a. The introduction of hydroxyl groups increase the intermolecular forces and hence the boiling points because the electronegative oxygen, and, more importantly the presence of hydrogen bonds.
Considering the observations above, we can match the boiling points as follows:
CH₄ - 162 ⁸C
CH₃CH₃ -88.5 ⁸C
(CH₃)₂ CHCH₂CH₃ 28 ⁸C
CH₃3(CH2)₃CH₃ 36 ⁸C
CH₃OH 64.5 ⁸C
CH₃CH₂OH 78.3 ⁸C
CH₃CHOHCH₃ 82.5 ⁸C
C₅H₉OH 140 ⁸C
C₆H₅CH₂OH 205 ⁸C
HOCH₂CHOHCH₂OH 290 ⁸C
Note: There was a mistake in the symbols used for the 162 and 88.5 values which are negative and correspond to the common gases methane and ethane
