Put the following experimental steps for the Acid-Base Equilibria (HCL and NaOH) in the correct order: a. Perform a scout colorimetric titration. of diluted vinegar with NaOH.
b. One pair performs a careful colorimetric titration of diluted vinegar with NaOH and the other pair performs a potentiometic titration of diluted vinegar with NaOH.
c. Perform a scout colorimetric titration of HCl with NaOH.
d. One pair performs a careful colorimetric titration of HCl with NaOH and the other pair performs a potentiometic titration of HCl with NaOH.
e. Dilute a solution of commercial vinegar to prepare it for titration.

e. Dilute a solution of commercial vinegar to prepare it for titration.

a. Perform a scout colorimetric titration of diluted vinegar with NaOH.

b. One pair performs a careful colorimetric titration of diluted vinegar with NaOH and the other pair performs a potentiometric titration of diluted vinegar with NaOH.

c. Perform a scout colorimetric titration of HCL with NaOH.

d. One pair performs a careful colorimetric titration of HCL with NaOH and the other pair performs a potentiometric titration of HCL with NaOH.

Explanation:

Titration is a process in which acid and base are introduced together to determine unknown concentration of acid or base. This is also known as neutralization reaction. The chemical process form salt and water as by products because acid molecules react with base molecules.

the energy to form plasma comes from heat or electricity in the atmosphere!

sunny~ ☺

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.