Answer:
[NaCH₃COO] = 2.26M
Explanation:
17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)
Let's determine the volume of solution, by density
Mass of solution / Volume of solution = Solution density
100 g / Volume of solution = 1.09 g/mL
100 g / 1.09 g/mL = 91.7 mL
17 grams of solute is contained in 91.7 mL
Molarity (M) = Mol of solute /L of solution
91.7 mL / 1000 = 0.0917L
17 g / 82 g/m = 0.207 moles
Molariy = 0.207 moles / 0.0917L → 2.26M

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.

Hydroxide ion is a strong and would react with H+ to form water
OH-+H+---->H2O
I think that work is being done on the books because they are being moved to their proper location and they will be sorted properly rather than lying on a table. Without lifting or carrying, you could sort the books by their genre or title name on the bookshelf so it will be sorted much more efficiently.
I’m not sure if this is the answer you are looking for but I hope it helps :)