Uranium-232 has a half-life of 68.8 years. After 344.0 years, how much uranium-232 will remain from a 375.0-g sample?

Chemistry

1 answer:

N76 [4]3 years ago

8 0

Uranium-232 will remain 11.72 g

<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:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}$

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

t 1/2 = half-life = 68.8 years

T = duration of decay = 344 years

No= initial sample = 375 g

Uranium-232 remain :

$\tt Nt=375.\dfrac{1}{2}^{344/68.8}\\\\Nt=375.\dfrac{1}{2}^5\\\\Nt=11.72~g$

You might be interested in

A balloon has a volume of 10,500 liters, and the temperature is 15°C. If the temperature were -25°C, what would the volume of th

harina [27]

Answer is: b) 9,042 L.
Ideal gas law: p·V = n·R·T.
p is the pressure of the gas.
V is the volume of the gas.
n is amount of substance.
R is universal gas constant.
T is temperature.
T₁ = 15°C = 288 K.
V₁ = 10,5 L.
T₂ = -25°C = 248 K.
V₂ = ?
V₁/T₁ = V₂/T₂
V₂ = 10,5 L · 248 K ÷ 288 K
V₂ = 9,042 L.

8 0

3 years ago

A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperat

stira [4]

Answer:

8,19 L

Explanation:

The combined gas law is a equation that can be used when the initial and final conditions -pressure,volume,amount of moles,temperature-, of a gas change during a process. It can be written:

$\frac{P1V1}{n1T1}=\frac{P2V2}{n2T2}$