<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:
Substance B, boiling point of 105 °C
Explanation:
Non volatile substances have high boiling points
Answer:
See explanation below
Explanation:
In this case, we have the equilibrium reaction which is:
H₂ + I₂ <------> 2HI Kp = 54
Now, we have the partial pressures of each element in equilibrium, therefore, we can use the expression of equilibrium in this case to calculate the remaining pressure:
Kp = PpHI² / PpH₂ * PpI₂
Solving for the partial pressure of iodine:
PpI₂ = PpHI² / PpH₂ * Kp
Replacing the given values, we have:
PpI₂ = (2.1)² / 0.933 * 54
PpI₂ = 4.41 / 50.382
PpI₂ = 0.088 atm
Answer: A dune is a mound of sand formed by the wind, usually along the beach or in a desert. Dunes can also be formed by strong currents beneath the water.
The Empirical Formula is
<span>a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.</span>
