Suppose an atom gains an e-. What will be the overall electrical charge? What if an atom loses one e-?

2 answers:

7 0

Depending on what the original charge of the atom is, when an e- is added, that charge will drop by one. If an atom loses an e- the charge will be one higher.

Tatiana [17]3 years ago

7 0

Answer:

Atoms do not always contain the same number of electrons and protons, although this state is common. When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral. In contrast, when an atom loses or gains an electron (or the rarer case of losing or gaining a proton, which requires a nuclear reaction), the total charges add up to something other than zero. The atom is then said to be electrically charged, or "ionized". There is a major difference between the neutral state and the ionized state. In the neutral state, an atom has little electromagnetic attraction to other atoms. Note that the electric field of a neutral atom is weak, but is not exactly zero because the atom is not a point particle. If another atom gets close enough to the atom, they may begin to share electrons. Chemically, we say that the atoms have formed bonds.

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Zanzabum

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8 0

3 years ago

Read 2 more answers

Consider these compounds:

Afina-wow [57]

Answer:

Without doing any calculations it is possible to determine that silver chromate is more soluble than C,D , and silver chromate is less soluble than none .

It is not possible to determine whether silver chromate is more or less soluble than A,B by simply comparing Kspvalues

Explanation:

For this kind of question you need to obtain the balanced equation for the solubility, first, you will write the dissociation equation for every molecule, obtaining the ions in which it dissociates. Then you substitute in the Kps equation

$A_{n}B_{m} -> nA^{m+}+mB^{n-} \\K_{s}=[A^{m}]^{n}[B^{n}]^{m}$

Considering the corresponding stoichiometry you'll substitute in your dissociation equations as follow.

$Ag_{2}CrO_{4}(s)->2Ag^{+}(aq)+CrO_{4} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\A) PbF_{2}(s)->Pb^{2+} (aq)+2F^{-}\\s->s+2s\\s->[s][2s]^{2}=4s^{3}\\\\\\B)Ag_{2}SO_{3}(s)->2Ag^{+}(aq)+SO_{3} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\\\C)NiCO_{3}(s)->Ni^{2+}(aq)+CO_{3} ^{2-}\\ s ->s+s\\s->[s][s]=s^{2} \\\\\\D)AgCl(s)->Ag^{+}(aq)+Cl^{-}(aq)\\ s ->s+s\\s->[s][s]=s^{2} \\$

IIn this case, the silver chromate has a Kps of $4sx^{3}$ , same as the compound in options A and B, comparing these numbers you can't determine which one is bigger. Finally, options C and D have a kps of $s^{2}$ , this value is smaller than silver chromate's kps.