B. Biomass
(I guess so cause other ones are already being used)
The transitions which fall to the lowest principle position release the greatest energies. In this case, this would be the transition from the 5p to the 3s orbital (a Paschen transition).
Hope this helps!
I believe it would be better to use an orbital designation than the written configuration, if the number of electrons in the ground state of the atom are quite high for the given element, as above 50, for instance.
This saves space and also one can see the discrete quantized energy levels associated with the subshells of the main energy levels if written in orbital designation.
A is the correct answer and this question should be in physics not chemistry
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
- <em>Volume of the unit cell</em>
Volume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
<h3> 1.756x10⁻²² cm³ → Volume of the unit cell in cm³</h3><h3 />
- <em>Mass of the unit cell:</em>
<em>There are 4 atoms of gold:</em>
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
<h3>1.296x10⁻²¹g is the mass of the unit cell</h3><h3 />
- <em>Density of the metal:</em>
1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
<h3>7.38 g/cm³ is the density of the metal</h3>
