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lukranit [14]
2 years ago
9

Consider this reaction: NH4 HPO42− → NH3 H2PO4− Which is the Bronsted-Lowry acid ? NH4 HPO42- HPO42– H2PO4–.

Chemistry
1 answer:
kvv77 [185]2 years ago
8 0

Bronsted - Lowry acid in the given reaction is NH₄, as it gives H⁺ ion.

<h3>What is Bronsted - Lowry acid?</h3>

According to the theory of Bronsted - Lowry, acids are those substances which gives H⁺ ion or proton in the aqueous medium.

Given chemical reaction is :

NH₄ + HPO₄²⁻ → NH₃ + H₂PO₄⁻

In the above reaction NH₄ is the Bronsted - Lowry acid as it gives H⁺ ion in the reaction and changes to NH₃ which is the conjugate base of NH₄. Whereas HPO₄²⁻ is the  Bronsted - Lowry base as it accepts the H⁺ ion to form H₂PO₄⁻ which is the conjugate acid of it.

Hence, option (1) is correct, i.e. NH₄ is the Bronsted - Lowry acid.

To know more about Bronsted - Lowry acid, visit the below link:

brainly.com/question/1435076

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A gas sample occupies a volume of 1 L at 200K. The temperature at which the gas would double its volume is
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Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β
Fantom [35]
<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

  • Neptunium-237
  • Thallium-205
  • Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

  • We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
  • When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

 x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0
2 years ago
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