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.

By definition, half-life is the amount of time it takes for a radioactive substance to decrease by one-half its original value. In this question, the isotope's half life is 20min, meaning after 20min, 1g of the isotope will be reduced to 0.5g. After another 20min, 0.5g of the isotope will be reduced to 0.25g. Adding up the two 20min time frames it took to go from 1g of substance to 0.25g gives 40min.

There is an equation to calculate half-life problems, but if the sample is cut into halves every half-life, there is a simple way to find the time passed. Let's start with the total mass of the sample at time . For every  the sample size will be cut in half:

After  there will be  of the isotope left.

For problems with half-life, we will use the equation

Where  is the initial mass of the sample,  is the amount left after time  passes and  is the half-life of the isotope. We have

To estimate the half-life we just see how long it takes for the sample to be cut in half. The green lines indicate when there is half the initial sample () which happens after  passes. Therefore the half-life is .

Check:  Will the sample be cut in half again after another ? The orange lines indicate the mass of the sample after another half-life and indeed there is now  which is half of .

First of all, it's important to understand terminology used in this kind of a concept question. A nuclide refers to a particular atom or nucleus with neutrons and protons. As we know, neutrons refer to the neutrally charged particles within a nucleus, while protons refer to those that are positively charged. 

The zone of stability is a roughly linear pattern that follows a positive correlation with number of protons and neutrons. As the number of protons increases in the nucleus, so does the number of neutrons to allow for stability. 

When a nuclide falls above or below the zone of stability, it creates a beta or alpha positron, respectively. These nuclides would not be considered stable, as they show degradation over time. 

The ideal gas law has some conditions that must be met, conditions that certainly cannot be met in the real world. These conditions include that the gases cannot interact with one another, gases must be moving in a random straight-line fashion, gas molecules must not take up any space, and gases must be in perfect elastic collisions with the walls of the container. These conditions minimize the effect that gas molecules have on one other, allowing a prediction based on completely random and unimpeded molecular movement. In reality, these conditions are impossible. All real gas molecules have a molecular volume and some degree of intermolecular attraction forces.

In this case, two variables are changed between the initial and final containers: volume and temperature. Since we are looking for the final pressure on the container, we can use the combined gas law in order to solve for the final pressure:

When using the ideal gas law, remember that temperature must be in Kelvin, not Celsius, so we will need to convert.

Use the given values to solve for the final pressure.

Once again, we can use the ideal gas law in order to solve for the unknown gas. We will start again with the ideal gas law:

Density has the units of mass over volume, which means that we can rearrange the ideal gas law so that density is on one side of the equation. However, we need to substitute one of the values so mass is also in the equation. Remember that moles are equal to mass divided by molar mass, so we can rewrite "n" in order to solve for density:

STP means that the container is at 1 atmosphere of pressure and 273 Kelvin. Knowing this, we can solve for molar mass:

Fluorine gas has a molar mass of 38.00 grams per mol. As a result, we determine that fluorine gas is the unknown gas in the container.