beryllium (Be) and calcium (Ca)
chlorine (Cl) and helium (He)
oxygen (O) and carbon (C)
Answer:
Detail is given below.
Explanation:
Given data;
Abundance of Berkmarium-95 = 70%
Abundance of Berkmarium-97 = 28%
Abundance of Berkmarium-94 = 2%
Average atomic mass closer to which isotope = ?
Solution:
1st of all we will calculate the average atomic mass of Berkmarium.
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (70×95)+(28×97)+(2×94) /100
Average atomic mass = 6650 + 2716+ 188 / 100
Average atomic mass= 9554 / 100
Average atomic mass = 95.54 amu
The average atomic mass is closer to the isotope Berkmarium-95 because it is present in abundance as compared to the other two isotope. So this isotope constitute most of the part of Berkmarium.
The [H+] in the solution is found using the Kw expression: Kw = [H+][OH-]. Since Kw is known and [OH-] is given, [H+] can be calculated.
The concentration of hydrogen ions ([H+]) in a solution can be determined using the equilibrium constant for water (Kw). Kw is the product of the concentrations of hydrogen ions and hydroxide ions ([OH-]) in the solution. At 25°C, Kw has a constant value of 1.0 x 10^-14. Thus, if the [OH-] concentration in the solution is known, the [H+] concentration can be calculated by dividing Kw by [OH-]. In this case, the [OH-] is given as 5.42 x 10^-5 M, so [H+] can be found by dividing 1.0 x 10^-14 by 5.42 x 10^-5, which gives a value of 1.84 x 10^-10 M. Therefore, the (Ht] in the solution is 1.84 x 10^-10 M.
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B) SrI2
C) SrO
D) SrS