Compute the theoretical density of zns given that the zn-s distance and bond angle are 0. 234 nm and 109. 5o, respectively? The atomic weights of zinc and sulfur are 65. 41 g/mol and 32. 06 g/mol.

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Answer 1

Density of ZnS given that the zn-s distance and bond angle are 0. 234 nm and 109. 5o, respectively if the atomic weights of zinc and sulfur are 65. 41 g/mol and 32. 06 g/mol. is 4.1109 g/cm³

For density In the crystal structure, we determine the angle's value for one set of ZnS bonds. θ θ + ∅ + 90° = 180° θ = 90° - ∅ θ = 90° - (109.5° / 2) After determining that = 35.25°, we derive the value of x from the geometry: distance angle d = 0.234 x = dsin = 0.234 sin35.25°) = 0.135 nm = 0.135 107 cm; following this, we get the length of the unit cell: a = 4x a = 4(0.135) a = 0.54 nm = 0.54 107 cm; and The formula for n' is (number of corner atoms in unit ell contribution of each corner atom in unit cell) + (number of face center atoms in unit cell contribution of each face center atom in unit cell). n' = 8 × 1/8) + (6 × 1/2) = 1 + 3 = 4

Density = P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)] = 389.88 / 94.84 = 4.1109 g/cm³

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Related Questions

Suppose that 25. 0 ml of 0. 10 m ch3cooh(aq) ( ka =1. 8 x 10-5) is titrated with 0. 10 m naoh(aq). (a)what is the initial ph of the 0. 10 m ch3cooh (aq) solution?

Answers

The initial ph of the 0. 10M ch3cooh  (aq) solution is  -0.64.

The salt sodium acetate is present at the equivalency point. It is a salt of a strong base and a weak acid. The expression "its pH" provides the information.

pH = 7 + 0.5p + 0.5logC

where C is the salt concentration. Monobasic acid is Acetic acid here.

25 ml of 0.10 M NaOH solution are needed for the titration of 25 ml of 0.10 M acetic acid.

Total volume = 25 + 25 = 50ml

The salt concentration = C = 0.10×  = 0.05

The pH for weak acid is given by,

pH = [tex]\frac{1}{2}[/tex] (p[tex]K_{a}[/tex] + logC)

pH =  [tex]\frac{1}{2}[/tex] (1.8× + log0.05)

pH =  [tex]\frac{1}{2}[/tex] [1.8× + (-1.30)]

pH =  [tex]\frac{1}{2}[/tex] (1.8× -1.30)

pH = -0.64

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The molar solubility of cui is 2. 26 × 10-6 m in pure water. Calculate the ksp for cui.

Answers

Ksp(the solubility product constant) = [Cu⁺] [I⁻]

So, the Ksp for Cui would be:

Ksp = (2.26 × 10⁻⁶) (2.26 × 10⁻⁶) = 5.11 x 10⁻¹²

Formula used:

K = [tex]K_{sp} = [A^+]^a [B^-]^b[/tex], where

Ksp = solubility product constant

A⁺    = cation in an aquious solution

B⁻ = anion in an aqueous solution

a, b = relative concentrations of a and b

Definition

The equilibrium constant for a solid material dissolving in an aqueous solution is the SOLUBILITY PRODUCT CONSTANT, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.

Take into account the general dissolving response (in aqueous solutions) below:

                           aA(s)⇔cC(aq)+dD(aq)

The molarities or concentrations of the products (cC and dD) must be multiplied in order to find the Ksp. Any product that has a coefficient in front of it must be raised to the power of that coefficient (and also multiply the concentration by that coefficient).

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