Without doing any calculations, determine which of the following solutions would be most acidic.

a. 0.0100 M in HCl and 0.0100 M in KOH
b. 0.100 M in NaCN and 0.100 M in CaCl2
c. 0.0100 M in NH4Cl and 0.0100 M in CH3NH3Br
d. 0.0100 M in HF and 0.0100 M in KBr

In the first combination neutralization takes place to give a salt. So, solution 'a' is neutral in nature.

In the solution 'c', both salts are resulted by the combination of weak base and strong acid. The combination of these salts suppresses the acidity.

In last combination basic nature is observed due to the presence of CN⁻ ions. Thus, the solution 'd' is basic in nature.

Out of the five given solutions, 0.0100 M in HF and 0.0100 M in KBr is most acidic.  Therefore, solution 'b' is most acidic in nature.

Explanation:

your answer is c) plasma

~*"auba14"*~

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole