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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 =
= 8.47 x 10⁻⁴
Hence,
[Ca²⁺] = 10<em>a</em> = 8.47 x 10⁻³ M
Answer:
The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.
Explanation:
The two poles North pole and South pole of the earth are distinct from the magnetic north pole and south pole of the earth. The magnetic north pole is the point from where the magnetic field of the Earth points downwards in the vertical direction. There is a gradual change in the magnetic poles with the passing of the time. The magnetic fields have an extension from the interior of the Earth to the outer space.
Answer:
It will not achieve the desired separation
Explanation:
Chromatography is a separation method that involves the use of a stationary phase and a mobile phase. The stationary phase is immobile, in the particular instance of this question, the stationary phase is paper. The mobile phase is the appropriate solvent, in this case, a salt-water solution.
If the level of solvent is above the dye spots, it will introduce error into the separation. The solvent (if volatile) may evaporate without drawing up and separating the solute. Secondly, the solvent may simply dissolve the spots without achieving any meaningful separation of the components in the system. This second reason is particularly why the salt solution must be below the dye spots in this chromatographic separation.
Qc < Kc, the reaction proceeds from left to right to reach equilibrium
<h3>Further explanation
</h3>
Given
K = 50.2 at 445°C
[H2] = [I2] = [HI] = 1.75 × 10⁻³ M At 445ºC
Reaction
H2(g) + I2(g) ⇔2HI(g)
Required
Qc
Solution
Qc for the reaction
Qc < Kc ⇒ reaction from left(reactants) to right (products) (the reaction will shift on the right) until it reaches equilibrium (Qc = Kc)