Consider the stable elements through lead (Z = 82). In how many instances are the atomic weights of the elements out of order relative to the atomic numbers of the elements?

Short Answer

Expert verified
There are \(3\) instances where the atomic weights are out of order relative to the atomic numbers of the elements up to lead (Z = 82), namely: 1. Argon (Z = 18, A = 39.95) and Potassium (Z = 19, A = 39.10) 2. Cobalt (Z = 27, A = 58.93) and Nickel (Z = 28, A = 58.69) 3. Tellurium (Z = 52, A = 127.60) and Iodine (Z = 53, A = 126.90)

Step by step solution

01

Identify the elements to consider

We need to consider the elements up to lead (Z = 82). This means we will consider all the elements from hydrogen (Z = 1) to lead (Z = 82).
02

List the atomic weights and atomic numbers

Obtain a list of atomic weights and atomic numbers for these elements. This information can be found on a periodic table or other reliable sources.
03

Compare the atomic weights and atomic numbers

Go through the list and compare each element's atomic weight to the atomic weights of elements with lower atomic numbers. If the atomic weight is less than that of an element with a lower atomic number, it's considered out of order.
04

Count the instances

Keep track of the number of instances where the atomic weights are out of order relative to the atomic numbers. After carefully analyzing the elements up to lead (Z = 82) and comparing their atomic weights and atomic numbers, it turns out that there are \(3\) instances in which the atomic weights are out of order relative to the atomic numbers of the elements: 1. Argon (Z = 18, A = 39.95) and Potassium (Z = 19, A = 39.10) 2. Cobalt (Z = 27, A = 58.93) and Nickel (Z = 28, A = 58.69) 3. Tellurium (Z = 52, A = 127.60) and Iodine (Z = 53, A = 126.90) Thus, there are \(3\) instances in which the atomic weights are out of order relative to the atomic numbers.

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