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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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If your options are among the following:
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The temperature of a 95.4 g piece of Cu increases from 25.0 °C to 48.0 °C when the Cu absorbs 849 J of heat. What is the specifc
melisa1 [442]
<h3>Answer:</h3>

0.387 J/g°C

<h3>Explanation:</h3>
  • To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
  • Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
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in our question we are given;

Mass of copper, m as 95.4 g

Initial temperature = 25 °C

Final temperature = 48 °C

Thus, change in temperature, ΔT = 23°C

Quantity of heat absorbed, Q as 849 J

We are required to calculate the specific heat capacity of copper

Rearranging the formula we get

c = Q ÷ mΔT

Therefore,

Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)

                                        = 0.3869 J/g°C

                                        = 0.387 J/g°C

Therefore, the specific heat capacity of copper is 0.387 J/g°C

3 0
2 years ago
How would you separate a mixture of water, sand, salt and iron filings into its four component parts
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3 0
3 years ago
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Answer:

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8 0
3 years ago
Use the following equation to answer question 1-4. Make sure you balance first.
worty [1.4K]
<h3>Answer:</h3>

5.2 mol H₂O

<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
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<u>Chemistry</u>

<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O

[Given] 10.4 mol HCl

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

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4 0
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