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dusya [7]
2 years ago
10

NEED HELP!!!

Chemistry
1 answer:
Anit [1.1K]2 years ago
3 0

Answer:

energy is released in the reaction.

Explanation:

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_________is a process in which o2 is released as a by-product of oxidation-reduction reactions
Kisachek [45]

Answer : The combustion is a process in which oxygen is released as a by-product of oxidation-reduction reactions.

Explanation :

Combustion reaction : It is defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water.

The chemical equation of combustion reaction is:

CH_4+2O_2\rightarrow CO_2+2H_2O

The combustion reaction is also a redox reaction.

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

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The combustion reaction is also a redox reaction in which the carbon shows oxidation by the addition of oxygen or removal of hydrogen and oxygen shows reduction by the addition of hydrogen or removal of oxygen.

Hence, the combustion is a process in which oxygen is released as a by-product of oxidation-reduction reactions.

6 0
2 years ago
Consider the following incomplete reaction. Mg + 2Y ---> MgCl₂ + H₂ Choose the formula for the missing substance Y.
vesna_86 [32]
It would be B as the answer
7 0
2 years ago
Movement of the ___<br> creates the London dispersion forces.
Tanzania [10]

Answer: electrons

Explanation: moving electrons cause momentarily charge

Distribution on molecule. This distribution induces similar distribution to

Adjacent molecule.

7 0
2 years ago
A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

-1.2465 * 10^9 \frac{atoms}{cm^3.cm}

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

8 0
2 years ago
What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final
krok68 [10]

<u>Answer:</u>

P2 = 778.05 mm Hg = 1.02 atm

<u>Explanation:</u>

We are to find the final pressure (expressed in atm)  of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

where P1 = 724 mm hg

V1 = 3.05 L

T1 = 298 K

P2 = ?

V2 = 2.6 L

T2 = 173 K

Substituting the given values in the equation to get:

\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}

P2 = 778.05 mm Hg = 1.02 atm

7 0
2 years ago
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