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name jayden colemun period 2 date 4/11/25 isotopes, ions, and atoms wor…

Question

name jayden colemun period 2 date 4/11/25 isotopes, ions, and atoms worksheet atomic # = # of protons. mass # = atomic # + neutrons. protons = electrons when charge is zero.

atomic #mass ## p⁺# e⁻# n⁰chargesymbol
2)18071109
3)403846
4) 9223886
5)$_{82}^{206}$pb⁴⁺
6)3445-2
7)1134849
8) 21420
9)$_{15}^{31}$p³⁻
10)8380126
11)$_{47}^{108}$ag
12)11649+3
13)12853-1
14) 7618872

Explanation:

Response

Let's solve each row step by step using the given formulas: \( \text{Atomic \#} = \# \text{ of protons} \), \( \text{Mass \#} = \text{Atomic \#} + \text{neutrons} \), and \( \text{Protons} = \text{electrons} \) when charge is zero (otherwise, \( \text{Charge} = \text{Protons} - \text{Electrons} \)).

Row 1:
  • Atomic \# = 17, so \( \# p^+ = 17 \).
  • Charge = 0, so \( \# e^- = \# p^+ = 17 \).
  • Mass \# = Atomic \# + \( \# n^0 = 17 + 19 = 36 \).
  • Symbol: Atomic \# 17 is Cl, so \( ^{36}_{17}\text{Cl} \).
Row 2:
  • \( \# p^+ = \# e^- = 71 \) (charge unknown, but \( \# p^+ = \) Atomic \# = 71).
  • Mass \# = 180, so \( \# n^0 = 180 - 71 = 109 \) (matches).
  • Charge = \( \# p^+ - \# e^- = 71 - 71 = 0 \).
  • Symbol: Atomic \# 71 is Lu, so \( ^{180}_{71}\text{Lu} \).
Row 3:
  • \( \# p^+ = 40 \), so Atomic \# = 40.
  • \( \# e^- = 38 \), so Charge = \( 40 - 38 = +2 \).
  • Mass \# = 40 + 46 = 86.
  • Symbol: Atomic \# 40 is Zr, so \( ^{86}_{40}\text{Zr}^{2+} \).
Row 4:
  • Atomic \# = 92, so \( \# p^+ = 92 \).
  • \( \# e^- = 86 \), so Charge = \( 92 - 86 = +6 \).
  • \( \# n^0 = 238 - 92 = 146 \).
  • Symbol: Atomic \# 92 is U, so \( ^{238}_{92}\text{U}^{6+} \).
Row 5:
  • Symbol \( \ce{^{206}_{82}Pb^{4+}} \): Atomic \# = 82, so \( \# p^+ = 82 \).
  • Charge = +4, so \( \# e^- = 82 - 4 = 78 \).
  • Mass \# = 206, so \( \# n^0 = 206 - 82 = 124 \).
Row 6:
  • \( \# p^+ = 34 \), so Atomic \# = 34.
  • Charge = -2, so \( \# e^- = 34 + 2 = 36 \).
  • Mass \# = 34 + 45 = 79.
  • Symbol: Atomic \# 34 is Se, so \( ^{79}_{34}\text{Se}^{2-} \).
Row 7:
  • \( \# p^+ = 48 \), so Atomic \# = 48.
  • \( \# e^- = 49 \), so Charge = \( 48 - 49 = -1 \).
  • \( \# n^0 = 113 - 48 = 65 \).
  • Symbol: Atomic \# 48 is Cd, so \( ^{113}_{48}\text{Cd}^{-} \).
Row 8:
  • Atomic \# = 21, so \( \# p^+ = 21 \).
  • Charge = 0, so \( \# e^- = 21 \).
  • \( \# n^0 = 42 - 21 = 21 \).
  • Symbol: Atomic \# 21 is Sc, so \( ^{42}_{21}\text{Sc} \).
Row 9:
  • Symbol \( \ce{^{31}_{15}P^{3-}} \): Atomic \# = 15, so \( \# p^+ = 15 \).
  • Charge = -3, so \( \# e^- = 15 + 3 = 18 \).
  • Mass \# = 31, so \( \# n^0 = 31 - 15 = 16 \).
Row 10:
  • \( \# p^+ = 83 \), so Atomic \# = 83.
  • \( \# e^- = 80 \), so Charge = \( 83 - 80 = +3 \).
  • Mass \# = 83 + 126 = 209.
  • Symbol: Atomic \# 83 is Bi, so \( ^{209}_{83}\text{Bi}^{3+} \).
Row 11:
  • Symbol \( \ce{^{108}_{47}Ag} \): Atomic \# = 47, so \( \# p^+ = 47 \).
  • Charge = 0, so \( \# e^- = 47 \).
  • \( \# n^0 = 108 - 47 = 61 \).
  • Mass \# = 108.
Row 12:
  • \( \# p^+ = 49 \), so Atomic \# = 49.
  • Charge = +3, so \( \# e^- = 49 - 3 = 46 \).
  • \( \# n^0 = 116 - 49 = 67 \).
  • Symbol: Atomic \# 49 is In, so \( ^{116}_{49}\text{In}^{3+} \).
Row 13:
  • \( \# p^+ = 53 \), so Atomic \# = 53.
  • Charge = -1, so \( \# e^- = 53 + 1 = 54 \).
  • \( \# n^0 = 128 - 53 = 75 \).
  • Symbol: Atomic \# 53 is I, so \( ^{128}_{53}\text{I}^{-} \).
Row 14:
  • Atomic \# = 76, so \( \# p^+ = 76 \).
  • \( \# e^- = 72 \), so Charge = \( 76 - 72 = +4 \).
  • \( \# n^0 = 188 - 76 = 112 \).
  • Symbol: Atomic \# 76 is Os, so \( ^{188}_{76}\text{Os}^{4+} \).
Final Table (Filled):
RowAtomic #Mass ## p⁺# e⁻# n⁰ChargeSymbol
27118071711090\( ^{180}_{71}\text{Lu} \)
34086403846+2\( ^{86}_{40}\text{Zr}^{2+} \)
4922389286146+6\( ^{238}_{92}\text{U}^{6+} \…

