Question

name jayden coleman period 2 date
isotopes, ions, and atoms worksheet
atomic # = # of protons. mass # = atomic # + neutrons. protons = electrons when charge is zero.

atomic #	mass #	# p⁺	# e⁻	# n⁰	charge	symbol
2)	180		71	109
3)		40	38	46
4) 92	238		86
5)						( ce{_{82}^{206}pb^{4+}} )
6)		34		45	-2
7)	113	48	49
8) 21	42				0
9)						( ce{_{15}^{31}p^{3-}} )
10)		83	80	126
11)						( ce{_{47}^{108}ag} )
12)	116	49			+3
13)	128	53			-1
14) 76	188		72

Explanation:

Problem 1:

Step1: Calculate Mass # (Atomic # + neutrons)

Atomic # = 17, neutrons = 19, so Mass # = \( 17 + 19 = 36 \)

Step2: # \( p^+ \) = Atomic # = 17

Step3: # \( e^- \) = # \( p^+ \) (charge 0) = 17

Step4: Symbol: Atomic # 17 is Cl, so \( ^{36}_{17}\text{Cl} \)

Step1: Atomic # = # \( p^+ \) = # \( e^- \) (charge? Let's check: Mass # = 180 = Atomic # + 109, so Atomic # = \( 180 - 109 = 71 \))

Step2: # \( p^+ \) = 71, # \( e^- \) = 71, so charge = \( 71 - 71 = 0 \)

Step3: Symbol: Atomic # 71 is Lu, so \( ^{180}_{71}\text{Lu} \)

Step1: Atomic # = # \( p^+ \) = 40

Step2: Mass # = Atomic # + neutrons = \( 40 + 46 = 86 \)

Step3: Charge = # \( p^+ - \) # \( e^- = 40 - 38 = +2 \)

Step4: Symbol: Atomic # 40 is Zr, so \( ^{86}_{40}\text{Zr}^{2+} \)

Answer:

Mass #: 36; # \( p^+ \): 17; # \( e^- \): 17; Symbol: \( ^{36}_{17}\text{Cl} \)

Problem 2: