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Gennadij [26K]
3 years ago
15

Hydroxyapatite, Ca 10 ( PO 4 ) 6 ( OH ) 2 Ca10(PO4)6(OH)2 , has a solubility constant of Ksp = 2.34 × 10 − 59 2.34×10−59 , and d

issociates according to Ca 10 ( PO 4 ) 6 ( OH ) 2 ( s ) − ⇀ ↽ − 10 Ca 2 + ( aq ) + 6 PO 3 − 4 ( aq ) + 2 OH − ( aq ) Ca10(PO4)6(OH)2(s)↽−−⇀10Ca2+(aq)+6PO43−(aq)+2OH−(aq) Solid hydroxyapatite is dissolved in water to form a saturated solution. What is the concentration of Ca 2 + Ca2+ in this solution if [ OH − ] [OH−] is fixed at 2.50 × 10 − 6 M 2.50×10−6 M ?
Chemistry
1 answer:
Margarita [4]3 years ago
5 0

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

<em />

<em>1.315x10⁻³M = [Ca²⁺]</em>

<em />

I hope it helps!

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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
How many liters of volume is one mole of gas at standard temperature and<br> pressure?
horsena [70]

22.7 liters

The molar volume of an ideal gas depends on the temperature and pressure. One mole of any ideal gas occupies 22.7 liters at 0 0C and 1 bar (STP).

Hope this helped

3 0
3 years ago
Lactose, C12H22O11, is a naturally occurring sugar found in mammalian milk. A 0.335 M solution of lactose in water has a density
Dmitry_Shevchenko [17]

The molality of the solution is 0.00037 m.

<h3>What is concentration?</h3>

The term concentration refers to the amount of solute in a solution.

We have the following information;

Molarity = 0.335 M

Density =  1.0432 g/mL

Temperature = 20 o C

The molality of the solution is obtained from;

m = 0.335 M ×  1.0432 g/mL/ 1000(1.0432 g/mL) - 0.335 M (342 g/mol)

m = 0.344/1043.2 - 114.57

m =  0.344/928.63

m = 0.00037 m

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5 0
2 years ago
A sample of magnesium is burned in oxygen to form magnesium oxide. What mass of oxygen is consumed if 74.62 g magnesium oxide is
lana [24]
74.62 g of magnesium oxide is formed from 45.00 g magnesium so 74.62-45.00= 29.62 g of oxygen is consumed or in other words a new compound is formed in the burning of magnesium in oxygen with a heavier mass than the pure magnesium.
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A sample of (NH4)2SO4 contains 0.750 mole. what is the mass of the sample​
steposvetlana [31]
The answer is 99.10.
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