Question

1. fill in the blanks with either positive or negative. a. fluorine atoms gain an electron to become fluoride ions. these ions have a negative charge. b. calcium atoms lose electrons to become ions. these ions have a positive charge. 7. fill in the blanks. mass # (p⁺ + n°) ¹⁶₈o ³¹₁₅p ⁶⁵₃₀zn atomic # (p⁺) p⁺ p⁺ p⁺ n n n e⁻ e⁻ e⁻ e⁻ same as p⁺ if atom is neutral 8. fill in the blanks in the table below. isotope name isotope symbol protons neutrons electrons atomic number mass number argon - 40 ⁴⁰₁₈ar 18 22 18 18 40 argon - 38 ³⁸ar 18 20 18 18 38 ⁵⁷₂₆fe 16 34 28 30 34 28 37 48 37 manganese - 55

Explanation:

Step1: Recall atomic number concept

Atomic number is equal to number of protons. For $^{16}_8O$, atomic number is 8, so number of protons $p^+=8$.

Step2: Calculate number of neutrons

Mass number = protons + neutrons. For $^{16}_8O$, mass number is 16 and protons are 8, so neutrons $n = 16 - 8=8$.

Step3: Determine number of electrons

In a neutral atom, number of electrons $e^-$ is equal to number of protons. So for $^{16}_8O$, $e^- = 8$.

Step4: For $^{31}_{15}P$

Atomic number gives number of protons, so $p^+=15$. Mass number is 31, so neutrons $n=31 - 15 = 16$. In neutral atom, $e^-=15$.

Step5: For $^{65}_{30}Zn$

Atomic number gives number of protons, so $p^+=30$. Mass number is 65, so neutrons $n=65 - 30=35$. In neutral atom, $e^- = 30$.

Step6: For $^{57}_{26}Fe$

Protons $p^+=26$, neutrons $n=57 - 26 = 31$, electrons $e^-=26$ (neutral atom).

Step7: For isotope with atomic number 16 and mass number 34

Isotope symbol is $^{34}_{16}S$, protons $p^+=16$, neutrons $n=34 - 16 = 18$, electrons $e^-=16$ (neutral atom).

Step8: For isotope with 28 protons and 30 neutrons

Atomic number is 28, mass number is $28+30 = 58$, isotope symbol is $^{58}_{28}Ni$, electrons $e^-=28$ (neutral atom).

Step9: For isotope with 34 electrons and 28 protons

This is an ion. Protons $p^+=28$, atomic number is 28, mass number is not given but assuming neutral - like calculation for now, neutrons $n$ (if neutral) would be based on some unknown mass number. Isotope symbol could be $^{x}_{28}Ni$ (where $x$ is mass number). Since it has 34 electrons and 28 protons, it has a charge of $28 - 34=- 6$. But if we assume it's a neutral - atom reference for filling table, mass number is not given. Let's assume it's a mis - write and we consider neutral atom, then mass number is not calculable with given info.

Step10: For isotope with 37 protons and 48 neutrons

Atomic number is 37, mass number is $37 + 48=85$, isotope symbol is $^{85}_{37}Rb$, electrons $e^-=37$ (neutral atom).

Step11: For Manganese - 55

Isotope symbol is $^{55}_{25}Mn$, protons $p^+=25$, neutrons $n=55 - 25 = 30$, electrons $e^-=25$ (neutral atom).

Isotope Name	Isotope Symbol	Protons	Neutrons	Electrons	Atomic Number	Mass Number
Phosphorus - 31	$^{31}_{15}P$	15	16	15	15	31
Zinc - 65	$^{65}_{30}Zn$	30	35	30	30	65
Iron - 57	$^{57}_{26}Fe$	26	31	26	26	57
Sulfur - 34	$^{34}_{16}S$	16	18	16	16	34
Nickel - 58	$^{58}_{28}Ni$	28	30	28	28	58
(Un - named with 28 protons and 34 electrons)	(Assumed $^{x}_{28}Ni$)	28	(Unknown)	34	28	(Unknown)
Rubidium - 85	$^{85}_{37}Rb$	37	48	37	37	85
Manganese - 55	$^{55}_{25}Mn$	25	30	25	25	55

Answer:

Isotope Name	Isotope Symbol	Protons	Neutrons	Electrons	Atomic Number	Mass Number
Phosphorus - 31	$^{31}_{15}P$	15	16	15	15	31
Zinc - 65	$^{65}_{30}Zn$	30	35	30	30	65
Iron - 57	$^{57}_{26}Fe$	26	31	26	26	57
Sulfur - 34	$^{34}_{16}S$	16	18	16	16	34
Nickel - 58	$^{58}_{28}Ni$	28	30	28	28	58
(Un - named with 28 protons and 34 electrons)	(Assumed $^{x}_{28}Ni$)	28	(Unknown)	34	28	(Unknown)
Rubidium - 85	$^{85}_{37}Rb$	37	48	37	37	85
Manganese - 55	$^{55}_{25}Mn$	25	30	25	25	55