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WARRIOR [948]
3 years ago
12

Which contributes to the polarity of a water molecule

Chemistry
1 answer:
Arada [10]3 years ago
8 0

This deflection of the two hydrogen atoms to one side of the molecule is because of two love pairs of electrons on the other side of the oxygen atoms. In addition, due to the high electronegativity of an oxygen atom (since it has more protons) , it attracts most of the electron cloud of the molecule. The oxygen side of the water molecule is partially negative (negative dipole) while the hydrogen sides are partially positive (positive dipole).

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If an element is on the 4th row of the periodic table...how many energy levels/rings would you need to draw a Bohr 's model of t
lana [24]

Answer:

4 energy levels/rings

Explanation:

By the atomic model proposed by Niels Bohr, the electrons orbit around the nucleus in shells having different energy level each with the ability to contain a given maximum amount of electrons.

The closest shell to the nucleus named the K shell, can have a maximum of two electrons, the next shell known as the L shell can have 8 electrons while the next shell, M, can have a maximum of 18 electrons

The arrangement of the periodic table is based on the number and location of electrons such that elements in the 4th row (period 4 elements) have three completed energy levels/rings before, their valence ring making the total number of rings = 4 energy levels/rings.

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3 years ago
We say that salts will dissociate but acids will react with water. We say that acids will
Paha777 [63]

Answer:

ionize

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For example, in the chemical reaction;

HX + H20 -------> X- + H30+

HX is the acid because it loses its electron to water and forms the anion, X-, which is the conjugate base. Hence, it can be said that acid HX ionizes in water.

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A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s).

  • Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
  • Secondly, addition of liquid ammonia would form a precipitate with silver: AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq); Silver hydroxide at higher temperatures decomposes into black silver oxide: 2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g).
  • Thirdly, we also know we have Hg_2^{2+} in the mixture, since addition of potassium chromate produces a yellow precipitate: Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s). The latter precipitate is yellow.
3 0
4 years ago
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