The question reads: Excess electrons are placed on a small lead sphere with mass

Question

The question reads: Excess electrons are placed on a small lead sphere with mass7.40 g so that its net chargeis            -2.30 * 10-9 C (a) Find the number of excess electrons on thesphere.

(b) How many excess electrons are there per lead atom? The atomicnumber of lead is 82, and its atomic mass is 207 g/mol.
***Please show work so i can undertandplease*** The question reads: Excess electrons are placed on a small lead sphere with mass7.40 g so that its net chargeis            -2.30 * 10-9 C (a) Find the number of excess electrons on thesphere.

(b) How many excess electrons are there per lead atom? The atomicnumber of lead is 82, and its atomic mass is 207 g/mol.
***Please show work so i can undertandplease*** (a) Find the number of excess electrons on thesphere.

(b) How many excess electrons are there per lead atom? The atomicnumber of lead is 82, and its atomic mass is 207 g/mol.
***Please show work so i can undertandplease***

Explanation / Answer

mass m = 7.4 g Net charge q = -2.3* 10^-9 C we know q = ne from this the number of excess electrons on the sphere n = q /e where e = charge of electron = -1.6 *10^ -19C plug the values weget   n = 1.4375 * 10 ^ 10 (b). the number of excess electrons in 207 g lead = ( 207g / 7.4 g ) * 1.4375 * 10 ^ 10                                                                             = 40.211 * 10^10 1 mole contain 6.0238 10 ^ 23 atoms So, the number of excess electrons on 6.023*10^23 atoms =40.211*10^10 Therefore the number of excess electrons on one lead atom = (40.211*10^10) / ( 6.023*10^23)                                                                                          = 6.676 * 10 ^ -13

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