Answer:
1.315x10⁻³M = [Ca²⁺]
Explanation:
Based in the reaction:
Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)
Solubility product, ksp, is defined as:
ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²
From 1 mole of hydroxyapatite are produced 10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:
6/10 Ca²⁺ = PO₄³⁻
Replacing in ksp formula:
ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²
As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:
2.34x10⁻⁵⁹ = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²
3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶
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<em>1.315x10⁻³M = [Ca²⁺]</em>
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I hope it helps!
Answer : Total molecules that will be needed to visualize a single egg will be 78500 molecules of dye.
Explanation : As a single egg cell has an approximately diameter of 100 μm.
We can use this formula to calculate area of the cell membrane;
A = π
;
We can take π as 3.14 and we get;
A = 3.14 X
Soving we get;
A = 7850 μ
Here we have to calculate the amount of dye molecules which will be needed for 10 fluorescent molecules / μ
but;
here 1 μ
= 7850 μ
dye molecules.
Therefore, 10 fluorescent molecules will need;
7850 X 10 = 78500 molecules of dye.
Therefore, the answer is 78500 molecules of dye.
Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.