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stiks02 [169]
3 years ago
5

When Iodine-131 emits a β particle (beta particle), what nuclide is produced? *

Chemistry
2 answers:
k0ka [10]3 years ago
5 0

When Iodine-131 emits a β particle will produce Xe-131

<h3>Further explanation </h3>

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,  

  • alpha α particles ₂He⁴
  • beta β ₋₁e⁰ particles
  • gamma particles ₀γ⁰
  • positron particles ₁e⁰
  • neutron ₀n¹

So for reaction Iodine-131 :

\tt _{53}^{131}I\rightarrow _{-1}^0\beta +_{54}^{131}Xe

erastovalidia [21]3 years ago
5 0

Question, Answer, and Explanation are within the picture of files:

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Read 2 more answers
2C3H7OH + 9O2 --&gt; 6CO2 + 8H2O
Ne4ueva [31]
<h3>Answer:</h3>

733 g CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given]   5.55 mol C₃H₇OH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})
  2. Multiply/Divide:                                                                                               \displaystyle 732.767 \ g \ CO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

732.767 g CO₂ ≈ 733 g CO₂

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