Answer:

Let's solve each row step by step using the given formulas: \( \text{Atomic \#} = \# \text{ of protons} \), \( \text{Mass \#} = \text{Atomic \#} + \text{neutrons} \), and \( \text{Protons} = \text{electrons} \) when charge is zero (otherwise, \( \text{Charge} = \text{Protons} - \text{Electrons} \)).

Row 1:
  • Atomic \# = 17, so \( \# p^+ = 17 \).
  • Charge = 0, so \( \# e^- = \# p^+ = 17 \).
  • Mass \# = Atomic \# + \( \# n^0 = 17 + 19 = 36 \).
  • Symbol: Atomic \# 17 is Cl, so \( ^{36}_{17}\text{Cl} \).
Row 2:
  • \( \# p^+ = \# e^- = 71 \) (charge unknown, but \( \# p^+ = \) Atomic \# = 71).
  • Mass \# = 180, so \( \# n^0 = 180 - 71 = 109 \) (matches).
  • Charge = \( \# p^+ - \# e^- = 71 - 71 = 0 \).
  • Symbol: Atomic \# 71 is Lu, so \( ^{180}_{71}\text{Lu} \).
Row 3:
  • \( \# p^+ = 40 \), so Atomic \# = 40.
  • \( \# e^- = 38 \), so Charge = \( 40 - 38 = +2 \).
  • Mass \# = 40 + 46 = 86.
  • Symbol: Atomic \# 40 is Zr, so \( ^{86}_{40}\text{Zr}^{2+} \).
Row 4:
  • Atomic \# = 92, so \( \# p^+ = 92 \).
  • \( \# e^- = 86 \), so Charge = \( 92 - 86 = +6 \).
  • \( \# n^0 = 238 - 92 = 146 \).
  • Symbol: Atomic \# 92 is U, so \( ^{238}_{92}\text{U}^{6+} \).
Row 5:
  • Symbol \( \ce{^{206}_{82}Pb^{4+}} \): Atomic \# = 82, so \( \# p^+ = 82 \).
  • Charge = +4, so \( \# e^- = 82 - 4 = 78 \).
  • Mass \# = 206, so \( \# n^0 = 206 - 82 = 124 \).
Row 6:
  • \( \# p^+ = 34 \), so Atomic \# = 34.
  • Charge = -2, so \( \# e^- = 34 + 2 = 36 \).
  • Mass \# = 34 + 45 = 79.
  • Symbol: Atomic \# 34 is Se, so \( ^{79}_{34}\text{Se}^{2-} \).
Row 7:
  • \( \# p^+ = 48 \), so Atomic \# = 48.
  • \( \# e^- = 49 \), so Charge = \( 48 - 49 = -1 \).
  • \( \# n^0 = 113 - 48 = 65 \).
  • Symbol: Atomic \# 48 is Cd, so \( ^{113}_{48}\text{Cd}^{-} \).
Row 8:
  • Atomic \# = 21, so \( \# p^+ = 21 \).
  • Charge = 0, so \( \# e^- = 21 \).
  • \( \# n^0 = 42 - 21 = 21 \).
  • Symbol: Atomic \# 21 is Sc, so \( ^{42}_{21}\text{Sc} \).
