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4 years ago

What is the molarity of a solution that is made by adding 57.3 g of MgO to 500.0 mL of solution?

Chemistry

1 answer:

kifflom [539]4 years ago

0 0

Answer:

2.843 M

Explanation:

Molarity = moles / volume

Milliliters to liters:

500 mL = .500 L

Grams to moles:

MgO molar mass = 16.00 + 24.31 = 40.31 g/mol

57.3 g x 1 mol / 40.31 g = 1.421 mol

Molarity:

1.421 mol / .500 L = 2.843 M

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Answer:

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

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

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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.

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

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

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

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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:

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

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

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