Döbereiner grouped the known elements into <em>triads</em> (sets of three) so that
• The <em>atomic mass of the middle element</em> was approximately the average of the other two
• The <em>chemical properties of the middle element</em> were between those of the other two
• The <em>physical properties of the middle element</em> were between those of the other two
One example of a triad is Li – Na – K.
(a) Atomic mass of Na = 23.0 u
Average atomic mass of Li and K = (6.9 u + 39.1 u)/2 = 46.0 u/2 = 23.0 u
(b) Li reacts slowly with water. Na reacts rapidly. Potassium reacts violently.
(c) Melting point of Na = 371 °C.
Average melting point of Li and K = (454 °C + 330 °C)/2 = 784 °C/2
= 392 °C
Answer:
(a) 7.11 x 10⁻³⁷ m
(b) 1.11 x 10⁻³⁵ m
Explanation:
(a) The de Broglie wavelength is given by the expression:
λ = h/p = h/mv
where h is plancks constant, p is momentum which is equal to mass times velocity.
We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.
v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s
m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg
λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m
(b) For this part we have to use the uncertainty principle associated with wave-matter:
ΔpΔx > = h/4π
mΔvΔx > = h/4π
Δx = h/ (4π m Δv )
Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.
Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )
= 0.045 m/s
Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )
= 1.11 x 10⁻³⁵ m
This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.
Given the mass of 
=25.6 g
The molar mass of =390.35g/mol
Converting mass of to moles:
Converting mol to mol S:
Converting mol S to atoms of S using Avogadro's number:
1 mol = 
