The mixture contains 62 % one isomer and 38 % the enantiomer.
Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.
Then % (<em>S</em>) = 100 % -62 % = 38 %
ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %
a. Organic: C₁₀H₁₆KNO₉S₂; (CH₃)₄As₂; C₆H₁₂O₆
b. Inorganic: NaAsO₂; HSiCl₃; (BiO)₂CO₃; H₂P₂O₇; H₂O; CO₂
Compounds containing <em>both C and H</em> are organic.
Compounds that are <em>not organic</em> are inorganic.
The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
For more information about the partial pressure, refer to the link:
brainly.com/question/14623719
Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:
When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:
1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid
