Answer:
³⁸₂₀Ca.
Explanation:
³⁸₁₉K –> __ + ⁰₋₁β
Let ʸₓA represent the unknown.
Thus the equation above can be written as:
³⁸₁₉K –> ʸₓA + ⁰₋₁β
Thus, we can obtain the value of y an x as follow:
38 = y + 0
y = 38
19 = x + (–1)
19 = x – 1
Collect like terms
19 + 1 = x
x = 20
Thus,
ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca
Therefore, the equation is:
³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β
Answer:
C2H6 OR C6H12O6 THE ANSWER
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