Row 9:
  • Symbol \( \ce{^{31}_{15}P^{3-}} \): Atomic \# = 15, so \( \# p^+ = 15 \).
  • Charge = -3, so \( \# e^- = 15 + 3 = 18 \).
  • Mass \# = 31, so \( \# n^0 = 31 - 15 = 16 \).
Row 10:
  • \( \# p^+ = 83 \), so Atomic \# = 83.
  • \( \# e^- = 80 \), so Charge = \( 83 - 80 = +3 \).
  • Mass \# = 83 + 126 = 209.
  • Symbol: Atomic \# 83 is Bi, so \( ^{209}_{83}\text{Bi}^{3+} \).
Row 11:
  • Symbol \( \ce{^{108}_{47}Ag} \): Atomic \# = 47, so \( \# p^+ = 47 \).
  • Charge = 0, so \( \# e^- = 47 \).
  • \( \# n^0 = 108 - 47 = 61 \).
  • Mass \# = 108.
Row 12:
  • \( \# p^+ = 49 \), so Atomic \# = 49.
  • Charge = +3, so \( \# e^- = 49 - 3 = 46 \).
  • \( \# n^0 = 116 - 49 = 67 \).
  • Symbol: Atomic \# 49 is In, so \( ^{116}_{49}\text{In}^{3+} \).
Row 13:
  • \( \# p^+ = 53 \), so Atomic \# = 53.
  • Charge = -1, so \( \# e^- = 53 + 1 = 54 \).
  • \( \# n^0 = 128 - 53 = 75 \).
  • Symbol: Atomic \# 53 is I, so \( ^{128}_{53}\text{I}^{-} \).
Row 14:
  • Atomic \# = 76, so \( \# p^+ = 76 \).
  • \( \# e^- = 72 \), so Charge = \( 76 - 72 = +4 \).
  • \( \# n^0 = 188 - 76 = 112 \).
  • Symbol: Atomic \# 76 is Os, so \( ^{188}_{76}\text{Os}^{4+} \).
Final Table (Filled):
RowAtomic #Mass ## p⁺# e⁻# n⁰ChargeSymbol
27118071711090\( ^{180}_{71}\text{Lu} \)
34086403846+2\( ^{86}_{40}\text{Zr}^{2+} \)
4922389286146+6\( ^{238}_{92}\text{U}^{6+} \)
5822068278124+4\( \ce{^{206}_{82}Pb^{4+}} \)
63479343645-2\( ^{79}_{34}\text{Se}^{2-} \)
748113484965-1\( ^{113}_{48}\text{Cd}^{-} \)
821422121210\( ^{42}_{21}\text{Sc} \)
91531151816-3\( \ce{^{31}_{15}P^{3-}} \)
10832098380126+3\( ^{209}_{83}\text{Bi}^{3+} \)
11471084747610\( \ce{^{108}_{47}Ag} \)
1249116494667+3\( ^{116}_{49}\text{In}^{3+} \)
1353128535475-1\( ^{128}_{53}\text{I}^{-} \)
14761887672112+4\( ^{188}_{76}\text{Os}^{4+} \)

(Note: For each row, the filled values follow the formulas for atomic number, mass number, protons, electrons, neutrons, charge, and symbol.)