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3 years ago

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]3 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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3 years ago

How many moles of copper are needed to react with sulfur to produce 0.25 moles Ol

Tems11 [23]

Answer:

0.50 moles of Cu

Explanation:

The balanced chemical equation for given synthetic reaction is,

2 Cu + S → Cu₂S

According to balance chemical equation,

1 mole of Cu₂S is produced by = 2 moles of Cu

So,

0.25 moles of Cu₂S will be produced by = X moles of Cu

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X = 0.25 mol × 2 mol / 1 mol

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2 years ago

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3 years ago

How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaC

sp2606 [1]

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

$3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2$

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

$PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L$

Best regards!

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2 years ago