.5 mol of A will be left over since 1.5 mol of A will be used for every 3 mol of B due to the 2:1 ratio established by the formula.
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = = 165°C
Depression in freezing point =
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
The 2nd one I think but I need some points
Moles of Hydrogen present: 100 / 2 = 50 moles
Moles of Nitrogen present: 200 / 28 = 7.14 moles
Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles
Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.
Molar ratio of Nitrogen : Ammonia = 1 : 2
Moles of ammonia = 7.14 x 2 = 14.28 moles
Answer:
Just Barely Base/Neutral
Explanation:
a pH of 8.0 is greater than Neutral (7.0) but is still neutral due to it being more neutral than a base