
As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its
can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.
Let the three equations with
given be denoted as (1), (2), (3), and the last equation (4). Let
,
, and
be letters such that
. This relationship shall hold for all chemicals involved.
There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance,
shall resemble the number of
left on the product side when the second equation is directly added to the third. Similarly
Thus
and

Verify this conclusion against a fourth species involved-
for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

Electrons in an atom can be classified as core electrons and valence electrons. Valence electrons are those electrons which are present in valence shell and participates in bond formation. While, Core electrons are all remaining electrons which are not present in valence shell, hence not take part in bonding.
Atomic number of Selenium (Se) is 34 hence it has 34 electrons with following electronic configuration;
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
From electronic configuration it is found that the valence shell is 4, and the number of electrons present in valence shell are 6. So,
Core Electrons = Total Electrons - Valence Electrons
Core Electrons = 34 - 6
Core Electrons = 28
Result:
There are 28 core electrons in Selenium.
Explanation:
elements are based on electrical conductivity
There are 4 moles of spectator ions that remain in solution.
The equation of the reaction is;
Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)
We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.
For Na2CO3
1 mole of Na2CO3 yields 2 moles of NaNO3
3 moles of Na2CO3 yields 3 × 2/1 = 6 moles of NaNO3
For Pb(NO3)2
1 mole of Pb(NO3)2 yields 2 moles of NaNO3
2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3
We can see that Pb(NO3)2 is the limiting reactant.
Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.
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