Chemistry, 12.10.2019 09:30, cycy24

Which does not describe the relationship between the nucleus and the electron cloud of an atom?

a. the electron cloud takes up much more space than the nucleus.
b. he masses of the nucleus and electron cloud are balanced.
c. the electrons move much faster than the particles in the nucleus.
d. the charges of the nucleus and electron cloud are balanced.

Answers: 2

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answered: jalenrussell321

b. the masses of the nucleus and electron cloud are balanced.

Explanation:

1) From J.J. Thompson plum pudding model of the atom, and thereafter, it is know the the mass of the electrons are much, much smaller than the mass of an atom.

2) Also, from Ernest Rutherford gold foil experiment, it was determined that the atom consisted of a small and heavy positvely charged nucleus surrounded by electrons.

3) As per the scientific evidence, the current atom model (quantum model) we know that:

a. It is true that the electron cloud takes up much more space than the nucleus: the electrons are in regions called orbitals which are around the tiny nucleus.

b. It is not true that the masses of the nucleus and electron cloud are balanced: the relative masses of protons, neutrons and electrons are:

1 : 1 : 1 / 1840.

Which means that the nucleous contains almost all the mass of the atom.

c. It is true that the electrons move much faster than the particles in the nucleus: the particles in the nucleous stand in a very reduced space whose motion is only vibrational, while the electrons move rapidly i n the orbitals.

d. It is true that the charges of the nucleus and electron cloud are balanced: the atom is neutral because the positive charges of the nuclous equals the negative charges of the electrons.

answered: Guest

Ithink that’s it’s b

answered: Guest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.