Science 9- Bohr models worksheet

The azide ion, n−3, is a symmetrical ion, all of whose contributing structures have formal charges. draw three important contrib

stich3 [128]

Explanation:

Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.

The azide ion that is $N^{-}_3$ , is a symmetrical ion, all of whose contributing structures have formal charges.

Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.

Contributing structures of azide ion are drawn in the image attached.

5 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

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

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

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

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

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago

3. How can a community help implement Ecological Solid Waste Management.

Setler79 [48]

Answer:

It help cause we compost food scraps and other organic wastes. We also reuse and recycle materials to organize for government and industry to develop community recycling materials.

It depend on your opinion as a student.

Explanation:

Hope it helps

4 0

2 years ago

What is the ph of a solution with h+ = 7.0* 10

Mekhanik [1.2K]

Do you mean h=7.0+10? If so your answer is 70.

8 0

3 years ago

Explain specifically how an electron gives off light in an atom.

exis [7]

Answer:

Then, at some point, these higher energy electrons give up their "extra" energy in the form of a photon of light, and fall back down to their original energy level.

Explanation:

When properly stimulated, electrons in these materials move from a lower level of energy up to a higher level of energy and occupy a different orbital.

3 0

3 years ago