All AP Chemistry Resources
Example Questions
Example Question #1 : Other Atomic Concepts
As atomic radius decreases, the force of attraction between the nucleus of the atom and its electrons __________.
increases
stays the same
decreases
cannot be determined
increases
As the atomic radius decreases, electrons are drawn closer to the nucleus. Since the electromagnetic force between the positively charged nucleus and negatively charged electrons is a function of distance, the force of attraction, or effective nuclear charge, exerted on each electron will be greater.
Example Question #1 : Elements, Ions, And Isotopes
What is the shorthand electron configuration for gallium?
To write out the shorthand electronic configuration for Ga, you simply write down the noble gas that that comes before , in this case , then write the rest of the electron configuration as normal.
Example Question #2 : Elements, Ions, And Isotopes
An electron falling to a lower energy level gives off a blue glow (λ=475nm). How much energy is emitted?
3.14x10-37J
4.184 x 10-19 J
4.184 x 10-17 J
Energy is not emitted
3.14x10-40J
4.184 x 10-19 J
E=hf, or E=hc/λ
h is Planck's constant with a value of 6.62606957 × 10-34 m2 kg / s. Converting 475nm to m gives 4.75x10-7m. c is the speed of light with 3x108m/s. Putting these variables into the second equation you get 4.174x10-19J
Example Question #3 : Elements, Ions, And Isotopes
The atomic number of an atom is equal to the number of __________.
protons
electrons + neutrons
electrons
protons + neutrons
neutrons
protons
The atomic number is equal to the number of protons, because no two elements have the same number of protons. It is however possible for different elements to have the same number of electrons (through the loss or gain of electrons) or neutrons (isotopes).
Example Question #1 : Radioactive Decay And Nuclear Chemistry
What kind of radiation has no charge or mass?
gamma
alpha
beta
delta
gamma
This is the definition of gamma radiation.
Example Question #581 : High School Chemistry
Consider the following isotope of thorium:
What is the identity of the product following three alpha decay reactions?
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).
Example Question #2 : Radioactive Decay And Nuclear Chemistry
Consider the following isotope:
What is the identity of the product after the following series of decay reactions?
alpha decay, alpha decay, electron emission, positron emission, positron emission
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.
Example Question #81 : Elements And Atoms
Uranium-238 undergoes alpha decay. Which nucleus is formed as a result of this decay?
Thorium-234
He-4
Uranium-234
Uranium-238
Thorium-234
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.
Example Question #2 : Radioactive Decay And Nuclear Chemistry
Am-242 undergoes beta-decay at a half-life of 16 hours. If a chemist starts with a sample of of Am-242 and allows it to undergo beta-decay, how much of the original sample remains after 4 days?
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.
Example Question #1 : Radioactive Decay And Nuclear Chemistry
An alpha particle can also be written as which of the following?
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.
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