Answer:
See explanation
Explanation:
According to Louis de Broglie, matter has an associated wavelength. He was the first scientist to establish the idea of wave-particle duality or wave- particle paradox.
The display of wavelike properties by objects in the universe is dependent on the magnitude of the of the mass of the body. Small objects have a large associated wavelength and can be described completely by quantum mechanics.
A buckyball with a mass of 1.2 x 10-21 g, 0.7 nm wide, moving at 38. m/s has a very small mass and significant associated wavelength hence the system can be completely described by quantum mechanics.
Answer:
Answer is 'B' I think
Explanation:
Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.
Answer:
The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance (the decimal associated with percent of atoms of that element that are of a given isotopе). An element does not have an absolute atomic mass.
<em>Hope</em><em> this</em><em> </em><em>helps</em><em> </em><em>:</em><em>)</em>
Answer:
Filtration is a method for separating an insoluble solid from a liquid. When a mixture of sand and water is filtered: the sand stays behind in the filter paper (it becomes the residue ) the water passes through the filter paper (it becomes the filtrate )
Explanation:
Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3
