Chemistry, 28.02.2020 19:20, vonteee2

The half-life of Erbium-165 is 10.4 hours. After 24 hours a sample has been reduced to a mass of 2 mg. What was the initial mass of the sample, and how much will remain after 3 days?

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answered: boopiee2349

$[A_0]=10\ mg$

$[A_t]=0.08245\ mg$

Explanation:

Given that:

Half life = 10.4 hours

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{10.4}\ hour^{-1}$

The rate constant, k = 0.06664 hour⁻¹

Time = 24 hours

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

[A_t] is the concentration at time t = 2 mg

[A_0] is the initial concentration = ?

So,

$2\ mg=[A_0]\times e^{-0.06664\times 24}$

$[A_0]=10\ mg$

Now, time = 3 days = 3*24 hours = 72 hours ( 1 day = 24 hours)

$[A_0]=10\ mg$

Thus,

$[A_t]=10\times e^{-0.06664\times 72}\ mg=0.08245\ mg$

answered: Guest

the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products

answered: Guest

explanation:

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