The composition consists of 62 % one isomer and 38 % its enantiomer.
Assuming that the mixture comprises 62 % of the (R)-isomer.
Then the percentage of the (S) is calculated as 100 % - 62 % = 38 %.
Enantiomeric excess = % (R) - % (S) = 62 % - 38 % = 24 %.
We assume that the stated 50% is measured by volume. Molarity defines the concentration in terms of moles of solute per volume of solution.
To find the moles of NaOH, use: (0.1 moles / L)(0.4 L)
n = 0.04 moles of NaOH
Assuming we start with 1 mL of 50% NaOH solution,
(1 mL solution)(1.525 g/mL)(0.50) = 0.7625 g
Then, the number of moles calculates as follows,[
0.7625 g NaOH x (1 mol / 40 g) = 0.01906 moles of NaOH
The volume of solution required can be determined by:(0.04 moles of NaOH)(1 mL solution / 0.01906 moles of NaOH)
Thus, the needed volume comes out to be 2.09 mL
Answer: 2.09 mL
Answer:
5.57 L
Explanation:
- Apply the combined gas law formula.
- Change Celsius to Kelvin.
- Convert mmHg into atm.
- Enter the variables into the equation.
- I hope this information was beneficial! If you want, I can guide you through the process step by step.
Assuming the water vapor behaves as an ideal gas,
PV = nRT
For conversions, 760 mmHg = 101325 Pa and 1,000 L = 1 m³
(187.5 mmHg)(101325 Pa/760 mmHg)(5 L)(1 m³/1,000 L) = n(8.314 m³Pa/molK)(65+273 K)
Calculating for n,
n = 0.0445 mole of water
Considering the molar mass of water is 18 g/mol,
The mass of the vaporized water = 0.0445 * 18 = 0.8 g of water evaporated
Therefore,
The fraction of water that vaporized = 0.8/1.2 * 100 = 66.7%
