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3 years ago

Iodine-131 is a radioactive isotope. after 9.00 days, 46.0% of a sample of 131i remains. what is the half-life of 131i?

1 answer:

6 0

For this problem, we use the integrated rate law for first order radioactive decay which is expressed as follows:

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant.

We need to calculate first the value of k from the given ration An/Ao and the time it reached that ration. We do as follows:

An = Aoe^-kt
0.46= e^-k(9)
k = 0.0863 / day

At half life, the remaining substance would be equal to one-half of the original so that An/Ao would be equal to 1/2 or 0.50. We calculate the half-life as follows:

An = Aoe^-kt
0.50 = e^-0.0863(t)
t = 8.03 days

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In a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B. 2 A ⟶ 3 B How many molecules

Harman [31]

Answer:

There will be produced 1.71 moles of B which contain 1.03×10²⁴ molecules

Explanation:

The example reaction is:

2A → 3B

2 moles of A produce 3 moles of B

If we have the mass of A, we convert it to moles and then, we make the rule of three: 29.2 g / 25.6g/mol = 1.14 moles

Therefore 2 moles of A produce 3 moles of B

1.14 moles of A will produce (1.14 . 3) / 2 = 1.71 moles of B are produced

Now we can determine, the number of molecules

1 mol has NA molecules (6.02×10²³)

1.71 moles have (1.71 . NA) = 1.03×10²⁴ molecules

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Answer:

true

Explanation:

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garri49 [273]

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2 years ago

Which letter shows the most direct sunlight?

timama [110]

Answer:c

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3 years ago

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

$E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}$ , where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

5 0

2 years ago