Answer:
Source, processing and distribution are the components of water system.
Explanation:
There are three parts of water system i. e. the source, the processing and distribution. Water is extracted from a source such as underground water, lake or river etc. After extraction this water is transported to the processing unit where it can be purified and after purification it is distributed to all places where it is needed. Potential energy is a form of energy that flows through this water system because the water is extracted from a depth and we know that depth and height refers to potential energy.
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Solution : Given,
Density of solution = 
Molar mass of sulfuric acid (solute) = 98.079 g/mole
98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.
Mass of sulfuric acid (solute) = 98.0 g
Mass of solution = 100 g
Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g
First we have to calculate the volume of solution.
Now we have to calculate the molarity of solution.
Now we have to calculate the molality of the solution.
Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Answer:
c
Explanation:
because the fire gives off radiation
We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.
150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo
1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo
Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
