What volume would 8.01×1022 molecules of an ideal gas occupy at STP?

Answer:

Employer

coworkers

Safety data sheet (SDS)

Training material

BLogs

Explanation:

The resources which can be used inside of the workplaces is as follows

a. Employer: The employer is the person who is working in an organization plus for every employee the health and safety matters first

b. Coworkers: The coworkers are the person who is at similar level or for the same role plus for every coworker the health and safety matters first

c. Safety data sheet: In this sheet, it maintains the data regarding health and safety

d. Training material: This one is also used inside the workplace

e. Blogs: The blog is an article in which the health and related data could be written and provided

Therefore these all the resources are used inside the organization

Answer:

Employer

Co-workers

SDSs

Training materials

Explanation:

Answer:

Metallic bonding

Explanation:

Ionic bonding involves the transfer of electrons from a highly electropositive metal to a highly electronegative nonmetal.

The metallic bond is somewhat similar to the ionic bond since there are also charged positive metal ions. The only difference is that there isn't any electronegative element that accepts the electrons.

In a metallic bond, the positively charged metal ions are bound together by a sea of mobile electrons. The attractive force between the metal ions and the mobile electrons hold the metallic crystal lattice together.

Answer:

1. C- Three.

2. A- Methionine

3. D- Translocation.

4. C- OH.

5. A - 5'

6. A - 3' carbon

7. A. adenine and guanine

Explanation:

1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.

2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.

3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.

4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).

5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.

6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.

7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).

Answer: An ion of a single pure element always has an oxidation number of

zero.

Explanation:

The question his incomplete, the complete question is;

Which of the following required Bohr's model of the atom to need modification

A. energies of electrons are quantized

b. electrons do not follow circular orbits around the nucleus

c. Quantized electrons energies are responsible for emission spectra lines

d. an electrons energy increases the farther it moves from the nucleus

Answer:

electrons do not follow circular orbits around the nucleus

Explanation:

Neils Bohr's model of the atom closely follows the planetary model of Rutherford, hence is it sometimes referred to as the Bohr-Rutherford's model of the atom. Its main points are that;

Electrons are found in circular orbits

Electronic transition within the atom leads to spectral lines

The energy of an orbit is related to its size.

One major refinement of the Bohr's model of the atom is the Bohr-sommerfield model. Sommerfeld added that electrons are found in elliptical rather than circular orbits. This Bohr-sommerfield model was able to explain atomic spectral effects, such as the Stark effect in spectral line splitting.

Hence, the fact that electrons are found in elliptical rather than circular orbits was a major failure of Bohr's original proposition.

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

Answer:

The correct answer is - Al(OH)3

Explanation:

At the point when a substance is blended in with a soluble, there are a few potential outcomes. The deciding variable for the outcome is the solubility of the substance, which is characterized as the maximum concentration of the solute. These rules help figure out which substances are solvent, and how much.

According to the 11 rules of solubility rules, the insoluble compound in water is - Al(OH)3

Answer:

Na2So4

Explanation:

If you consult a table of solubility rules, like the one below, you will see that sodium sulfate (Na2SO4) is soluble in water.

Answer:

[tex]\Large \boxed{\mathrm{C_3 H_8 \ (g)+5O_2 \ (g) \Rightarrow 3CO_2 \ (g)+4 H_2 O \ (l)}}[/tex]

[tex]\rule[225]{225}{2}[/tex]

Explanation:

[tex]\sf C_3 H_8 +O_2 \Rightarrow CO_2 + H_2 O[/tex]

Balancing Carbon atoms on the right side,

[tex]\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 + H_2 O[/tex]

Balancing Hydrogen atoms on the right side,

[tex]\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 +4 H_2 O[/tex]

Balancing Oxygen atoms on the left side,

[tex]\sf C_3 H_8 +5O_2 \Rightarrow 3CO_2 +4 H_2 O[/tex]

[tex]\rule[225]{225}{2}[/tex]

The combustion reaction of propane is :C₃H₈ [tex]_(g)[/tex]+5 O₂ [tex]_(g)[/tex][tex]\rightarrow[/tex]3 CO₂[tex]_(g)[/tex] + 4 H₂O[tex]_(l)[/tex].

Combustion is defined as a high temperature exothermic reaction  between a fuel which acts as a reductant  and an oxidant which is usually atmospheric oxygen.It does not result in fire as the flame is visible only when substance which undergoes combustion vaporizes.

Activation energy must be overcome so that combustion is initiated.Solid fuels such as wood and coal initially undergo endothermal pyrolysis  which results in gaseous fuels. It is widely used in industry  and the application is based on concepts of chemistry, physics and mechanics.

It is a complex chemical process involving many steps which depend on properties of combustible material. It is a type of redox reaction.

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Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

Answer:

C. Potential energy

Explanation:

Kinetic energy and gravitational potential energy are both forms of potential energy. Potential energy is stored energy, when an object is not in motion it has stored energy. When an object is an motion it has kinetic energy. An object posses gravitational potential energy when it is above or below the zero height.

