<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:
1.208x10⁻³M and 392.5ppm La(NO3)3
Explanation:
The reaction that occurs is:
La2O3 + 6HNO3 → 2La(NO3)3 + 3H2O
Molarity is defined as the moles of solute (In this case, LaO3) per liter of solution. And ppm, are mg of solute per liter of solution.
To solve this question we must find the moles of La(NO3)3 produced and its mass in milligrams to find molarity and ppm:
<em>Moles La2O3 -Molar mass: 325.81g/mol-</em>
0.1968g * (1mol / 325.81g) = 6.04x10⁻⁴ moles La2O3
<em>Moles La(NO3)3:</em>
6.04x10⁻⁴ moles La2O3 * (2mol La(NO3)3 / 1mol La2O3) = 1.208x10⁻³ moles La(NO3)3
<em>Molarity:</em>
1.208x10⁻³ moles La(NO3)3 / 1L =
<h3>1.208x10⁻³M</h3>
<em>Mass La(NO3)3 -Molar mass: 324.92g/mol-</em>
1.208x10⁻³ moles La(NO3)3 * (324.92g / mol) = 0.392.5g La(NO3)3
In mg:
392.5mg La(NO3)3 / 1L =
392.5ppm La(NO3)3
0.25 moles of H3PO4 is needed to
neutralize 1 liter of a 0.75M solution of NaOH. I am hoping that this answer
has satisfied your query and it will be able to help you in your endeavor, and
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