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:
a)
⇒
⇒
b)
⇒
⇒
Explanation:
A)
Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:
⇒
The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:
⇒
b)
Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:
⇒
Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:
⇒
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
Answer:
both will be at liquid state. the particles will move rapidly in all directions and will collide with other particles in random motion
Answer:
ΔHreaction = 263.15 kJ/mol
Explanation:
The reaction is as follow:
OH + CF₂Cl₂ → HOF + CFCl₂
You need to calculate the enthalpy of reaction and for this it is necessary to know the standard enthalpies for each of the compounds. These enthalpies are as follows and can be found in your textbook or on the Internet.
ΔHreaction = ∑ΔHproducts - ∑ΔHreactants
