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KATRIN_1 [288]
3 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]3 years ago
7 0

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

☆ Dont forget to mark brainliest ☆

Dahasolnce [82]3 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.

You might be interested in
The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at
horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol  

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0
3 years ago
Select all that are true
kipiarov [429]
I think it’s B not quite sure ! Sorry
8 0
2 years ago
A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0
3 years ago
PLS HELP QUICK ALOTTT OF POINTS
timofeeve [1]

Answer:

\boxed {\boxed {\sf 0.80 \ mol\ F}}

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

Flip the ratio so the units of atoms of fluorine cancel each other out.

4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}

4.8 \times 10^{23}  *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }

Condense into 1 fraction.

\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F

Divide.

0.7970773829 \ mol \ F

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

0.80 \ mol \ F

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

6 0
3 years ago
Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac
GalinKa [24]
<h2>a) The rate at which NO_2 is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen O_2 is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt} = 0.066 M/s

Rate in terms of disappearance of O_2 = -\frac{1d[O_2]}{dt}

Rate in terms of appearance of NO_2= \frac{1d[NO_2]}{2dt}

1. The rate of formation of NO_2

-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}

\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s

2. The rate of disappearance of O_2

-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}

-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s

Learn more about rate law

brainly.com/question/13019661

https://brainly.in/question/1297322

7 0
3 years ago
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