Answer:
Explanation:
Utilizing Rydber's equation:
ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have
n=1 to n= infinity
ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)
n= 3 to n= infinity
ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J
b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .
1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹
1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m
λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm
Molar mass of solute is 30 g/mol
molality is defined as the number of moles of solute in 1 kg of solvent
number of moles of solute - 5.0 g / 30 g/mol = 0.17 mol
molality of solution is 0.70 mol/kg
the mass of solvent when there are 0.7 mol of solute - 1 kg
then the mass of solvent when there's 0.17 mol of solute - 0.17 mol / 0.7 mol/kg
= 0.242 kg
therefore mass of solvent is 0.242 kg
I believe the answer is Pangaea
I hope this helps :)
Explanation:
As you move across the periodic table, the number of protons and neutrons increases but the number of orbital levels of the period remains the same. The atomic radii therefore decrease, across the period, because the increase in proton number causes an increased pull of the orbital electrons bringing them closer to the nucleus.
As you move down a group in a periodic table, the number of orbital levels increase. The effective nuclear charge of the nucleus of the atoms decreases due to the increased number of orbital levels that shield the valence electrons from the attractive force nucleus.
