3. each box has 2 isotopes of the same element. fill in the isotope names and any missing information. use the isotope name convention, e.g. carbon - 14. assume neutral atom in ground state.

| # of n | 17 | 15 | # of e- | -10 | -12 |
| # of p+ | 25 | | # of e- | 6 | |
| # of e- | | | # of p+ | | 6 |
| | | | # of n | | |

| | iron- | iron- | | -54 | -56 |
| # of p+ | | | # of n | | |
| # of e- | | | # of e- | | |
| # of n | 27 | 30 | # of p+ | 24 | |

4. fill in any missing information. assume neutral atom in ground state.

| isotope name | mass number | atomic number | # of p+ | # of e- | # of n° | full nuclear notation |
| chromium - 50 | | | | | | 50 24 c |

Question

3. each box has 2 isotopes of the same element. fill in the isotope names and any missing information. use the isotope name convention, e.g. carbon - 14. assume neutral atom in ground state.

| # of n | 17 | 15 | # of e- | -10 | -12 |
| # of p+ | 25 |  | # of e- | 6 |  |
| # of e- |  |  | # of p+ |  | 6 |
|  |  |  | # of n |  |  |

|  | iron- | iron- |  | -54 | -56 |
| # of p+ |  |  | # of n |  |  |
| # of e- |  |  | # of e- |  |  |
| # of n | 27 | 30 | # of p+ | 24 |  |

4. fill in any missing information. assume neutral atom in ground state.

| isotope name | mass number | atomic number | # of p+ | # of e- | # of n° | full nuclear notation |
| chromium - 50 |  |  |  |  |  | 50 24 c |

Accepted Answer

# Explanation:
## Step1: Recall atomic - related concepts
For a neutral atom in the ground state, the number of electrons ($e^-$) is equal to the number of protons ($P^+$), and the mass number ($A$) is the sum of the number of protons and neutrons ($N$). The atomic number ($Z$) is equal to the number of protons.

## Step2: First table (left - top)
Given $N = 17$ and $P^+=25$. For a neutral atom, $e^-=P^+=25$. The mass number $A = P^++N=25 + 17=42$. The element with $P^+=25$ is manganese. The isotope name is Manganese - 42.
For the second isotope in the first box, if $N = 15$ and $P^+=25$, then $e^-=25$ and the mass number $A=25 + 15 = 40$, isotope name is Manganese - 40.

## Step3: Second table (right - top)
Given $e^- = 6$ and $P^+=6$, then $N$ can be calculated from the isotope name convention. If the isotope name is not given directly, assume we are dealing with carbon (since $P^+=6$). For a neutral carbon atom with $e^- = 6$ and $P^+=6$. If the first isotope has $e^-=6$ and $P^+=6$, and assume mass number $A_1$ and $A_2$ from the given data. If $e^- = 6$ and $P^+=6$, for a neutral atom it is carbon. If $N_1$ and $N_2$ are the number of neutrons. We know $A=P^++N$.
For the first carbon isotope, if $e^-=6$ and $P^+=6$, assume mass number $A_1$ is not given directly. For the second carbon isotope, if $e^- = 6$ and $P^+=6$, and $N_2$ is related to the mass - number. Since $P^+=6$, the element is carbon. If $e^-=6$ and $P^+=6$, for a neutral atom, and if we assume the mass numbers from the context of isotopes, if $N_1$ and $N_2$ are the number of neutrons, $A_1=P^++N_1$ and $A_2=P^++N_2$.

## Step4: Third table (left - bottom)
For iron, the atomic number of iron $Z = 26$. For Iron - $x$ isotopes, $P^+=26$ for both. For the first isotope with $N = 27$, $e^-=26$ (neutral atom) and mass number $A_1=26 + 27=53$, isotope name is Iron - 53. For the second isotope with $N = 30$, $e^-=26$ and mass number $A_2=26+30 = 56$, isotope name is Iron - 56.

