How do ionization energies provide evidence for the quantization of the energies of electrons?

damaskus [11]

A reducing agent is one which is oxidised in the reaction itself. When you take into account the oxidation numbers you will see that the Cl- ions are oxidised from an oxidation number of -1 to 0 in Cl2. Therefore Cl- ions are the reducing agent.

3 0

3 years ago

Muscles, bones, skin, and

ziro4ka [17]

Answer:

The answer is Protein.

8 0

3 years ago

What is the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form sol

rusak2 [61]

Answer:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

Explanation:

Hello!

In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

$Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)$

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

$3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)$

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

Best regards!

7 0

3 years ago

When a solution produces equal numbers of hydrogen and hydroxyl ions, it is said to be neutral. Select one: True False

ioda

Answer:

true

Explanation:

The hydroxyl group is a functional group formed by an oxygen atom and a hydrogen atom.

3 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