The half-life of phosphorus-32 is 14.26 days. calculate its decay constant. express the decay constant numerically in inverse da
2 answers:
The decay constant(λ) = <u>0.0486/day</u>
<h3>Further explanation</h3>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:
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
The half-life can be expressed in a decay constant( λ)
The half-life of phosphorus-32 is 14.26 days
Then : t 1/2 = 14.26 days
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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Answer:
Mass of 205 mL of the liquid is 164 g
Explanation:
We know,
Here density of liquid is 0.798 g/mL and volume of liquid is 205 mL
So, mass of liquid =
=
= 164 g
hence mass of 205 mL of the liquid is 164 g.
a=b=33400 J
c=4180 J
Further explanation
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
The heat to change the phase can be formulated :
Q = mLf (melting/freezing)
Q = mLv (vaporization/condensation)
Lf=latent heat of fusion
Lv=latent heat of vaporization
The heat of fusion for water at 0 °C : 334 J/g and the heat of vaporization at 100 °C : 2,230J/g
a. melting
b. freezing
c.raise the temperature(c water = 4.18 J/g C)