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ss7ja [257]
3 years ago
14

What is a limited resource in the use of nuclear fission for energy?

Chemistry
1 answer:
lys-0071 [83]3 years ago
5 0
Uranium is the answer
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Sloan [31]
Membranes are barriers and “gatekeepers”, they only let certain molecules pass through them.

they also transport nutrients to the cell
5 0
3 years ago
Read 2 more answers
A student jumps of a sled toward the west after it stops at the bottom of an icy hill. Based on the law of action-reaction, in w
Delicious77 [7]

Answer:

East

Explanation:

Given Newton's third law of motion; "Action and reaction are equal and opposite", when a student jumps off a sled toward the west after it stops at the bottom of an icy hill, the sled will move in the East direction.

This is because, the force exerted on the sled is a reaction force and is opposite in direction to the force that thrusts the boy westward though equal in magnitude with the former.

3 0
3 years ago
All of the following are exothermic reactions except,
monitta
Exothermic reactions release heat the surrounding. So Freezing? 
8 0
3 years ago
. how many lone pairs of electrons are present in the lewis structure of calcium sulfide?
miv72 [106K]
Answer:
            <u>Four</u>

Explanation:
                     Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,

                                   Ca    →    Ca²⁺  +  2 e⁻

These two electrons are accepte by Sulfur as,

                                     S  +  2 e⁻    →    S²⁻

So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,

7 0
3 years ago
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
3 years ago
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