## Step5: Fourth table (right - bottom)
Given $P^+=24$, the element is chromium. For a neutral atom, $e^-=24$. For the first isotope with mass number $A_1 = 54$, $N_1=A_1 - P^+=54 - 24=30$. For the second isotope with mass number $A_2 = 56$, $N_2=A_2 - P^+=56 - 24=32$.

## Step6: Fifth table (question 4)
For Chromium - 50:
The atomic number of chromium $Z = 24$, so $P^+=24$. For a neutral atom, $e^-=24$. The mass number $A = 50$, and $N=A - P^+=50 - 24=26$. The full nuclear notation is $_ {24}^{50}Cr$.
For the second isotope, given $P^+=14$, the element is silicon. The atomic number $Z = 14$, for a neutral atom $e^-=14$. Given $N = 17$, the mass number $A=P^++N=14 + 17=31$. The isotope name is Silicon - 31, and the full nuclear notation is $_ {14}^{31}Si$.

# Answer:
| Isotope Name | Mass Number | Atomic Number | # of $P^+$ | # of $e^-$ | # of $N$ | Full Nuclear Notation |
|---|---|---|---|---|---|---|
| Manganese - 42 | 42 | 25 | 25 | 25 | 17 | $_ {25}^{42}Mn$ |
| Manganese - 40 | 40 | 25 | 25 | 25 | 15 | $_ {25}^{40}Mn$ |
| Carbon - (unknown mass - 1) | - | 6 | 6 | 6 | (unknown) | $_ {6}^{A_1}C$ |
| Carbon - (unknown mass - 2) | - | 6 | 6 | 6 | (unknown) | $_ {6}^{A_2}C$ |
| Iron - 53 | 53 | 26 | 26 | 26 | 27 | $_ {26}^{53}Fe$ |
| Iron - 56 | 56 | 26 | 26 | 26 | 30 | $_ {26}^{56}Fe$ |
| Chromium - 54 | 54 | 24 | 24 | 24 | 30 | $_ {24}^{54}Cr$ |
| Chromium - 56 | 56 | 24 | 24 | 24 | 32 | $_ {24}^{56}Cr$ |
| Chromium - 50 | 50 | 24 | 24 | 24 | 26 | $_ {24}^{50}Cr$ |
| Silicon - 31 | 31 | 14 | 14 | 14 | 17 | $_ {14}^{31}Si$ |

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QUESTION IMAGE

Question

Explanation:

Step1: Recall atomic - related concepts

Step2: First table (left - top)

Step3: Second table (right - top)

Step4: Third table (left - bottom)

Step5: Fourth table (right - bottom)

Step6: Fifth table (question 4)

Answer:

# of n	17	15	# of e-	-10	-12
# of p+	25		# of e-	6
# of e-			# of p+		6
			# of n

	iron-	iron-		-54	-56
# of p+			# of n
# of e-			# of e-
# of n	27	30	# of p+	24

isotope name	mass number	atomic number	# of p+	# of e-	# of n°	full nuclear notation
chromium - 50						50 24 c

Isotope Name	Mass Number	Atomic Number	# of \(P^+\)	# of \(e^-\)	# of \(N\)	Full Nuclear Notation
Manganese - 40	40	25	25	25	15	\(_ {25}^{40}Mn\)
Carbon - (unknown mass - 1)	-	6	6	6	(unknown)	\(_ {6}^{A_1}C\)
Carbon - (unknown mass - 2)	-	6	6	6	(unknown)	\(_ {6}^{A_2}C\)
Iron - 53	53	26	26	26	27	\(_ {26}^{53}Fe\)
Iron - 56	56	26	26	26	30	\(_ {26}^{56}Fe\)
Chromium - 54	54	24	24	24	30	\(_ {24}^{54}Cr\)
Chromium - 56	56	24	24	24	32	\(_ {24}^{56}Cr\)
Chromium - 50	50	24	24	24	26	\(_ {24}^{50}Cr\)
Silicon - 31	31	14	14	14	17	\(_ {14}^{31}Si\)