Download How many electrons are present in a chromium

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Periodic table wikipedia , lookup

Livermorium wikipedia , lookup

History of molecular theory wikipedia , lookup

Tennessine wikipedia , lookup

Valley of stability wikipedia , lookup

Extended periodic table wikipedia , lookup

Ununennium wikipedia , lookup

Unbinilium wikipedia , lookup

Transcript
How many electrons are present in
a chromium-52 atom?
24 electrons
How many protons are present in a
chromium-52 atom?
24 protons
How many neutrons are present in a
chromium-52 atom?
28 neutrons
Give the number of protons,
neutrons, and electrons in the
106Pd2+ ion.
46 protons
60 neutrons
44 electrons
+2 ion indicates 2 more positives
than negatives meaning that two
electrons have been lost
You have a neutral atom that has
atomic number 77 and a mass number
of 192. How many protons,
neutrons, and electrons does it
have?
77 protons
115 neutrons
77 electrons
since neutral, #protons =
#electrons
How many paired electrons are in a
germanium atom? How many unpaired
electrons are in this atom?
30 paired, 2 unpaired
[Ar] 4s2 3d10 4p2
Hund’s Rule says that in 4p, the
electrons will enter each orbital
one at a time until each orbital
has at least one electron
Give the orbital filling diagram
for germanium (shorthand okay).
[Ar]
4s
3d
4p
Electrons will go to the place of
________ energy first. This is
attributed to the _______
_________.
Lowest; Aufbau principle
No two electrons in an atom can
have the same set of 4 quantum
numbers. This statement is an
illustration of ??
The Pauli exclusion principle
Give the number of protons,
neutrons, and electrons in the
127I-1 ion.
53 protons
74 neutrons
54 electrons
-1 indicates that this atom has
gained 1 electron since there is a
negative charge indicating one
more electron than proton
In order for an electron to move
away from the nucleus to an outer
energy level, it must _____
energy.
Gain; in lab, we provided energy
with the Bunsen burner flame
permitting electrons to escape
ground state and move to excited
state
If an electron falls back toward
the nucleus, we observe a change
in energy that corresponds to
colors of different wavelength.
What happens to the energy of the
electron?
As the electron moves back toward
the nucleus, it gives off energy
falling back to the original
energy level.
The visible spectrum ranges in
color from red to violet and in
wavelength from ??
Red 700 nm – highest wavelength
Violet 400 nm – lowest wavelength
A photon of light has a frequency
of
6.3 x 1015 Hz. What is its
wavelength, in cm? The speed of
light = 3 x 108 m/sec.
 = c/ = 3 x 108 m/sec = 4.76 x
10-8 m
6.3 x 1015 /sec
4.76 x 10-8 m ( 100 cm ) = 4.76 x
10-6 cm
A particle is found to have a
wavelength of 450 nm. What is its
frequency? The speed of light = 3
x 108 m/sec.
 = c/ =
1014 Hz
3 x 108 m/sec
= 6.7 x
4.5 x 10-7 m
Determine the energy that a
particle with frequency of 9.6 x
1014/sec has. Planck’s constant is
6.63 x 10-34J-sec.
E = h = 6.63 x 10-34J-sec(9.6 x
1014/sec)
E = 6.37 x 10-19J
What is the energy of a particle
that has a wavelength of 0.000075
cm? Planck’s constant is 6.63 x
10-34 J-sec.
E = h
However, we don’t have  and cm:m
issue
 = c/ = 3 x 108 m/sec = 4 x
1014/sec
7.5 x 10-7 m
Then E = h
6.63 x 10-34 J-sec(4 x 1014/sec)
2.65 x 10-19J
The energy of a particle is found
to be
6.2 x 10-19 J. What frequency is
associated with this particle?
Planck’s constant is 6.63 x 10-34
J-sec.
E = h
You have energy and Planck’s
constant so rearrange to solve for
.
E/h = 
6.2 x 10-19 J
= 9.35 x 1014/sec
6.63 x 10-34 J-sec
It was determined that the energy
of a photon is 7.3 x 10-19 J. What
is the wavelength of the photon?
Planck’s constant is 6.63 x 10-34
J-sec and the speed of light = 3 x
108 m/sec.
E = h rearranging, we have
E/h = 
7.3 x 10-19 J
= 1.1 x
1015/sec
6.63 x 10-34 J-sec
then
 = c/ = 3 x 108 m/sec = 2.73 x
10-7m
1.1 x 1015/sec
Lithium in nature occurs as a
mixture of 6Li atoms with a mass
of 6.015 amu and 7Li with a mass
of 7.016 amu. The atomic mass of
lithium is 6.941. What is the
percent abundance of each of the
two isotopes?
x + y = 1
x = 1 – y
6.941
6.941
6.941
6.941
0.926
0.925
=
=
=
–
=
=
6.015 (x) + 7.016 (y)
6.015 (1 – y) + 7.016y
6.015 – 6.015y + 7.016y
6.015 = 7.016y – 6.015y
1.001y
y
therefore, 6Li occurs 7.5% of the
time and 7Li occurs 92.5% of the
time. This answer makes sense as
the atomic mass is much closer to
the mass of the heavier isotope
indicating that it should have a
much greater abundance than 6Li.
Atoms of the same element that
vary in number of neutrons and
therefore also in mass are called
??
Isotopes
Atoms of an element that carry a
charge due to either a gain or
loss of electrons are known as ??
Ions
When an atom gains electrons, it
will have a ___ charge.
Negative because it has more
electrons (-) than protons (+)
Which particle always remains the
same for the same element?
Protons remain the same because
protons define the element.
The number of protons is given by
the ??
atomic number
The sum of the number of protons
plus the number of neutrons is
called the ??
Mass number
If an element consists of 60.1% of
atoms with a mass of 68.926 amu
and the remainder of the atoms
have a mass of 70.925 amu, what is
the atomic mass of the element?
Based on this mass, what do you
think the identity of the element
is?
100% - 60.1% = 39.9% of isotope 2
AM = 68.926 (0.601) + 70.925
(0.399)
AM = 69.723 amu
Looking at the periodic table,
this element may be gallium as it
seems to be closest.
Give the quantum numbers for the
valence electrons in Mg.
Magnesium has the configuration
1s2 2s2 2p6 3s2. The valence
electrons are the two in the 3s
sublevel.
3s1
3s2
n = 3, l = 0, ml = 0, ms = ½
n = 3, l = 0, ml = 0, ms = -½
Give the orbital filling diagram
and the full electron
configuration for molybdenum.
Tungsten (W) consists of two
isotopes. 183W has a mass of
183.26 amu and 184W has a mass of
184.72 amu. Determine the
abundance of each isotope of
tungsten.
First, go to the periodic table to
obtain the atomic mass of tungsten
(183.84 amu). Then, set up your
equation.
183.84 = (183.26)(x) + 184.72(y)
We know that x + y = 1; thus, y =
1-x
Substitute for y in the equation.
183.84 = 183.26x + 184.72(1-x)
183.84 = 183.26x + 184.72 –
184.72x
183.84 – 184.72 = 183.26x –
184.72x
-0.88 = -1.46x
0.603 = x
0.391 = y
Thus, 183W occurs 60.3% of the time
and 184W occurs 39.1% of the time.