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KATRIN_1 [288]
4 years ago
8

Which identifies an oxidation-reduction reaction? a double replacement reaction a neutralization reaction a reaction in which ox

idation numbers change a reaction in which no electrons are transferred
Chemistry
2 answers:
OLga [1]4 years ago
7 0

A reaction in which oxidation numbers change is the answer! :D

☆ Dont forget to mark brainliest ☆

Dahasolnce [82]4 years ago
5 0

Answer: Option (c) is the correct answer.

Explanation:

In an oxidation reaction there is loss of electrons whereas in a reduction reaction there is gain of electrons.

So, overall there is exchange of electrons in an oxidation-reduction reaction.

For example, CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O is an oxidation-reduction reaction.

As there is change in oxidation number of oxygen from 0 to -2 and change in oxidation number of carbon from -4 to +4.

Thus, we can conclude that a reaction in which oxidation numbers change identifies an oxidation-reduction reaction.

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The half-life of radium-226 is 1,600 years. It decays into radon-222. What fraction of the original amount of radium-226 in a sa
Mnenie [13.5K]

Answer:

Your answer will be C. 1/32

6 0
3 years ago
Read 2 more answers
Balance the following chemical equation Na + Cl2 -> NaCl and explain how the balanced equation models the law of conservation
Alexus [3.1K]

Answer:

2 Na + 1 Cl2 -> 2 NaCl

Explanation:

The answer is really simple, because if you have 1 nonmetal element that has a subscript of 2, you need to multiply the product and the first reactant by 2 to balance it.

7 0
4 years ago
The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
4 years ago
ILL MARK BRAINIEST WHOEVER ANSWERS CORRECTLY
natulia [17]

Answer:

I think its A

sorry if im wrong

Explanation:

4 0
3 years ago
What property of matter does not change even though the appearance may change?
IceJOKER [234]

Answer:

The law of the conservation of mass states that matter is neither created nor destroyed, only converted to other forms. Therefore, the mass never changes, even if its appearance does.

Explanation:

4 0
3 years ago
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