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2 years ago

Write the balanced nuclear equation for the alpha decay of each isotope.

Chemistry

1 answer:

brilliants [131]2 years ago

8 0

Explanation:

First, a quick revision of radioactive decay:

During alpha decay, an alpha particle is emitted from the nucleus —- it is the equivalent of a helium atom (i.e. it has a mass of 4 and an atomic number of 2). So, let's take the following question:

Polonium-210 is a radioisotope that decays by alpha-emission. Write a balanced nuclear equation for the alpha decay of polonium-210.

In symbols, the equation becomes

210/84Po--->?+4/2HE

The sums of the superscripts and of the subscripts must be the same on each side of the equation.

Take 4 away from the mass number (210-4 = 206)

Take 2 away from the atomic number (84-2 = 82). Lead is element number 82.

So, the equation is

210/84 Po--->206/82Pb+4/2He

Now let's try one for beta decay — remember that, in beta decay, a neutron turns into a proton and emits an electron from the nucleus (we call this a beta particle)

Write a balanced nuclear equation for the beta decay of cerium-144)

In nuclear equations, we write an electron as 0^-1e.

144/58Ce-->144/59Pr+^0-1e

Here's a fission reaction.

A nucleus of uranium-235 absorbs a neutron and splits in a chain reaction to form lanthanum-145, another product, and three neutrons. What is the other product?

We write a neutron as 1/0n, so the equation is

235/92U +1/0n--->145/57La+X+3 1/0n

Sum of superscripts on left = 236. Sum of superscripts on right = 148. So X

must have mass number = 236 – 148 = 88.

Sum of subscripts on left = 92. Sum of subscripts on right = 57. So X

must have atomic number = 92 – 57 = 35. Element 35 is bromine.

The nuclear equation is

235/92U+1/0n--->145/57La+88/35Br+31/0N

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pH = pKa - log [HA]/[A-]
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The pka of a carboxylic can be affected greatly by the presence of various functional groups within its structure. An example of an inductive effect changing the pka can be shown with trichloroacetic acid, Cl3CCOOH. This molecule has a pka of 0.7. The decrease in pka relative to formic acid is due to the presence of the Cl3C- group, and more specifically the presence of the chlorine atoms. The electronegative chlorine atoms are able to withdraw the electron density away from the oxygen atoms and towards themselves, thus helping to stabilize the negative charge and stabilize the conjugate base. This results in an increase in acidity and decrease in pka.

The same Cl3CCOOH example can be used to explain how dipoles can effect the acidity of carboxylic acids. Compared to standard acetic acid, H3CCOOH with a pka of 4.76, trichloroacetic acid is much more acidic. The difference between these structures is the presence of C-Cl bonds in place of C-H bonds. A C-Cl bond is much more polar than a C-H bond, due the large electronegativity of the chlorine atom. This results in a carbon with a partial positive charge and a chlorine with a partial negative charge. In the conjugate base of the acid, where the molecule has a negative charge localized on the oxygen atoms, the dipole moment of the C-Cl bond is oriented such that the partial positive charge is on the carbon that is adjacent to the oxygen atoms containing the negative charge. Therefore, the electrostatic attraction between the positive end of the C-Cl dipole and the negative charge of the anionic oxygen helps to stabilize the entire species. This level of stabilization is not present in acetic acid where there are C-H bonds instead of C-Cl bonds since the C-H bonds do not have a large dipole moment.

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Dispersion forces, which are also called London forces, usually increase with molar mass because molecules with larger molar mass tend to have more electrons, and dispersion forces increase in strength with the number of electrons. Furthermore, larger molar mass often means a bigger atom whose electron distribution is more easily disturbed because the outer electrons are less tightly held by the nuclei.

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