Identify the product of radioactive decay and classify the given nuclear reactions accordingly.
2 answers:
The product of radioactive decay :
- ₂He⁴
- ₋₁e⁰
- γ
- ₁e⁰
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually radioactive elements have an unstable atomic nucleus.
The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles
General formulas used in decay:
Information:
T = duration of decay
t1 / 2 = elemental half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
At the core, the reaction applies the law of eternity
• 1. energy
the energy before and after the reaction is the same
• 2. atomic number
the number of atomic numbers before and after the same
• 3. mass number
the number of mass numbers before and after the same
Particles that play a role in core reactions include
• alpha α particles ₂He⁴
atomic number decreases by 2, mass number decreases by 4
• beta β ₋₁e⁰ particles
atomic number remains the same , mass number decreases by -1
• gamma particles γ
Core reactions that occur usually release high energy
• positron particles ₁e⁰
atomic number remains the same , mass number decreases by +1
In general, the core reaction equation can be written:
<h3>X + a ---> Y + b + Q </h3>
X = target core
a = particle fired
Y = new core
b = the particle produced
Q = heat energy
or can be written simply
X (a, b) Y
<h3>Learn more </h3>the identity of the missing nucleus
the process of radioactive decay
nuclear fusion and nuclear fission reactions
the product of radioactive decay
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f(x)=x+2−x^2
df/dx=1–2x
and that is zero (indicating a maximum) at x=1/2
So the maximum distance is f(1/2)=2.5–0.25=2.25
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If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>