Balancing Decay Reactions.

Fill in the blank with the correct piece to balance the decay equation.

1. (184/74) W (4/2) He + _____
2. (210/82) Pb______+ (206/80) Hg
3. (35/16) S  (0/-1) e + _______
4. ________  (0/-1) e + (60/28) Ni

My work:
5. (184/74) W (4/2) He + (180/72) Hf
6. (210/82) Pb(4/2) He + (206/80) Hg
7. (35/16) S  (0/-1) e + (35/17) Cl
8. (60/27) Co  (0/-1) e + (60/28) Ni

The boxes are post to arrows to the right.

Are you'r answers right??

Yees

To balance decay reactions, we need to ensure that the number of protons and the number of nucleons (protons + neutrons) are conserved on both sides of the equation. Let's go through each of the given reactions and see if your answers are correct.

1. (184/74) W -> (4/2) He + _____

In this decay reaction, an alpha particle (helium nucleus) is emitted from tungsten (W). The mass number on the left side is 184, and the atomic number is 74.

The atomic number and mass number on the right side should be equal to those on the left side for the reaction to be balanced.

Your answer: (184/74) W -> (4/2) He + (180/72) Hf

To check if it's correct, we need to verify the atomic numbers and the mass numbers.

On the left side:
Atomic number = 74
Mass number = 184

On the right side:
Atomic number = 2 (From He)
Mass number = 4 (From He)

The atomic number is conserved since 74 = 2. However, the mass number is not conserved since 184 ≠ 4. Therefore, your answer is incorrect.

Let's continue to the next reaction,

2. (210/82) Pb -> ______ + (206/80) Hg

In this decay reaction, a beta particle (electron) is emitted from lead (Pb). The mass number on the left side is 210, and the atomic number is 82.

The same rules apply. The atomic number and mass number on the right side should be equal to those on the left side for the reaction to be balanced.

Your answer: (210/82) Pb -> (4/2) He + (206/80) Hg

On the left side:
Atomic number = 82
Mass number = 210

On the right side:
Atomic number = 2 (From He)
Mass number = 4 (From He) + 206 (From Hg)

Again, the atomic number is conserved since 82 = 2. However, the mass number is not conserved since 210 ≠ 4 + 206. Therefore, your answer is incorrect.

Now let's move to the next reaction,

3. (35/16) S -> (0/-1) e + _______

In this decay reaction, an electron (beta particle) is emitted from sulfur (S). The mass number on the left side is 35, and the atomic number is 16.

Again, the atomic number and mass number on the right side should be equal to those on the left side for the reaction to be balanced.

Your answer: (35/16) S -> (0/-1) e + (35/17) Cl

On the left side:
Atomic number = 16
Mass number = 35

On the right side:
Atomic number = -1 (From e)
Mass number = 0 (From e) + 35 (From Cl)

The atomic number is conserved since 16 = -1. However, the mass number is not conserved since 35 ≠ 0 + 35. Therefore, your answer is incorrect.

Lastly,

4. ________ -> (0/-1) e + (60/28) Ni

In this decay reaction, an electron (beta particle) is emitted and produces nickel (Ni). We need to determine the element that decays and produces the electron and nickel.

Your answer: (60/27) Co -> (0/-1) e + (60/28) Ni

To check if it's correct, we need to verify the atomic numbers and the mass numbers.

On the left side:
Atomic number = 27
Mass number = 60

On the right side:
Atomic number = -1 (From e)
Mass number = 0 (From e) + 28 (From Ni)

The atomic number is conserved since 27 = -1. Additionally, the mass number is conserved since 60 = 0 + 28. Therefore, your answer is correct.

To summarize, the correct balanced decay equations are:
1. (184/74) W -> (4/2) He + (180/72) Hf
2. (210/82) Pb -> (0/-1) e + (206/80) Hg
3. (35/16) S -> (0/-1) e + (35/17) Cl
4. (60/27) Co -> (0/-1) e + (60/28) Ni

You correctly balanced equation 4, but equations 1, 2, and 3 are still incorrect.