AP Chemistry : AP Chemistry
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Example Questions
Example Question #611 : Ap Chemistry
What differentiates a homogeneous mixture from a pure substance?
A mixture has a defined composition.
A homogeneous mixture is uniform throughout.
A pure substance can be separated into elements by physical means.
The composition of a pure substance is fixed.
A pure substance is always a compound while mixtures can be elements or compounds.
The composition of a pure substance is fixed.
A heterogeneous mixture is a mixture of two or more components that do not appears uniform throughout. A pure substance contains only one component that has a fixed composition.
Example Question #612 : Ap Chemistry
A slice of watermelon would best be described as?
A pure compound
None of the above
A heterogeneous mixture of compounds
A heterogeneous mixture of elements and compounds
A homogeneous mixture of compounds
A heterogeneous mixture of elements and compounds
A heterogeneous mixture is a mixture of two or more components that do not appears uniform throughout, but contains components that are elemental (single atom type) and compounds (contain two or more different elements).
Example Question #3 : Atomic Structure And Properties
Using the PES spectra above, what element is illustrated?
C
Ne
Na
N
Li
N
The peak at ~ 40 MJ binding energy corresponds to the 1s orbital. Since there are peaks at lower binding energy, it is implied that this orbital is filled (by 2 electrons). There are two peaks at lower binding energy corresponding to the 2s and 2p orbitals, respectively. The 2s peak has the same intensity as the 1s peak indicating it has the same number of electrons (2). The 2p peak is 3/2 as tall as the 2s peak and indicates that indicates it has 3 electrons in the orbital. The electron configuration of 1s22s22p3 makes the element N.
Example Question #1 : Atomic Structure And Properties
Using the PES spectra above, what answer explains the differences in the position and intensity of the 3s peaks between Na and Mg?
Mg experiences a larger effective nuclear charge and has more electrons in the 3s orbital.
None of the above
Mg is larger in size making the electrons in the 3s orbital farther from the nucleus and thus easier to remove
Na experiences a larger effective nuclear charge on the 3s electrons and are thus harder to remove.
Na experiences a greater shielding effect from the 2s and 1s electrons making the 3s electrons easier to remove.
Mg experiences a larger effective nuclear charge and has more electrons in the 3s orbital.
Mg has a Z=12 while Na is Z=11. The increased number of protons in Mg gives rise to a greater effective nuclear charge. This greater effective nuclear charge binds the electrons in their respective orbitals more tightly than an element with a lower effective nuclear charge. The peak height of Mg is 2x that of Na for the 3s orbital because it has twice the electrons present.
Example Question #6 : Atomic Structure And Properties
Using the spectra above, answer the following: Why is the oxygen 1s electrons further to the right than the nitrogen 1s orbital?
Oxygen has more electrons and thus a greater amount of electron-electron repulsion present in the atom.
Nitrogen has 3 p-electrons and are thus unpaired leading to lower electron-electron repulsions.
There is a greater degree of electron shielding occurring in the oxygen 2p orbital.
Nitrogen is larger than oxygen thus making the 1s electrons easier to remove.
The effective nuclear charge on oxygen is greater than that of nitrogen.
The effective nuclear charge on oxygen is greater than that of nitrogen.
Oxygen has a Z=8 vs nitrogen’s Z=7. This greater effective nuclear charge binds the electrons in their respective orbitals more tightly than an element with a lower effective nuclear charge.
Example Question #1 : Photoelectron Spectroscopy
Using the spectra above, answer the following: What is the electron configuration of the element shown above?
The binding energy at ~ 53 Mj represents the 1s orbital. With other peaks present, this 1s orbital is fully occupied by 2 electrons. The peaks at lower energy would correspond to the 2s and 2p, respectively. Since the 2p peak is twice the intensity of the 2s or 1s peak, it has twice the electrons (4). This gives the electron configuration 1s22s22p4.
Example Question #2 : Atomic Structure And Properties
Using the spectra above, answer the following: What is the wavelength required, in m, to remove a valence electron from the element shown above?
None of the above
Step 1. Convert the Binding Energy from MJ/mol to J
Step 2:
Example Question #611 : Ap Chemistry
How many yellow spheres and green spheres are there per unit cell shown?
1 yellow, 1 green
8 yellow, 1 green
2 yellow, 1 green
None of the above
4 yellow, 1 green
1 yellow, 1 green
For a unit cell, the corners count as 1/8 of a sphere and atoms completely within the unit cell count as 1. This gives us 1 yellow and 1 green sphere in the unit cell shown above.
Example Question #612 : Ap Chemistry
Based upon the above image, what type of packing is this?
Tetragonal close packing (tcp)
Body centered packing (bcp)
Hexagonal close packing (hcp)
Simple cubic packing (scp)
Cubic close packing (ccp)
Cubic close packing (ccp)
The figure above represents ABCABC packing which is cubic close packing.
Example Question #613 : Ap Chemistry
Based upon the above image, what type of packing is this?
Hexagonal close packing (hcp)
Tetragonal close packing (tcp)
Body centered packing (bcp)
Simple cubic packing (scp)
Cubic close packing (ccp)
Hexagonal close packing (hcp)
The figure above represents ABABA packing which is hexagonal close packing.
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