Every mole of MgCl2 reacts with 2 moles of KOH, therefore the 4 moles of KOH will only react with 2 moles of MgCl2, making it the limiting reagent and therefore KOH determines how much Mg(OH)2 is produced.
Answer:
A) 100 mL
Explanation:
At constant temperature and number of moles, Using Boyle's law
Given ,
V₁ = 200 mL
V₂ = ?
P₁ = 60 kPa
P₂ = 120 kPa
Using above equation as:
In the so called rain shadow effect we have interaction between all of the four major Earth spheres. When we have a coastal region where there's a high mountain range, the part of the mountain that is facing the sea will differ a lot from the part of the mountain that is on the other side. The water from the sea evaporates. The water vapor makes the air wet. The warm and wet air masses from the sea will come to the coastline, once they reach the mountain they will start to accumulate as they can not pass through it. As they accumulate rainfall appears. The rainfall contributes to a lush vegetation on this side of the mountain (windward side). The rain shadow effect appears on the leeward side of the mountain, and it mostly gets dry, strong, downward winds. These conditions result in drier climate, much less vegetation, and much increased erosion. Thus we can easily see that we have in this case interaction between the hydrosphere (the sea and the rainfall), the geosphere (the ground, soil, rocks), biosphere (the vegetation), and atmosphere (the winds, the clouds).
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
