Answer:
An ionic bond is the bonding between a non-metal and a metal, that occurs when charged atoms (ions) attract.
Explanation:
Here I put the function of Iconic Bond.
- <em>Ionic bonds form so that the outermost energy level of atoms are filled. Ion. an atom or group of atoms that bring out a positive or negative electric charge as a result of having lost or gained one or more electrons.</em>
<em>Therefore, I hope this helps!</em>
When Sodium and Chlorine come together they transfer an electron.
- Source: google
Hopefully this was clear and you understood!
The mole fraction of KCl in the solution is 0.1051
calculation
mole fraction of KCl in solution = moles of KCl / total number of moles(moles of KCl +moles of H2O)
moles=mass/molar mass
mass of KCl=32.7g
molar mass of KCl= 39 +35.5
moles of KCl is therefore= 32.7g/74.5 g/mol=0.439 moles
find the moles of H2O= mass of H2O/molar mass
mass of H2O=100-32.7=67.3g
molar mass of H2O=( 1 x2) +16=18 g/mol
moles = 67.3/18 =3.739 moles
total moles=3.739+0.439=4.178 moles
mole fraction is therefore=0.439/4.178=0.1051
Answer: Tin (Sn)
Explanation: The electron configuration for tin (Sn) is shown in the picture. It's last electrons are:
5s^2 4d^10 5p^2
The valence electrons are in the 5th electron shell and include 2 each in the 5s and 5p orbitals.
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
