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3 years ago

Describe an oxidation-reduction reaction.

Chemistry

1 answer:

DaniilM [7]3 years ago

3 0

Answer: Oxidation reaction occurs when a reactant losses electrons in the reaction.

Reduction reaction occurs when a reactant gains electrons in the reaction.

Explanation: Reduction reaction : It is defined as the reaction in which a substance gains electrons. That means, the gain of electrons takes place.

Or we can say that, reduction reaction occurs when a reactant gains electrons in the reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

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These statements describe three different reactions.

Naily [24]

An atom gains an electron from another atom. Hence, option B is the correct answer.

An atom is a particle of matter that uniquely defines a chemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons.

When an atom shares electrons with another atom then it results in the formation of a covalent bond.

Whereas when an atom transfer electrons from one atom to another then it results in the formation of an ionic bond.

When the nucleus of an atom splits then it represents a nuclear fission reaction and energy is released during this process.

Hence, option B is the correct answer.

Learn more about the atom here:

brainly.com/question/1566330

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5 0

2 years ago

Which of the following is a statement of Hess's law?

attashe74 [19]

Answer:

Explanation:

the enthalpy of reaction is independent of the reaction path

7 0

3 years ago

A husband and a wife have two sons one boy has a straight thumb while the other has a bent thumb what causes the difference

Usimov [2.4K]

Answer:

Double/Triple joints

Explanation:

Double/Triple joints is when you can move a body part (thumb) farther than normal. (I think)

3 0

3 years ago

Chemistry: Gas Laws Problems!

Ronch [10]

Answer : The final volume at STP is, 1000 L

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 1520 mmHg = 2 atm (1 atm = 760 mmHg)

$P_2$ = final pressure at STP = 1 atm

$V_1$ = initial volume = 500.0 L

$V_2$ = final volume at STP = ?

Now put all the given values in the above formula, we get:

$2atm\times 500.0L=1atm\times V_2$

$V_2=1000L$

Therefore, the final volume at STP is, 1000 L

7 0

3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

5 0

3 years ago