Answer:
Elemental gold to have a Face-centered cubic structure.
Explanation:
From the information given:
Radius of gold = 144 pm
Its density = 19.32 g/cm³
Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:
a = 407 pm
In a unit cell, Volume (V) = a³
V = (407 pm)³
V = 6.74 × 10⁷ pm³
V = 6.74 × 10⁻²³ cm³
Recall that:
Net no. of an atom in an FCC unit cell = 4
Thus;
density d = 19.41 g/cm³
Similarly; For a body-centered cubic structure
where;
r = 144
a = 332.56 pm
In a unit cell, Volume V = a³
V = (332.56 pm)³
V = 3.68 × 10⁷ pm³
V 3.68 × 10⁻²³ cm³
Recall that:
Net no. of atoms in BCC cell = 2
∴
density =17.78 g/cm³
From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.
This makes the elemental gold to have a Face-centered cubic structure.
Answer:
Chlorine
Explanation:
Each arrow represents one electron. Most of the boxes are filled, meaning they have two electrons. The last box only has one arrow, so it only has one electron. If you add it up, you get 17, which is Chlorine.
Answer:
900 K
Explanation:
Recall the ideal gas law:
Because only pressure and temperature is changing, we can rearrange the equation as follows:
The right-hand side stays constant. Therefore:
The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.
Substitute and solve for <em>T</em>₂:
Hence, the temperature must be reach 900 K.
Answer : The partial pressure of 
at equilibrium is, 1.0 × 10⁻⁶
Explanation :
The partial pressure of 
= 
The partial pressure of = 
The partial pressure of 
=
The balanced equilibrium reaction is,
Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴
At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)
The expression of equilibrium constant 
for the reaction will be:
Now put all the values in this expression, we get :
The partial pressure of at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶
Therefore, the partial pressure of at equilibrium is, 1.0 × 10⁻⁶
