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Alexeev081 [22]
3 years ago
9

What is observed when a solution of potassium iodide is added to silver nitrate solution ? a) No reaction takes place (b) White

precipitate of silver iodide is formed (c) yellow precipitate of Agl is formed (d) Agl is soluble in water.
Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer:

c) yellow precipitate of Agl is formed

Explanation:

KI + AgNO3 = AgI (s)+ KNO3

c) yellow precipitate of Agl is formed

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A certain radioactive isotope decays at a rate of 0.2​% annually. Determine the ​half-life of this​ isotope, to the nearest year
pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau}

Where:

m - Current isotope mass, measured in kilograms.

t - Time, measured in years.

\tau - Time constant, measured in years.

The solution of this differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%

1 - \frac{m(t+1)}{m(t)} = 0.002

1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002

1 - e^{-\frac{1}{\tau} } = 0.002

e^{-\frac{1}{\tau} } = 0.998

-\frac{1}{\tau} = \ln 0.998

The time constant associated to the decay is:

\tau = -\frac{1}{\ln 0.998}

\tau \approx 499.500\,years

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

t_{1/2} = \tau \cdot \ln 2

If \tau \approx 499.500\,years, the half-life of the isotope is:

t_{1/2} = (499.500\,years)\cdot \ln 2

t_{1/2}\approx 346.227\,years

The half-life of the radioactive isotope is 346 years.

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