Answer:
Option C (nuclear binding energy) is the appropriate choice.
Explanation:
- At either the nuclear scale, the nuclear binding energy seems to be the energy needed to remove and replace a structure of the atom itself into the characterize elements (to counteract the intense nuclear arsenal).
- Nuclear warheads (bargaining power) bind everything together neutrons as well as protons within an elementary particle.
Some other options in question aren't relevant to the particular instance. So that the option preceding will also be the right one.
Answer:
B.
Explanation:
One mole is the amount of substance that contain the Avogadro number which is equal to 6.022×10^23 atom, molecules or ions.
The mean kinetic energy per molecule is , where is the Boltzmann constant and T is the absolute temperature.
So at 1000°C, the T = 1273.15 K, kB=1.38 × 10-23, therefore the mean kinetic energy is 2.635 × 10⁻²⁰J.
<h3><u>
What is Kinetic energy ?</u></h3>
The energy an item has as a result of motion is known as kinetic energy.
A force must be applied to an item in order to accelerate it. We must put forth effort in order to apply a force. After the job is finished, energy is transferred to the item, which then moves at a new, constant speed. Kinetic energy is the type of energy that is transmitted and is dependent on the mass and speed attained.
Kinetic energy may be converted into other types of energy and transported between things. A flying squirrel may run into a chipmunk that is standing still, for instance. Some of the squirrel's original kinetic energy may have been transferred to the chipmunk or changed into another kind of energy after the impact.
To view more about kinetic energy, refer to;
brainly.com/question/2972267
#SPJ4
The temperature, pressure and volume of a gas are all related
Answer:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)
∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
∴ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
∴ Kc = Kc = 1 / PO2∧6
Explanation:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)
∴ O / Al: 0 → +2 ≡ 2e-
Na: +1 → +2
∴ R / H: +1 → 0
2 - Al - 2
2 - Na - 1
8 - O - 8
14 - H - 14
⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
1 - S - 1
4 - O - 4
2 - H - 2
⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
8 - P - 8
12 - O - 12
⇒ Kc = 1 / PO2∧6
