Each element is____than___

A weighed piece of magnesium ribbon is added to a dried crucible, which is reweighed and heated in air to form the compound MgO.

olga nikolaevna [1]

Answer:

One of the errors for low percentage of magnesium could be because not all the magnesium may have reacted.

Explanation:

During the heating process, if the magnesium have not reacted completely, it can lead to low percentage of magnesium in the oxide formed. The product may still look a bit greyish rather than whitish after the heating process.

4 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

2 years ago

A student has two substances at a lab table. One substance is iron pyrite (fool's gold) and the other is real gold. After placin

blagie [28]

The answer is A) 1 mL

6 0

3 years ago

Read 2 more answers

Why is it difficult to accurately represent ammonia and methane on paper?

andrew-mc [135]

Answer:

The correct answer is because the molecular structure.

Explanation:

The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.

In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.

Have a nice day!

6 0

2 years ago

Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr

lions [1.4K]

Charles's Law i think

8 0

2 years ago

Read 2 more answers

Each element is____than____element.

Each element isthanelement.