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1 year ago

Give the systematic names for the formulas or give the formulas for the names:

Chemistry

1 answer:

lilavasa [31]1 year ago

4 0

Lead (II) acetate trihydrate is the systematic name for the formula Pb(C₂H₃O₂)₂ . 3H₂O.

<h3>What is Molecular Formula ?</h3>

The chemical formula that gives total number of atoms of each element in one molecule of a compound is called Molecular Formula.

<h3>What is Oxidation State ?</h3>

Oxidation state is also known as oxidation number. It is defined as the atom is equal to the total number of electrons which have been removed from the element in order to form chemical bond with other atom.

Now find the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O

Assume the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O be x

x + 2 × (-1) + 3 × 0 = 0

x - 2 + 0 = 0

x = 2

Oxidation state of Pb is +2 or (II)

Thus from the above conclusion we can say that The systematic name for the formula Pb(C₂H₃O₂)₂ .3H₂O is Lead (II) acetate trihydrate.

Learn more about the Molecular Formula here: brainly.com/question/15960587

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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]

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

<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)

<u>For Iron:</u>

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 %

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