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 :

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

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7 0

3 years ago

How many molecules of co2 in 97.3 grams of co2

s2008m [1.1K]

Answer:

1.33 × 10²⁴ molecules CO₂

General Formulas and Concepts:

Chemistry - Stoichiometry

Reading a Periodic Table
Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

97.3 g CO₂

Step 2: Define conversions

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

$97.3 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} )$ = 1.33138 × 10²⁴ molecules CO₂