Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Answer:
Option (D)
Explanation:
Weathering is usually defined as the disintegration of rocks at the surface of the earth. This break down of rocks is mainly caused by the geological processes that occur on the earth's surface. This process results in the formation of sediments that are transported and deposited in a new environment.
This weathering process primarily takes place in three different ways such as-
- Physical weathering- Here, the rocks are broken down by the physical agents such as wind, water, ice.
- Chemical weathering- Here the rocks are broken down when interacts with the chemical containing water.
- Biological weathering- here, the rocks are broken down due to the activities done by organisms such as plants and animals.
In the given condition, Quincy can see a real example of rock weathering in the high mountainous region, as in the mountainous region the rocks are frequently weathered and eroded by the agents such as wind, water, and ice.
Thus, the correct answer is option (D).
They are all transioning in states of matter
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23
Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles
Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol
Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams
Answer:
Explanation:
Common Examples of the Law of Definite Proportions
Water, written as the chemical compound H20, is made up of atoms of hydrogen and oxygen. If one oxygen atom is combined with two hydrogen atoms, water is created.
