Answer:
The concentration of [Ca²⁺] is 8.47 x 10⁻³ M
Explanation:
We consider the solubility of hydroxyapatite,
Ca₁₀(PO₄)₆(OH)₂ ⇔ 10Ca²⁺ + 6PO₄³⁻ + 2 OH⁻
Assumed that there is <em>a</em> mol of hydoxyapatite disolved in water, yielding <em>10a</em> mol Ca²⁺ of and <em>6a</em> mol of PO₄³⁻
We also have Ksp equation,
Ksp = [Ca²⁺]¹⁰ x [PO₄³⁻]⁶ x [OH⁻]² = 2.34 x 10⁻⁵⁹
⇔ 10a¹⁰ x 6a⁶ x (5.30 x 10⁻⁶)² = 2.24 x 10⁻⁵⁹
⇔ 60a¹⁶ = 2.24 x 10⁻⁵⁹ / 5.30 x 10⁻¹²
⇔ a¹⁶ = 0.007 x 10⁻⁴⁷ = 7 x 10⁻⁵⁰
⇔ a = ![\sqrt[16]{7 . 10^{-50} }](https://tex.z-dn.net/?f=%5Csqrt%5B16%5D%7B7%20.%2010%5E%7B-50%7D%20%7D)
= 8.47 x 10⁻⁴
Hence,
[Ca²⁺] = 10<em>a</em> = 8.47 x 10⁻³ M
Answer: heat
Explanation: if solid ice is heated it becomes in liquid and then to gaseous steam.
The question is incomplete, the complete question is;
The heat of vaporization ΔHv of acetic acid HCH3CO2 is 41.0 /kJmol. Calculate the change in entropy ΔS when 954.g of acetic acid condenses at 118.1°C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits.
Answer:
-1.67 JK-1
Explanation:
Since Heat of vaporization of acetic acid = 41.0 kJ/mol
Therefore:
Heat of condensation of acetic acid = -41.0 kJ/mol
Mass of acetic acid = 954 g
Temperature of condensation = 118.1 °C or 391.1 K
Number of moles of acetic acid = 954 g/60g/mol = 15.9 moles
Heat evolved during condensation = 15.9 moles * -41.0 kJ/mol = -651.9 KJ
Entropy change (ΔS) = Heat evolved/ Temperature = -651.9 KJ/391.1 K
Entropy change (ΔS) = -1.67 JK-1
Answer:
la esperanza ayuda
Explanation:
Cuando se utiliza el mol deben especificarse las entidades elementales de que se trata. ... ¿Cuántas partículas (moléculas) de oxígeno (O2) hay en una mol de dicho gas? ... 20 g de Fe; 20 g de S; 20 g de oxígeno; 20 g de Ca; 20 g de CaCO3.
7.38mol Fe(6.022x10^23 over 1 mol)= 44.4x10^23 but now you need to move the decimal because 44 isn’t between 1 & 9.
4.44x10^24
