When 0.300 g of a diprotic acid was titrated with 0.100 m lioh, 40.0 ml of the lioh solution was needed to reach the second equivalence point. identify the formula of the diprotic acid. h2seh2teh2c2o4h2c4h4o6h2s

H₂C₄H₄O₆

Explanation:

The clue of this question is to find the molar mass of the diprotic acid and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.

The procedure is:

Find the number of moles of the base: LiOHUse stoichionetry to infere the number of moles of the acid.Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.Compare and conclude.

Solution:

1. Number of moles of the base, LiOH:

M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.

2. Stoichiometry:

Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.

Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.

3. Molar mass of the acid:

molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol

4. Molar mass of the possible diprotic acids:

a. H₂Se: 2×1.008 g/mol + 78.96  g/mol = 80.976 g/mol

b. H₂Te: 2×1.008 g/mol + 127.6  g/mol = 129.616 g/mol

c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol

d. H₂C₄H₄O₆ = 6×1.008 g/mol +  4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.

Conclusion: since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic 

16- counting by fours