What is the de broglie wavelength of an electron (m = 9. 11 × 10-31 kg) moving at a velocity of 3. 0 × 107 m/s?
1 answer:
the de Broglie wavelength of the electron moving at a velocity of 3. 0 × 107 m/s is 2.4×10∧-11nm
lambda = h/mv
The de Broglie wavelength is defined as follows: lambda = h/mv, wherein the greek letter lambda represents the wavelength, h is Planck's consistent, m is the particle's mass and v is its pace. One may also explicit mv as the particle's momentum
according to wave-particle duality, the De Broglie wavelength is a wavelength manifested in all the gadgets in quantum mechanics which determines the chance density of locating the item at a given factor of the configuration area. The de Broglie wavelength of a particle is inversely proportional to its momentum
Learn more about de Broglie wavelength here: brainly.com/question/19052579
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Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.
Explanation: Reaction to form alum from Aluminium is given as:
We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.
By stoichiometry,
2 moles of Al is producing 2 moles of Alum
Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol
Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol
54 g/mol of aluminium will produce 948 g/mol of alum, so
Amount of Alum produced = 17.34 grams
Theoretical yield of alum = 17.34 grams.