1 kg =1000 g = 10³ g,
1 m = 100 cm ,
(1m)³ = (100 cm)³= (10²)³ cm³ = 10⁶ cm³ or 10⁶ mL
5,427 kg/m³ = <span>5,427 kg/ 1m³ = (5427 * 10³ g)/ 10⁶ mL=5427/10³ g/mL=
=5427/1000 g/mL = 5.427 g/mL
</span>5.427 g/mL is density of Mercury, and 1.0 g/mL is density of water.
Density of Mercury is more then density of the water, so
mercury will sink in the water.
First we find for the wavelength of the photon released due
to change in energy level. We use the Rydberg equation:
1/ʎ = R [1/n1^2 – 1/n2^2]
where,
ʎ is the wavelength
R is the rydbergs constant = 1.097×10^7 m^-1
n1 is the 1st energy level = 1
n2 is the higher energy level = infinity, so 1/n2 = 0
Calculating for ʎ:
1/ʎ = 1.097×10^7 m^-1 * [1/1^2 – 0]
ʎ = 9.1158 x 10^-8 m
Then calculate the energy using Plancks equation:
E = hc/ʎ
where,
h is plancks constant = 6.626×10^−34 J s
c is speed of light = 3x10^8 m/s
E = (6.626×10^−34 J s * 3x10^8 m/s) / 9.1158 x 10^-8 m
E = 2.18 x 10^-18 J = 2.18 x 10^-21 kJ
This is still per atom, so multiply by Avogadros number =
6.022 x 10^23 atoms / mol:
E = (2.18 x 10^-21 kJ / atom) * (6.022 x 10^23 atoms /
mol)
E = 1312 kJ/mol
Light PLEASE Thanks me I need points to ask questions
H2 is known to exist. For dihydrogen, H2, we can identify the frontier molecular orbitals (FMOs). The highest occupied molecular orbital (or HOMO) is the σ (sigma) 1s MO. The lowest unoccupied MO (LUMO) is the σ* (sigma star) 1s MO which is antibonding.
