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vivado [14]
3 years ago
10

Complete the following nuclear equations by entering the missing isotope:

Chemistry
1 answer:
Margarita [4]3 years ago
3 0

The nuclear equation :

₈₂²¹⁴Pb ⇒ ₈₃²¹⁴Bi + ₋₁⁰e

<h3>Further explanation </h3>

Given

₈₂²¹⁴Pb

beta β ₋₁e⁰ particles

Required

Nuclear equation

Solution

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

The principle used is the sum of the atomic number and mass number before and after the decay reaction is the same

The reaction

₈₂²¹⁴Pb ⇒ X + ₋₁⁰e

The element X has

-the atomic number = 82 + 1 = 83

-the mass number = 214

In the periodic system, the element with atomic number 83=Bismuth

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Answer:

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If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>

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Dissolve

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When an acid reacts with a base, salt and water is formed! Why??​
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How many grams of NaOH needed to completely neutralize 3L of 1.75M HCL
Sedaia [141]

Answer:

209.98 g of NaOH

Explanation:

We are given;

  • Volume of HCl as 3 L
  • Molarity of HCl as 1.75 M

We are required to calculate the mass of NaOH required to completely neutralize the acid given.

First, we write a  balanced equation for the reaction between NaOH and HCl

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NaOH + HCl → NaCl + H₂O

Second, we determine the number of moles of HCl

Number of moles = Molarity × Volume

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Third, we use the mole ratio to determine the moles of NaOH

From the reaction,

1 mole of NaOH reacts with 1 mole of HCl

Therefore;

Moles of NaOH = Moles of HCl

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Fourth, we determine the mass of NaOH

Molar mass of NaOH = 39.997 g/mol

Mass of NaOH = 5.25 moles × 39.997 g/mol

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Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl

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3 years ago
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