Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in oxidation-reduction reactions.

2 answers:

4 0

The process of oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom or ion and reduction is the gain of electrons or a decrease in oxidation state. The oxidation number is the total number of electrons that an atom gains or loses in order to form a chemical bond with another atom. Redox reactions are electron transfer processes and include all chemical reactions in which atoms have their oxidation state changed.

adell [148]4 years ago

4 0

Answer:Oxidation is the loss of electrons.

Reduction is the gain of electrons.

Oxidation increases the oxidation number.

Reduction decreases the oxidation number.

In an oxidation-reduction reaction, the electrons from the oxidized atom or ion are transferred to the atom or ion that is reduced.

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When n = 4, l = 2, and ml = -1, to what orbital type does this refer?

Leya [2.2K]

will tell you the subshell type, so you are dealing with a d orbital.

ml will tell you the orientation of the orbital, there is no distinction between the orbitals so -1 doesnt specifically mean anything.

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3 years ago

P4O10 -> 4P+5O2 How many moles of phosphorus would be produced if 5.3 mol of P4O10 reacted?

viktelen [127]

Answer:

21.2 moles.

Explanation:

Hello!

In this case, for the given chemical reaction, we can see there is a 1:4 mole ratio between tetraphosphorous decaoxide and phosphorous; therefore, the following proportional factor provides the requested moles of phodphorous:

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3 years ago

1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced

koban [17]

Answer:

$\large \boxed{\text{200 g CO}_{{2}}}$

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ: 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol: 1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol CO₂:1 mol C₃H₈

$\rm \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}$

2. Calculate the mass of CO₂.

$\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}$