has two sigma bonds between oxygen and hydrogen. It also has two pair of lone electrons on oxygen. The number of bonds and lone pairs contribute to the hybridization. For example, if a molecule consists of one sigma bond and one electron pair, the hybridization would be sp. If the molecule consists of two sigma bonds and one electron pair, the hybridization would be , and so forth.    

This is the definition of gamma radiation.

During alpha decay, an element emits a helium nucleus with 2 neutrons and 2 protons. Thus, the atomic mass of the new element is decreased by four, and the atomic number is decreased by two.

Three subsequent alpha decays result in a new element with an atomic mass of 232 - 3(4) = 220, and a new atomic number of 90 - 3(2) = 84.

Using the periodic table, we find the element with this atomic number is polonium (Po).

In alpha decay, a helium nucleus is emitted, and thus the isotope loses 2 protons and 2 neutrons.

In electron emission, a neutron in the nucleus is converted into a proton and an emitted electron.

In positron emission, a proton in the nucleus is converted into a neutron and an emitted positron.

The given isotope will lose 4 protons and 4 neutrons via alpha emission, gain 1 proton and lose 1 neutron via electron emission, and lose 2 protons and gain 2 neutrons via positron emission. The result is a loss of 5 protons and 8 mass units.

Accounting for the changes in atomic mass and number, we find that the final element is 141-praseodymium.

In alpha decay, an element loses the equivalent of a helium nucleus, or 2 protons and 2 neutrons. Uranium-238 has an atomic number of 92 and a mass of 238 units, so the product of alpha decay will have an atomic number of 90 and a mass of 234 units. This corresponds to thorium-234.

With each half-life, the amount of Am-242 is halved. 4 days is 96 hours, which is 6 half-lives. For half-life questions use the formula:

Where  is the number of half lives. 

Thus  of Am-242 is left after 4 days.

An alpha particle can be written as  or . The latter is used frequently in nuclear physics because it is more helpful when calculating resulting nuclei after radioactive decay. In alpha emission, the nucleus loses two protons and two neutrons.

A beam of electrons is known also as β radiation. An electron has atomic number of  and zero mass number. Hence, the reaction can be written:

Where  can be either, an electron or β radiation. The mass number is balanced because    and the atomic number is balanced because  . This kind of nuclear reactions are called beta decay reactions.

In order to answer this question, we'll need to employ the equation for first-order decay. The reason we know that this is a first order decay process is because we are told that the compound in question is radioactive. It's important to recall that all radioactive decay processes occur via a first order decay mechanism.

Before jumping into the math, let's first make a quick ballpark prediction of where our answer should be. We know that the half-life for any compound is the amount of time it takes for half of that compound to decay. Furthermore, we're given the value for a compound's half-life. Since the amount of time that has passed in the question is less than the half-life, we would expect to still have over half of the compound left. Thus, our answer should be above .

Where  is the amount of compound that exists at a specified time

 is the amount of compound that exists initially

 is the decay constant for the compound

 is the specified amount of time that has passed

To solve for the fraction of compound that exists at a given time relative to the initial amount, we can make the following changes to the above equation.

Thus far, we have everything we need except for the decay constant. However, as long as we know the half life of the compound, we can calculate the decay constant using the following formula that applies to all first order reactions.

Notice that the above decay constant has units of inverse-time. This is what we would expect, since all first order reactions have rate constants with these units.

Now, we are ready to plug in the value we calculated for the decay constant into the rate equation to find our answer.

Going back to the quick prediction we made, we can see that this calculated value agrees with our prediction.

By comparing the oxidation number of an atom as a reactant and its oxidation number as a product, we can determine if the atom has been oxidized or reduced. In increase in oxidation number indicates a loss of electrons, or oxidation. A decrease in oxidation number signals a gain of electrons, or reduction.

For electrochemistry, you should familiarize yourself with the traditional oxidation states of hydrogen , halogens , oxygen , and elemental atoms . 

Carbon is initially in the form of methane, meaning that it is attached to four hydrogen atoms. The molecule is neutral, and each hydrogen has an oxidation number of . Carbon must have an initial oxidation state of in order to balance the molecular charge.

In the a product, carbon is attached to two oxygens, each with a charge of . Again, the molecule is neutral, so carbon must balance these charges. This means that carbon's final oxidation state is .

Since carbon went from an oxidation state of  to , we can conclude that carbon has been oxidized in the reaction.