stevensonjacmp
Answered

Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

*An element has three naturally occuring isotopes with the following masses and abundances
Isotopic mass(amu):
27.977
28.976
29. 974
Fractional abundance:
0.9221
0.0470
0.0309

a. calculate the atomic weight of this element
b.What is the identity of the element?​

Sagot :

dsdrajlin

Answer:

a. 28.08 amu

b. Silicon

Explanation:

Step 1: Given data

Isotope         Isotopic mass (mi)         Fractional abundance (abi)

 ²⁸X               27.977 amu                   0.9221

 ²⁹X               28.976 amu                  0.0470

 ³⁰X               29. 974 amu                 0.0309

Step 2: Calculate the atomic weight (aw) of this element

We will use the following expression.

aw = ∑mi × abi

aw = 27.977 amu × 0.9221 + 28.976 amu × 0.0470 + 29. 974 amu × 0.0309

aw = 28.08 amu

Looking at the periodic table, the element with atomic weight = 28.08 amu is silicon.

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.