Question

for questions 8 - 12, balance the nuclear equation, then indicate if the equation represents a. 8. $_{91}^{231}pa
ightarrow_{77}^{192}ir+\frac{39}{14}s$ 9. $_{2}^{4}he + _{1}^{2}h
ightarrow\frac{6}{3}?$ 10. $_{4}^{9}be+_{6}^{12}c
ightarrow\frac{21}{10}ne$ 11. $_{92}^{238}u+_{0}^{1}n
ightarrow_{52}^{128}te+\frac{111}{40}zr$ 12. $_{105}^{262}db
ightarrow_{42}^{96}mo+_{14}^{28}si+\frac{139}{99}n$ 13. why cant we use fusion reactions in a nuclear power plant? 14. why is it so hard to dispose of nuclear waste? 15. what are the benefits to having a nuclear power plant nearby? 16. what are the hazards to having a nuclear power plant nearby?

Explanation:

Step1: Balance mass - number

For nuclear equations, the sum of mass - numbers on the left - hand side must equal the sum of mass - numbers on the right - hand side, and the sum of atomic numbers on the left - hand side must equal the sum of atomic numbers on the right - hand side.

Step2: Determine missing particles for each equation

Equation 8:

The mass - number of $^{231}_{91}Pa$ is 231 and of $^{192}_{77}Ir$ is 192. The difference in mass - number is $231 - 192=39$. The atomic number of $Pa$ is 91 and of $Ir$ is 77. The difference in atomic number is $91 - 77 = 14$. So the missing particle is $^{39}_{14}Si$.

Equation 9:

The mass - number of $^{4}_{2}He$ is 4 and of $^{2}_{1}H$ is 2, so the mass - number of the product is $4 + 2=6$. The atomic number of $He$ is 2 and of $H$ is 1, so the atomic number of the product is $2+1 = 3$. The product is $^{6}_{3}Li$.

Equation 10:

The mass - number of $^{9}_{4}Be$ is 9 and of $^{12}_{6}C$ is 12, so the mass - number of the product is $9 + 12=21$. The atomic number of $Be$ is 4 and of $C$ is 6, so the atomic number of the product is $4 + 6=10$. The product is $^{21}_{10}Ne$.

Equation 11:

The mass - number of $^{238}_{92}U$ is 238 and of $^{1}_{0}n$ is 1, so the total mass - number on the left is $238+1 = 239$. The mass - number of $^{128}_{52}Te$ is 128. The mass - number of the missing particle is $239 - 128 = 111$. The atomic number of $U$ is 92, of $n$ is 0 and of $Te$ is 52. The atomic number of the missing particle is $92-52 = 40$. The missing particle is $^{111}_{40}Zr$.

Equation 12:

The mass - number of $^{262}_{105}Db$ is 262. The mass - number of $^{96}_{42}Mo$ is 96 and of $^{28}_{14}Si$ is 28. Let the number of neutrons ($^{1}_{0}n$) be $x$. Then $262=96 + 28+x$, so $x=\frac{262-(96 + 28)}{1}=138$.

for questions 13 - 16:

Question 13:

Fusion reactions require extremely high temperatures (millions of degrees Celsius) and pressures to overcome the electrostatic repulsion between positively - charged atomic nuclei. Currently, we lack the technology to contain and control these reactions in a cost - effective and stable manner for power generation.

Question 14:

Nuclear waste is hard to dispose of because it is highly radioactive and has a long half - life. It can remain hazardous for thousands to millions of years. It requires special storage facilities that can prevent radiation leakage and protect the environment and human health.

Question 15:

Benefits of having a nuclear power plant nearby include providing a large amount of baseload electricity (constant power supply), reducing dependence on fossil fuels (lowering greenhouse gas emissions compared to coal or gas plants), and potentially creating jobs in the local area.

Question 16:

Hazards of having a nuclear power plant nearby include the risk of nuclear accidents (such as meltdowns) which can release large amounts of radioactive material, causing health problems (cancer, genetic mutations) and environmental contamination. There is also the issue of nuclear waste storage in the vicinity.

Answer:

1. $^{231}_{91}Pa

ightarrow^{192}_{77}Ir + ^{39}_{14}Si$

1. $^{4}_{2}He+^{2}_{1}H

ightarrow^{6}_{3}Li$

1. $^{9}_{4}Be + ^{12}_{6}C

ightarrow^{21}_{10}Ne$

1. $^{238}_{92}U+^{1}_{0}n

ightarrow^{128}_{52}Te+^{111}_{40}Zr$

1. $^{262}_{105}Db

ightarrow^{96}_{42}Mo+^{28}_{14}Si + 138^{1}_{0}n$