Calculate the volume of concentrated HCl (12 M) needed to convert the sodium benzoate back to benzoic acid.

The question is incomplete, the complete question is;

Calculate the volume of concentrated HCL (12 M) needed to convert 1.5 g of sodium benzoate back to benzoic acid.

8.33 * 10^-4 L

Explanation:

Equation of the reaction;

C6H5COONa + HCl >>>> C6H5COOH + NaCl

Number of moles of  sodium benzoate = mass/molar mass

molar mass of sodium benzoate =144.11 g/mol

mass of sodium benzoate = 1.5 g

Number of moles of  sodium benzoate = 1.5g/144.11 g/mol

Number of moles of  sodium benzoate = 0.01 moles

Since the reaction is 1:1,  0.01 moles of HCl reacted.

but;

n = CV

n = number of moles

C = concentration

V = volume

V = n/C

V =  0.01 moles/12 M

V = 8.33 * 10^-4 L

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.