A 25.00 mL sample of unknown concentration of HNO3 solution requires 22.62 mL of 0.02000 M NaOH to reach the equivalence point. What is the concentration of the unknown HNO3 solution

The concentration of the unknown HNO3 solution = 0.01809 M

Explanation:

For the acid-base reaction,  HNO3 + NaOH> NaN03 + H20

we have that

C1 V1 = C2 V2

Where ,

C1 = concentration of HNO3=?

V1 = volume of HNO3 = 25.00 mL,

V2 = volume of NaOH = 22.62 mL,  

C2 = concentration of NaOH = 0.02000 M

Therefore ,

25.00 mL x C1 = 22.62 mL x 0.02000 M    

 = (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M

The concentration of the unknown HNO3 solution = 0.01809 M

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o