ME 2733 Chapter : HW1 Solutions
Document Summary
2. 2 knowing the weight wi of each isotope and its atomic mass one can evaluate the average atomic mass by using the following expression: (cid:2157)=(cid:2205)(cid:2778)(cid:2157)(cid:2778)+(cid:2205)(cid:2779)(cid:2157)(cid:2779)+(cid:2205)(cid:2780)(cid:2157)(cid:2780)+(cid:2205)(cid:2781)(cid:2157)(cid:2781) 2. 8 the na ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the same as neon . The cl ion is a chlorine atom that has acquired one extra electron; therefore, it has an electron configuration the same as argon. 4s electron configuration is that of a transition metal because of an incomplete d subshell. (b) the 1s. 3s and 3p subshells filled. electron configuration is that of a rare gas because it has both (c) the 1s. 5 electron configuration is that of a halogen because it is one electron deficient from having a filled l shell. (d) the 1s. 3s electron configuration is that of an alkaline earth metal because of two s electrons. (e) the 1s.
