Answer:
4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.
Explanation:
Deviations from ideality are due to intermolecular forces and to the nonzero volume of the molecules themselves. At infinite volume, the volume of the molecules themselves is negligible compared with the infinite volume the gas occupies.
However, the volume occupied by the gas molecules must be taken into account. Each <u>molecule does occupy a finite, although small, intrinsic volume.</u>
The non-zero volume of the molecules implies that instead of moving in a given volume V they are limited to doing so in a smaller volume. Thus, the molecules will be closer to each other and repulsive forces will dominate, resulting in greater pressure than the one calculated with the ideal gas law, that means, without considering the volume occupied by the molecules.
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:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
Answer:
The composition of 1-propanol in vapor phase = 11.54 %
The composition of 2-propanol in vapor phase = 88.46 %
Explanation:
Mole fraction of components in liquid phase:
1-propanol = 
2-propanol = 
Partial pressure of 1-propanol =
Partial pressure of 2-propanol =
According to Raoults law:
Mole fraction of components in vapor phase:
1-propanol = 
The composition of 1-propanol in vapor phase:
100 × 0.1154= 11.54 %
2-propanol = 
The composition of 2-propanol in vapor phase:
100 × 0.8846 = 88.46 %
