Two naturally occurring isotopes exist for the hypothetical element dippium. 25.0 percent of dippium atoms have a relative mass of 17.0 amu and 75.0 percent a relative mass of 21.3 amu. What is the atomic mass of dippium?

Please show the work

thanks

17(0.25) + 21.3(0.75) = ?

44.7

Ah, dippium, a mysterious and hypothetical element! Let's dig into the math and uncover the atomic mass while having some fun along the way.

To calculate the atomic mass, we need to take into account the percentages and relative masses of each isotope.

Let's assume we have 100 atoms of dippium:

25% of 100 atoms = 25 atoms with a relative mass of 17.0 amu
75% of 100 atoms = 75 atoms with a relative mass of 21.3 amu

To calculate the atomic mass, we'll multiply the relative masses by the percentages, sum them up, and divide by 100.

(25 atoms * 17.0 amu) + (75 atoms * 21.3 amu)
-----------------------------------------------
100

(425 amu + 1,597.5 amu)
----------------------
100

2,022.5 amu
------------
100

Atomic mass of dippium is 20.225 amu.

So, dippium's atomic mass is 20.225 amu. Now you know that even in the world of hypothetical elements, math can be a real clown sometimes!

To find the atomic mass of dippium, we need to calculate the weighted average of the isotopes based on their relative abundances and relative masses.

Step 1: Convert the percentages to decimal form:
The given percentages are 25.0% and 75.0%. We need to divide these values by 100 to convert them to decimal form:
25.0% = 25.0/100 = 0.25
75.0% = 75.0/100 = 0.75

Step 2: Calculate the weighted average:
Now, we calculate the weighted average by multiplying the relative abundance of each isotope by its relative mass and then summing up the results.

Weighted average = (Relative abundance of isotope 1 × Relative mass of isotope 1) + (Relative abundance of isotope 2 × Relative mass of isotope 2)

Weighted average = (0.25 × 17.0 amu) + (0.75 × 21.3 amu)

Weighted average = (4.25 amu) + (15.975 amu)

Weighted average = 20.225 amu

Therefore, the atomic mass of dippium is approximately 20.23 amu.

Note: It is important to round the answer to the appropriate number of significant figures based on the given data. In this case, the relative abundance values are given to two significant figures, so the final answer should also be rounded to two significant figures.