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Elza [17]
4 years ago
5

How many formula units are in 12.5 moles of calcium phosphate?

Chemistry
1 answer:
dedylja [7]4 years ago
7 0
A formula unit is to an ionic compound as an atom is to an element. That is, for calcium phosphate, one formula unit contains 3 atoms of calcium and 2 ions of phosphate PO4, as shown in the chemical formula Ca3(PO4)2. To determine the number of formula units given the moles, we multiply the number of moles by Avogadro's number:
12.5 moles*(6.02x10^23 formula units/mole) = 7.525 x 10^24 formula units
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With thermodynamics, one cannot determine ________. With thermodynamics, one cannot determine ________. the temperature at which
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6 0
4 years ago
90 grams of glucose, C6H12O6, is dissolved in 2 liters of solution. Determine the molarity (M).
Masja [62]

Answer:

0.25 M

Explanation:

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7 0
4 years ago
When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons a
qaws [65]

Answer:

The edge of the length is \mathbf{L = 8.54 \times 10^{-10} \ m}

Explanation:

From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)

here;

h = planck's constant = 6.626 \times 10^{-34} \ Js

n_i = the quantum no in a specified direction

m = mass (of particle)

L = length of the box

At the ground state n_x = n_y = n_z=1

The energy at the ground state can be calculated by using the formula:

E_1 =\dfrac{3h^2}{8mL^2}

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

∴

E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}

E_2 =\dfrac{(4+1+1)h^2}{8mL^2}

E_2 =\dfrac{(6)h^2}{8mL^2}

The transition energy needed to move from the ground to the excited state is:

\Delta E= E_2 - E_1

\Delta E= \dfrac{6h^2}{8mL^2}-  \dfrac{3h^2}{8mL^2}

\Delta E= \dfrac{3h^2}{8mL^2}} ----- (1)

Recall that:

the  wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0  × 10⁻⁷ m

However, the energy-related with the electronic transition is:

\Delta E =\dfrac{hc}{\lambda}

\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }

\Delta E =2.48 \times 10^{-19}  \ J

Replacing the value of \Delta E in (1); then:

2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}

Making the edge length L the subject of the formula; we have:

L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }

L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }

\mathbf{L = 8.54 \times 10^{-10} \ m}

Thus, the edge of the length is \mathbf{L = 8.54 \times 10^{-10} \ m}

5 0
3 years ago
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