Answer:
Momentum of first train car will reduce
Explanation:
When the moving care collides with the stationary car, it will increase the momentum of the stationary car. However, its own momentum will reduce.
It is so because the speed of the first train car will reduce after collision due to loss of energy in the collision while the stationary car may gain some momentum due to rise in velocity from zero (velocity at stationary position).
Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
There are only 2 atoms in an Oxygen molecule
Answer:
4
Explanation:
Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.
The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low
The given 2.6 µC of charge is due to a buildup of electrons, each of which has a charge of 1.6 x 10^-19 C. The 2.6 <span>µC is equivalent to 2.6 x 10^-6 C, so we can divide this by the individual charge of an electron:
</span>2.6 x 10^-6 C / 1.6 x 10^-19 (C/electron) = 1.625 x 10^13 electrons
