Imagine our class has discovered a new element and named it Berkmarium. Berkmarium is known to have 3 naturally occurring isotopes. In nature, 70% of the element is Berkmarium-95, 28% is Berkmarium-97, and 2% is Berkmarium-94. Which isotope will the average atomic mass of Berkmarum be closest to, and why? Support your answer with a calculation.

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.

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

c)red bank

explanation:

red bank is a borough in monmouth county, new jersey, incorporated in 1908 and located on the navesink river, the area's original transportation route to the ocean and other ports.