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Reika [66]
3 years ago
6

An atom has the same number of wich two particles?

Chemistry
1 answer:
Flura [38]3 years ago
6 0
Protons and electrons
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Need help asap with this chemistry if someone could help me
Burka [1]

Answer:

<h3>1)</h3>

Structure One:

  • N: -2
  • C: 0
  • O: +1

Structure Two:

  • N: 0
  • C: 0
  • O: -1

Structure Three:

  • N: -1
  • C: 0
  • O: 0.

Structure Number Two would likely be the most stable structure.

<h3>2)</h3>
  • All five C atoms: 0
  • All six H atoms to C: 0
  • N atom: +1.

The N atom is the one that is "likely" to be attracted to an anion. See explanation.

Explanation:

When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:

\text{Formal Charge} \\ = \text{Number of Valence Electrons in Element} \\ \phantom{=}-\text{Number of Chemical Bonds} \\\phantom{=} - \text{Number of nonbonding Lone Pair Electrons}.

For example, for the N atom in structure one of the first question,

  • N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
  • This N atom is connected to only 1 chemical bond.
  • There are three pairs, or 6 electrons that aren't in a chemical bond.

The formal charge of this N atom will be 5 - 1 - 6 = -2.

Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.

<h3>1)</h3>

Structure One:

  • N: -2
  • C: 0
  • O: +1

Structure Two:

  • N: 0
  • C: 0
  • O: -1

Structure Three:

  • N: -1
  • C: 0
  • O: 0.

In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.

Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.

<h3>2)</h3>

Similarly,

  • All five C atoms: 0
  • All six H atoms to C: 0
  • N atom: +1.

Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.

Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.

5 0
3 years ago
I need help on balacing equations
nevsk [136]

Answer: sorry hun can’t help with this I was asked different questions when I learned this :,)

Explanation: have a nice day

8 0
2 years ago
Do you think a colony on Mars should be nearer the northern or southern<br> polar cap? Explain.
Ksju [112]

Answer:

nearer to the north pole

Explanation:

because there is more water, which is basically gone everywhere else on Mars except for the south pole which has less water than the north pole

7 0
3 years ago
All organic compounds contain ______, hydrogen, and usually oxygen. nitrogen. phosphorus. sulfur carbon.
Murrr4er [49]
Carbon is the answer. all hydrocarbures have to contain carbon.
4 0
3 years ago
CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above.
kupik [55]

Answer:

The correct answer is option a.

Explanation:

CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)

Equilibrium concentration cadmium ions = [Cd^{2+}]=0.0585 M

Equilibrium concentration fluoride ions = [F^{-}]=0.117 M

Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.

The molar solubility of the solid cadmium fluoride = 0.0585 M

CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)..[1]

NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.

8 0
3 years ago
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