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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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That would be hydrogen and helium! :)

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3 years ago
What is the Molar Mass of methylammonium bromide: Use the Periodic Table and round to the nearest hundredths.
egoroff_w [7]

The molar mass of methylammonium bromide is 111u.

<h3>What is molar mass?</h3>

The molar mass is defined as the mass per unit amount of substance of a given chemical entity.

Multiply the atomic weight (from the periodic table) of each element by the number of atoms of that element present in the compound.

Add it all together and put units of grams/mole after the number.

Atomic weight of H is 1u

Atomic weight of N is  14u

Atomic weight of C is  12u

Atomic weight of Br is  79u

Calculating molar mass of  2H_3NCH_3Br =2(1 x3+ 14+12+ 1 x 3 +79) = 111u

Hence, the molar mass of methylammonium bromide is 111u.

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8 0
2 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
3 years ago
In a reaction that occurs in solution, the volume will not change. What happens to the concentration of the reactants? What happ
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Answer:

the reaction will increase

Explanation:

6 0
3 years ago
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