<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.
Answer:
D
Explanation:
The answer is D...and if you need explanation you can comment
Answer:
3.0 × 10²⁰ molecules
Explanation:
Given data:
Mass of ethanol = 2.3 × 10⁻²°³ g
Number of molecules = ?
Solution:
Number of moles of ethanol:
Number of moles = mass/ molar mass
Number of moles = 2.3 × 10⁻²°³ g / 46.07 g/mol
Number of moles = 0.05 × 10⁻²°³ mol
Number of molecules:
One mole = 6.022 × 10²³ molecules
0.05 × 10⁻²°³ mol × 6.022 × 10²³ molecules / 1 mol
0.30 × 10²⁰°⁷ molecules
3.0 × 10¹⁹°⁷ molecules which is almost equal to 3.0 × 10²⁰ molecules.
Answer:
V₂ = 155.2 mL
Explanation:
Given data:
Initial volume =175.6 mL
Initial pressure =0.9648 atm
Initial temperature = 298.1 K
Final temperature = 273 K
Final volume =?
Final pressure = 1 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂ /T₁P₂
V₂ = 0.9648 atm × 175.6 mL × 273 k / 298.1 K× 1 atm
V₂ = 46251.4 atm. k. mL/ 298.1 K× 1 atm
V₂ = 155.2 mL
Answer:
Option D is correct = 58 g
Explanation:
Data Given:
mass of LiOH = 120 g
Mass of Li3N= ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)
1 mol 3 mol
Convert moles to mass
Molar mass of LiOH = 24 g/mol
Molar mass of Li₃N = 35 g/mol
So,
Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)
1 mol (35 g/mol) 3 mol (24 g/mol)
35 g 72 g
So if we look at the reaction 35 g of Li₃N react with water and produces 72 g of LiOH , then how many g of Li₃N will be react to Produce by 120 g of LiOH
For this apply unity formula
35 g of Li₃N ≅ 72 g of LiOH
X of Li₃N ≅ 120 g of LiOH
By Doing cross multiplication
Mass of Li₃N = 35 g x 120 g / 72 g
mass of Li₃N = 58 g
120 g of LiOH will produce from 58 g of Li₃N
So,
Option D is correct = 58 g
