Explanation:
The required concentration of 
M1 =0.222 M.
The required volume of is V1 =225 mL.
The standard solution of is M2 =16 M.
The volume of standard solution required can be calculated as shown below:
Since the number of moles of solute does not change on dilution.
The number of moles 
Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.
Explanation:
Radium (atomic number 88) has similar properties to barium and is also in the Group 2 category. However, radium is a radioactive element and is generally under the category of radioisotopes in addition to being an alkaline earth metal, because it is not a stable element.
Answer:
C2H5O
Explanation:
In a 100 g sample we would have
53.31 g of C
11.18g of H
35.51g of O
First, we find the relative number of atoms of each element by dividing the number of grams the element has in the compound by its atomic mass.
Atomic mass of carbon is 12.011
Relative number of carbon atoms = 53.31 / 12.011 = 4.4
Atomic mass of hydrogen = 1.007
Relative number of hydrogen atoms : 11.18/1.007 = 11.1
Atomic mass of oxygen : 15.999
Relative number of oxygen atoms : 35.51 / 15.999 = 2.2
Now we find a ratio of the relative number of atoms by dividing the # of relative atoms of each element by the element's relative number of atoms that had the lowest number. ( oxygen which had 2.2 ) The outcome of each will be the subscript or number of atoms of each element.
Carbon : 4.4 / 2.2 = 2
Hydrogen : 11.1 / 2.2 = 5
Oxygen : 2.2 / 2.2 = 1
The answer is C2H5O
Answer:
There will be weight, there will be volume, there will be height
Explanation:
Answer:
15.5 gm
Explanation:
What is the mass of phosphorus that contains twice the number of atoms found in 14 g of iron?
[Relative atomic mass : P = 31; Fe = 56]
14 gm Fe = 14gm/ 56 gm/mole = 14 mole gm/56gm = 14/56 mole
0.25 moles
2 X 0.25 = 0.5 moles
1 mole P = 31 gm
so
0.5 moles P =31/2 =15.5 gm
