1. What volume in milliliters of 0.100 M HClO₃ is required to neutralize 40.0 mL of 0.140 M KOH? 2. A 25.0 mL solution of HNO₃ is neutralized with 15.7 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?

The correct answer is 1) 56 ml and 2) 0.314 M

Explanation:

1. The reaction taking place in the given case is,  

HClO₃ + KOH ⇒ KClO₃ + H2O, the molarity of HClO₃ given is 0.100 M, the molarity of KOH given is 0.140 M and the volume of KOH given is 40 ml, there is a need to find the volume of HClO₃.  

Therefore, the mole of HClO₃ = mole of KOH

= MHClO₃ × VHClO₃ = MKOH × VKOH

= 0.100 M × VHClO₃ = 0.140 M × 40 ml

VHClO₃ = 0.140 M × 40 ml/0.100 M

VHClO₃ = 56 ml.  

2. The reaction taking place is,  

2HNO₃ + Ba(OH)₂ ⇒ Ba(NO₃)₂ + 2H₂O

The volume of HNO₃ given is 25 ml, the molarity of Ba(OH)2 is 0.250 M, the volume of Ba(OH)2 is 15.7 ml, the n or the number of moles of HNO₃ is 2, and the n of Ba(OH)2 is 1, the concentration or M of HNO₃ is,  

M₁V₁/n₁ = M₂V₂/n₂

M₁ × 25/ 2 = 0.25 × 15.7/1

M₁ or molarity of HNO₃ = 0.314 M

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

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.