Answer:
what are the questions???
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) =
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).
Answer:
The environment of space is lethal without appropriate protection: the greatest threat in the vacuum of space derives from the lack of oxygen and pressure, although temperature and radiation also pose risks. The effects of space exposure can result in ebullism, hypoxia, hypocapnia, and decompression sickness.
Answer:
7.07
Explanation:
HA = weak acid = 0.053
A+ = conjugate base = 0.045
Ka = 7.2x10^-8
Ka = [H+][A-]/HA
7 2x10^-8 = [H+][0.045]/0.053
[H+] = 7.2x10^-8 x 0.053/0.045
= 8.48x10^-8
PH = -log[H+]
= -log[8.48x10^-8]
PH = -[login.48 + log10^-8]
PH = -0.928 - (-8)log10
= 7.07
Elements can be defined by their unique properties and atomic.
