Examples of coulombs in the following topics:
-
- From the perspective of the voltage source and circuit outside the electrodes, the flow of electrons is generally described in terms of electrical current using the SI units of coulombs and amperes.
- It takes 96,485 coulombs to constitute a mole of electrons, a unit known as the faraday (F).
- current (in amperes) is the rate of charge transport: 1 amp = 1 $\frac {Coulombs}{second}$.
- $1.5\ hours \times \frac {3600\ seconds}{1\ hour} \times \frac {.22\ Coulombs}{second} \times \frac {1\ mole\ e^-}{96485\ Coulombs} \times \frac {1\ mole\ Cu^{2+}}{2\ mole\ e^-} \times \frac {63.54\ grams\ Cu}{1\ mole\ Cu} =$
- $0.39\ g\ Cu \times \frac {1\ mole\ Cu}{63.54\ g\ Cu} \times \frac{2\ moles\ e^-}{1\ mole\ Cu^{2+}} \times \frac {96485\ Coulombs}{1\ mole\ e^-} = 1184\ Coulombs$
-
- Here, n is the number of moles of electrons and F is the Faraday constant (96,485$\frac {Coulombs}{mole}$).
- One volt is $1\frac {Joule}{Coulomb}$.
- $\Delta G^o = -2 \ moles\ e^- \times 96485\frac {Coulombs}{mole} \times 0.12 \frac {Joules}{Coulomb}$
-
- A hydrogen atom is electrically neutral, containing a single proton and a single electron bound to the nucleus by the Coulomb force.
- The solution of the Schrödinger equation (wave equations) for the hydrogen atom uses the fact that the Coulomb potential produced by the nucleus is isotropic (it is radially symmetric in space and only depends on the distance to the nucleus).
- It is only here that the details of the 1/r Coulomb potential enter (leading to Laguerre polynomials in r).
-
- Since the coulombic forces that bind ions and highly polar molecules into solids are quite strong, we might expect these solids to be insoluble in most solvents.
- The electrically-charged ions undergo ion-dipole interactions with water to overcome strong coulombic attraction, and this produces an aqueous solution.
- The greater the value of a compound's lattice energy, the greater the force required to overcome coulombic attraction.
-
- There are two kinds of attractive forces shown in this model: Coulomb forces (the attraction between ions) and Van der Waals forces (an additional attractive force between all atoms).
-
- Here, n is the number of moles of electrons, F is the Faraday constant ($\frac {Coulombs}{mole}$), and ΔE is the cell potential.
-
- In this problem, 2 moles of electrons are being transferred, F is 96485 $\frac {Coulombs}{mole}$, R is 8.31$\frac {Joules}{K \times mole}$, T is 298 K, and Eo is 1.10$\frac {Joules}{Coulomb}$.
-
- The origin of the energy released in fusion of light elements is due to an interplay of two opposing forces: the nuclear force that draws together protons and neutrons, and the Coulomb force that causes protons to repel each other.
- The nuclear force is stronger than the Coulomb force for atomic nuclei smaller than iron, so building up these nuclei from lighter nuclei by fusion releases the extra energy from the net attraction of these particles.
-
- The rationale for this peak in binding energy is the interplay between the coulombic repulsion of the protons in the nucleus, because like charges repel each other, and the strong nuclear force, or strong force.
- As the size of the nucleus increases, the strong nuclear force is only felt between nucleons that are close together, while the coulombic repulsion continues to be felt throughout the nucleus; this leads to instability and hence the radioactivity and fissile nature of the heavier elements.
-
- The solution of the Schrödinger equation for the hydrogen atom uses the fact that the Coulomb potential produced by the nucleus is isotropic—it is radially symmetric in space and only depends on the distance to the nucleus.
- It is only here that the details of the 1/r Coulomb potential enter (leading to Laguerre polynomials in r).