Answer : The % of (+) limonene isomer = 79%
The % of (-) limonene isomer = 0%
The % of enantiomeric excess = 58%
Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.
Given,
% of pure limonene enantiomer = The % of (+) limonene isomer = 79%
Therefore, The % of (-) limonene isomer = 0%
Formula used :

Where, ee → enantiomeric excess
Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


= 58%
<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We calculate as follows:
</span>1.40x10^23 molecules of N2 ( 1 mol / 6.022 x 10^23 molecules ) ( 28.02 g / mol ) = 6.51 g N2
Answer:
The partial pressure of CO2 is 712,8 in torr
Explanation:
Molar fraction = Pressure in a compound / Total Pressure
Molar fraction H20 = 21,2 / 734 = 0,0288
Sum of molar fraction in a sample = 1
1 - 0,0288 = 0,9712 (molar fraction of CO2)
Molar fraction CO2 = Pressure CO2 / Total pressure
0,9712 . 734 = Pressure CO2
712,8 =Pressure CO2
Answer:
No, it is not.
Explanation:
Most solutions do not behave ideally. Designating two volatile substances as A and B, we can consider the following two cases:
Case 1: If the intermolecular forces between A and B molecules are weaker than those between A molecules and between B molecules, then there is a greater tendency for these molecules to leave the solution than in the case of an ideal solution. Consequently, the vapor pressure of the solution is greater than the sum of the vapor pressures as predicted by Raoult’s law for the same concentration. This behavior gives rise to the positive deviation.
Case 2: If A molecules attract B molecules more strongly than they do their own kind, the vapor pressure of the solution is less than the sum of the vapor pressures as predicted by Raoult’s law. Here we have a negative deviation.
The benzene/toluene system is an exception, since that solution behaves ideally.
Combustion of any hrdrocarbon yields carbon dioxide and water such that the hydrogen and carbon are derived from the hydrocarbon.
1 mole of carbon dioxide has a mass out of which 12 g is the mass of carbon, therefore, 3.38 g of carbon will contain (3.38 × 12)/44 = 0.922 g of carbon.
Thus, since the hydrocarbon has a mass of 1 g, then the mass of hydrogen will be (1-0.922 g) = 0.078 g.
To get the empirical formula we divide the mass of each element by the atomic mass to get the number of moles.
Carbon= 0.922/12 = 0.0768 moles
Hydrogen = 0.078/1 = 0.078 moles
Then we get the ratio of the moles of carbon : hydrogen
= 0.0768 : 0.078
= 1 : 1.016
≈ 1: 1
Therefore the empirical formula of the hydrocarbon will be CH