stevensonjacmp
Answered

At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields 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.

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.