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3 years ago

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State the three rules that define how electrons are arranged in atoms

2 answers:

Dmitry_Shevchenko [17]3 years ago

8 0

3 rules that define how electrons<span> can be </span>arranged<span> in an </span>atom's<span> orbital. All orbitals related to an energy level are of equal energy. Single </span>electrons<span> with the same spin must occupy each equal-energy orbital before additional </span>electrons<span> with opposite spins can occupy the same orbitals.</span>

Hunter-Best [27]3 years ago

6 0

The three correct rules or principles are the Aufbau Principle, Hund’s Rule, and the Pauli Exclusion Principle. Aufbau’s Principle states that each electron orbits the lowest energy orbital. It also states that all orbitals related to an energy level are of equal energy. Hund’s Rule states that single electrons with the same spin must occupy each equal-energy orbital before additional electrons with opposite spins can occupy the same orbitals. And finally, the Pauli Exclusion Principle states that a maximum of two electrons may occupy a single orbital, but only if the electrons have opposite spins. Hope this helps.

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Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

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<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

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