g While performing a neutralization reaction, Jonna added 21.73 mL of 0.142 M H2SO4 to 48.93 mL of 0.377 M KOH. How many moles of OH- are unreacted in the solution after the neutralization is complete?

0.01228 moles OH⁻

Explanation:

The reaction of H₂SO₄ with KOH is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Thus, we need to determine the moles of H2SO4 and KOH that reacts:

Moles H₂SO₄:

0.02173L * (0.142mol/L) = 3.086x10⁻³mol

Moles KOH:

0.04893L * (0.377mol/L) = 0.01845 mol KOH

The moles of KOH that reacts are:

3.086x10⁻³mol H₂SO₄ * (2 mol KOH / 1 mol H₂SO₄) = 6.172x10⁻³ moles KOH

And will remain in solution:

0.01845 mol KOH - 6.172x10⁻³ moles KOH = 0.01228 moles of KOH =

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.