The mixture of rock particle sand humus is called the soil.
If soil contains greater proportion of big particles it is called sandy soil. If the proportion of fine particles is relatively higher, then it is called clayey soil. If the amount of large and fine particles is about the same, then the soil is called loamy.
Answer: 2.4 ml
Solution :
Molar mass of 
= 17 g/mole
Given,: 28% w/w of solution means 28 g of ammonia in 100 g of solution.
Mass of solution = 100 g
Now we have to calculate the volume of solution.
Molarity : It is defined as the number of moles of solute present in one liter of solution.
where,
n = moles of solute 
= volume of solution in liter = 0.11 L
Now put all the given values in the formula of molarity, we get
Using molarity equation:
Answer:
The answer to your question is: V = 6.93 L
Explanation:
Data
N₂ = 5.6 g
Volume of NH₃ = ?
14 g of N ---------------- 1 mol
5.6 g ----------------------- x
x = (5.6 x 1) / 14 = 0.4 mol of N
Reaction
N₂ + 3H₂ ⇒ 2NH₃
1 mol of N₂ ---------------- 2 moles of NH₃
0.4 mol of N₂ -------------- x
x = (0.4 x 2) / 1
x = 0.8 mol of NH₃
Formula
PV = nRT
P = 5200 torr = 6.84 atm
V = ?
n = 0.8
R = 0.082 atm L/ mol °K
T = 450°C = 723°K
Substitution
V = (0.8)(0.082)(723) / 6.84
V = 6.93 L
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 %
