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Fittoniya [83]
3 years ago
7

(c) ammonium carbonate + lead nitrate ? Step-by-step solution Step 1 of 3 v Step 2 of 3 v Step 3 of 3 ^ c) The ions formed in so

lution are NH4+, Co2, Pb2+ and NO3, combine and form soluble compound NH.NO, so the ions NH,* and NO3 remain in solution, whereas Pb2+ and CO2-form insoluble compound PbCO; Total ionic equation is: 2 NH,+ (aq) + C0,- (aq) + Pb* (aq) + 2 N0,- (aq) → 2 NH,+ (aq) + 2 NO,- (aq) + PbCo () Net ionic equation is: Pb2+ (aq) + CO2-(aq) → PbCo(s)
Chemistry
1 answer:
kenny6666 [7]3 years ago
3 0

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium carbonate and lead nitrate is given as:

(NH_4)_2CO_3(aq.)+Pb(NO_3)_2(aq.)\rightarrow PbCO_3(s)+2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^+(aq.)+CO_3^{2-}(aq.)+Pb^{2+}(aq.)+2NO_3^-(aq.)\rightarrow PbCO_3(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow PbCO_3(s)

Hence, the net ionic equation is written above.

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What is the fourth largest element on the periodic table?
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Explain the two parts of a scientific name?
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Suppose a laboratory wants to identify an unknown pure substance. The valence electrons of the substance's atoms feel an effecti
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Answer:

  • The answer is the third option in the list:<em> It would have smaller atomic radii than Si and higher ionization energies than Si.</em>

Explanation:

The<em> effective nuclear charge</em> is that portion of the total nuclear charge that a given electron in an atom feels.

Since, the inner electrons repel the outer electrons, t<em>he effective nuclear charg</em>e of a determined electron is the sum of the positive charge (number of protons or atomic number) that it feels from the nucleus less the number of electrons that are in the shells that are are closer to the nucleus than the own shell of such (determined) electron.

Mathematically, <em>the effective nuclear charge (Zeff)</em> is equal to the atomic number (Z) minus the amount (S) that other electrons in the atom shield the given (determined) atom from the nucleus.

  • Zeff = Z - S.

Since, the valence electrons are the electrons in the outermost shell of the atom, you can find certain trend for the value Zeff.

Let's look at the group to which Si belongs, which is the group 14. This table summarizes the relevant data:

Element   Z   Group   # valence electrons     S                      Zeff = Z - S

C              6      14                      4                     6 - 4 = 2             6 -  2 = +4

Si             14     14                      4                     14 - 4 = 10         14 - 10 = +4

Ge           32     14                     4                     32 - 4 = 28       32 -28 = +4

Sn           50     14                     4                     50 - 4 = 46       50 - 46 = +4

Pb           82     14                     4                     82 - 4 = 78        82 - 78 = +4  

With that, you have shown that the valence electrons of the unknown substance's atoms feel an effective nuclear charge of +4 and you have a short list of 4 elements which can be the unknown element: C, Ge, Sn or Pb.

The second known characteristic of the unknown substance's atoms is that it has a <em>higher electronegativity than silicon (Si)</em><em>.</em>

So, you must use the known trend of the electronegativity in a group of the periodic table: the electronegativity decreases as you go down in a group. So, three of the elements (Ge, Sn, and Pb) have lower electronegativity than Si, which has left us with only one possibility: the element C. The valence electrons of carbon (C) atoms feel an effective nuclear charge of +4 and it carbon has a higher electronegativity than silicon.

Other two periodic trends attending the group number are the <em>atomic radii and the ionization energy</em>.

The atomic radii generally increases as you go from top to bottom in a group. This is because you are adding electrons to new higher main energy levels. So, you can conclude that the originally unknwon substance (carbon) has a smaller atomic radii, than Si.

The ionization energies generally decreases as you go from top to bottom in a group. This os due to the shielding effect: as seen, the effective nuclear charge of the atom's valence electrons remains constant, while the distance of the electrons from the nucleus increases (the valence electrons are farther away from the nucleus), which means the upper the element in a given group, the larger the ionization energy of the atoms.

With this, our conclusions about the unnkown substance are:

  • Since it has a higher electronegativity value than silicon (Si), it is right up of Si, and there is on only element possible element than can be (C).

  • Since, it is upper than silicon (Si), it would have smaller atomic radii.

  • Due to the shielding effect, it would have larger ionization energies.

  • The answer is the third option in the list: It would have smaller atomic radii than Si and higher ionization energies than Si.

6 0
2 years ago
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