<u>Answer:</u> The mass of second isotope of indium is 114.904 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the mass of isotope 2 of indium be 'x'
Mass of isotope 1 = 112.904 amu
Percentage abundance of isotope 1 = 4.28 %
Fractional abundance of isotope 1 = 0.0428
Mass of isotope 2 = x amu
Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %
Fractional abundance of isotope 2 = 0.9572
Average atomic mass of indium = 114.818 amu
Putting values in equation 1, we get:
![114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu](https://tex.z-dn.net/?f=114.818%3D%5B%28112.904%5Ctimes%200.0428%29%2B%28x%5Ctimes%200.9572%29%5D%5C%5C%5C%5Cx%3D114.904amu)
Hence, the mass of second isotope of indium is 114.904 amu
Answer:
The both compounds are different.
Explanation:
In order to confirm weather both compounds are same we will check the mole ration. If it is same the compounds will be same.
Given data:
For compound 1.
Mass of hydrogen = 15 g
Mass of oxygen = 120 g
Moles of hydrogen and oxygen = ?
Number of moles of hydrogen = 15 g/ 1g/mol = 15 mol
Number of moles of oxygen = 120 g/ 16 g/mol = 7.5 mol
Total number of moles = 22.5 mol
% of hydrogen = 15 /22.5 × 100 = 66.7%
% of oxygen = 7.5 / 22.5× 100 = 33.3%
For compound 2:
Mass of hydrogen = 2 g
Mass of oxygen = 32 g
Moles of hydrogen and oxygen = ?
Number of moles of hydrogen = 2 g/ 1g/mol = 2 mol
Number of moles of oxygen = 32 g/ 16 g/mol = 2 mol
Total number of moles = 4 mol
% of hydrogen = 2 /4 × 100 = 50%
% of oxygen = 2 / 4× 100 = 50%
The mass of magnesium should be less than 0.09g to enable a faster reaction rate. Magnesium reacts to form a white coating around it which stops the reaction. The lesser the gram the faster the reaction before the coating is formed. It is also advisable to use magnesium fillings to increase the rate of reaction.
Answer:
23.8
Explanation:
Formula
weight % = weight of solute/ weight of solution x 100
weight of solution = weight of salt + weight of water
weight of solution = 1.62 lb + 5.20 lb = 6.82 lb
weight % = 1.62 / 6,82 x 100
weight % = 0.238 x 100
weight % = 23.8