Mod 7

A. You have 2 objects that are positively charged, will they repel or attract or each other?

 

B. You have 2 objects that are negatively charged, will they repel or attract or each other

 

 

A.They will repel each other

 

B. They will be attracted to each other

(remember “opposites attract”)

 

 

A. A substance has 108 positive charges and 108 negative charges. What is the overall charge?

 

B. Suppose this same substance now has 58 negative charges and 50 positve charges, what is the overall charge now?

A. It is electrically neutral

 

B. Now it is electrically negatively charged

What did William Crookes discover about the atomic structure?
William Crookes discovered cathrode rays which were later determined to be electrons.

How many protons, neutrons, and electrons are in the following atoms:

 

A. Plutonium

 

B. Kypton

 

C. Silicon

 

Turn to page 205 in your book if you need to review on how to determine the number of protons, electrons, and nuetrons in an atom.

A. Plutonium has 94 electrons, 94 protons, and 150 neutrons. 

 

B. Kypton: 36 protons, 36 electrons, and 48 neutrons.

 

C. Silicon: 14 protons, 14 electrons, and 14 neutrons.

 

A. What is the symbol for for the atom that has 21 protons, 21 electrons, and 24 neutrons?

 

B. What is the symbol for the atom that has 10 protons, 10 electrons, and 10 neutrons?

A. 45Sc or Scandium-45

 

B. 20Ne or Neon-20

What is the definition of isotopes?
Isotopes are atoms with the same number of protons but with different number of neutrons.
You have 2 special holiday lights. One is a red light and the other is a green light. The red light is brighter that the green light. What can you say about their amplitudes? Which emits light of a higher frequency?
Brightness is determined by amplitude, therefore the red light has a higher amplitude than the green light. According to the mnemonic “ROY G. BIV, the greenlight has a shorter wavelength than the red light. A shorter wavelength means a higher frequency so the green light emits a higher frequency.

What is the frequency of light that has a wavelength of 20nm?

 

 

If you forgot how to solve, study equation 7.1 on page 217

First, the units disagree so we must change the “nm” to “m”.

 

20nm x 1 x 10-9m

    1        1 nm          =   2.00 x 10-9m

 

 

Now, the units agree and we can use equation 7.1

 

                  3.0 x 108m/s            

  f=  v    =     2.00 x 10-9m    =  1.5 x 1016 l/s

 

 

l/s is the same as hertz (Hz) so the frequency is:

 

1.5 x 1016 Hz

 

 

What is the wavelength of light that has an energy of 1.50 x 10-18J?

There is no direct relationship between energy and wavelength. However, given the energy, we can get the frequency (equation 7.2 on page 221) and then we can use the equation 7.1 from page 217 to go from frequency to wavelength.

 

We will start with frequency: E = h*f

 

Turn the equation around to solve for frequency:

 

               1.50 x 1018J

f= E/h =   6.63 x 1034 J/Hz  = 2.26 x 1015Hz

 

 

Now we can solve for wavelength:

 

             3.0 x 108m/s 

λ= f    =    2.26 x 1015l/s   = 1.3 x 10-7m

 

The wavelength is 1.3 x 10-7m

 

 

 

 

The _________ is the heaviest part of the atom
neutron
When an electron is “excited” and needs to get rid of excess energy, it will jump back to the orbit that is closest to the neutron, we say the atom is ________ and in this process, an atom will ________ light.

de-excited

 

emit

List the orbitals from lowest energy to higher energy

S-orbitals (lowest energy)

 

P-orbitals (next lowest energy)

 

d-orbitals (highest of the three orbitals)

What is wrong with the following electron configurations?

 

A. 1s22s63p2

 

B. 1s22s22p63s23p64s23d84p65s1

 

C. 1s22s22p63s23p64s23d104p64d105s2

A. There are too many electrons in the 2s orbital. There should only be 2 there.

 

B. The 3d orbital is not filled. You cannot go to the next orbital until it is filled up by the one below.

 

C. The 4d orbital should be after 5s orbital, not before it.

Give the full electron configuration of the following atoms:

 

A. Sc

 

B. Cl

 

C. Ca

A. To get to element Sc, we must go through row 1, which has 2 boxes in the s orbital block (1s2). We then go through all of row 2 which has 2 boxes in the s orbital block and six boxes in the p orbital block (2s22p6). We also go through row 3, which has 2 boxes in the s orbital and 6 in the p orbital block (3s23p6). We then go to the fourth row where we pass through both boxes of the s orbital block (4s2). Finally, we go through one box in the d orbital block. Since we subtract one from the row number for d orbitals, this gives us 3d1. Thus, or final electron configuration is :

 

1s22s22p63s23p64s23d1

 

 

B. To get to element Cl, we must go through row one, which has 2 boxes in the s orbital block (1s2). We then go through all of row 2 which has 2 boxes in the s orbital block and six in the p orbital block (2s22p6). We also go through both boxes in the s orbital block of row 3, (3s2). Finally, we go through 5 boxes in the p orbital block of row 3, giving us 3p5. Thus, our final configuration is :

 

1s22s22p63s23p5

 

C. To get Ca, we must go through 1, which has 2 boxes in the s orbital block (1s2). We then go through all of row 2 which has 2 boxes in the s orbital block and 6 boxes in the p orbital block (2s22p6). we also through row 3, which has 2 boxes in the s orbital and six in the p orbital block (3s23p6). We then go to the final row where we go through the first 2 boxes, giving us 4s2. Thus, our final electron configuration is:

 

1s22s22p63s23p64s2

Give abbreviated electron configurations for the following atoms:

 

 

A. P

 

B. Mo

 

C. Ba

 

A. The nearest 8A element that has a lower atomic number than P is Ne. The only difference between P and Ne is that there are two boxes in the row 2, s orbital group and three boxes in the row 3, p orbital group. Thus, the abbreviated electron configuration for P is:

 

[Ne]3s23p3

 

B. The neares 8A element that has a lower atomic number than Mo is Kr. The only difference between Mo and Kr is that there are 2 boxes in the row 5, s orbital group and 4 boxes int he row 4, d orbital group. Thus, the abbreviated electron configuration for Mo is:

 

[Kr]5s24d4

 

C. The nearest 8A element that has a lower atomic number than Ba is Xe. The only difference between Ba and Xe is that there are 2 boxes in the row 6, s orbital group. Thus, the abbreviated electron configuration for Ba is:

 

[Xe}6s2

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