Chemistry question. Image attached.

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

Explanation:

When lead (II) nitrate, Pb(NO3)2 undergoes a double displacement reaction with aqueous sodium chloride, NaCl, the following products are obtained:

Pb(NO3)2(aq) + NaCl(aq) —>

Aqueous Pb(NO3)2 will dissociate in solution as follow:

Pb(NO3)2(aq) —> Pb2+(aq) + 2NO3-(aq)

On the other hand, aqueous NaCl will dissociate as follow:

NaCl(aq) —> Na+(aq) + Cl-(aq)

The double displacement reaction will take place as follow:

Pb(NO3)2(aq) + NaCl(aq) —>

Pb2+(aq) + 2NO3-(aq) + Na+(aq) + Cl-(aq) —> Pb2+(aq) Cl-(aq) + Na+(aq) 2NO3-(aq)

Pb(NO3)2(aq) + NaCl(aq) —> PbCl2(s) + NaNO3(aq)

We simply balance the equation by putting 2 in front of NaCl and 2 in front of NaNO3 as shown below:

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

a) the particles of the chair are strongly held together.