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464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>
Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
D, since the anther is a male organ
Answer:
Al₂(SO₄)₃ and Na
Explanation:
Al has a charge of +3, Na has a charge of +1 and SO₄ has a charge of -2. Since cations and anions will bond we know that Al will bond with SO₄ leaving Na by itself (since this is a single replacement reaction). When Al bonds with SO₄ it makes aluminum sulfate which is Al₂(SO₄)₃ and Na will be left by itself.
Explanation:
Atomic Number = Number of protons
Mass Number = Number of protons + Number of neutrons
Isotopes are simply atoms of an element with the same number of protons and different number of neutrons.
First Isotope -- 238U
Number of neutrons = Mass Number - Atomic Number
Number of neutrons = 238 - 92 = 146
Second Isotope -- 235U
Number of neutrons = Mass Number - Atomic Number
Number of neutrons = 235 - 92 = 143