Answer:

0.583M NaCl

Explanation:

Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

In the puddle, the solute is sodium chloride that is dissolved in water and you have 0.449 moles of NaCl per liter of water

When the solution begins to evaporate, amount of water decreases whereas moles of NaCl remain constant.

As 23% of the water evaporates, amount of water that remains is 100-23 = 77%, that means now you have 0.449 moles of NaCl per 77% of a liter, 0.770L. The new concentration is:

0.449 moles NaCl / 0.770L =

Answer:

Fermentation

Explanation:

Fermentation is the general term used to describe the process by which sugars such as glucose, starch or cellulose are converted to ethanol and carbon (iv) oxide. It is anaerobic process meaning that it occurs in the absence of air or in very low oxygen concentrations.

Yeast and other microorganisms ferment glucose into ethanol and carbon (iv) oxide with the help of the enzyme zymase. Polysaccharides such as starch and cellulose are first broken down into glucose by enzymes such as diastases, maltase and cellulase, before it is then converted into ethanol and carbon (iv) oxide.

The equation for the conversion of glucose to ethanol and carbon (iv) oxide is as follows:

C₆H₁₂O₆(aq) -----> 2C₂H₅OH(aq) + 2CO₂(g)

Answer:

The answer is C love.

Explanation:

Answer:

a) It is when a solid is going into the liquid stage. The molecules are really close together.

b) The water could have impurities, altitude could effect the boiling point or the temperature could of been measured wrong.

c) The gas was colorless and odorless therefore it is Hydrogen gas.

d) In a Scientific setting finding the boiling point of a new unknown substances can help identify and organize them. In a commercial setting chemicals need to be stored a certain way according to their boiling point. Stored incorrectly could cause the substance to evaporate.

Answer:

Explanation:

1 molecule contains 1 carbon atom.

9.837 * 10^24 molecules contains 9.837 * 10^24  atom of carbon.

It's a 1 to 1 ratio.

Answer:

E) Al³⁺

Explanation:

A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.

In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;

A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.

K + 2e⁻  ----> K⁻²

B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.

Fe⁺ + 2e⁻  ----> Fe

C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.

O²⁻ + 2e⁻  ---->  O⁴⁻

D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.

Ne + 2e⁻   ---->   Ne²⁻

E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.

Al³⁺ + 2e⁻   ---->   Al⁺

Answer:

C. Grinding the solute down into tiny particles.

Explanation:

The dissolution of a solute has something to do with particle size. The size of solute particles usually determines how quickly a solute dissolves in a solvent. When large solute particles are introduced into the solvent, the large solute particles do not easily interact with solvent particles hence preventing easy dissolution in the solvent.

However, when the solute is ground into tiny particles, smaller solute particles interact more effectively with solvent particles hence dissolution is faster.

Therefore, tiny solute particles will dissolve faster in a solvent than a lump of solute. Summarily, small particle size enhances dissolution of a solute in the appropriate solvent.

Answer: stirring the solution vigorously

Grinding the solute down into tiny particles

gently heating the solution

Explanation:

A dissolution will proceed more readily when heated . Breaking up the solute as much as possible will aid in overcoming the solute-solute interaction, as will stirring the solution

Answer:

The answer is option A.

I hope this helps you.

Answer:

Macro Algae

Explanation:

probz

Answer:

Explanation:

a. change of colour:

A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products. The products have different molecular structures than the reactants. Different atoms and molecules radiate different colours of light. Hence, there usually is a change in colour during a chemical reaction.

Eg: copper reactions with the elements

b. Evolution of gas:

A gas evolution reaction is a chemical reaction in which one of the end products is a gas such as oxygen or carbon dioxide.

Eg: ammonium hydroxide breaks down to water and ammonia gas.

c. Change of smell :

Production of an Odor Some chemical changes produce new smells.  ... The formation of gas bubbles is another indicator that a chemical change may have occured.

Eg: The chemical change that occurs when an egg is rotting produces the smell of sulfur.

d. Change of state:

A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.

Eg: candle wax (solid) melts initially to produce molten wax (liquid)

plz mark as brainliest!!!!

Answer:

Hydrogen, carbon reaction

Answer:

molality

Explanation:

The SI unit for molality is moles per kilogram of solvent. A solution with a molality of 3 mol/kg is often described as "3 molal", "3 m" or "3 m". hope this helps you :)

Answer:

[tex]V_2 = 4.87 * 10^3[/tex]

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

[tex]PV = nRT[/tex]

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]

[tex]\frac{V}{T} = \frac{nR}{P}[/tex]

Represent [tex]\frac{nR}{P}[/tex] with k

[tex]\frac{V}{T} = k[/tex]

[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

At this point, we can solve for the required parameter using the following;

[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]

Multiply both sides by 323

[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]

[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]

[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]

[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]

[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]

[tex]V_2 = 4.87 * 10^3[/tex]

Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]

Answer:

[tex]\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}[/tex]

Explanation:

Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

[tex]n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}[/tex]

2. Moles of NH₃

[tex]n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}[/tex]

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

[tex]c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}[/tex]

(c) [NH₃]

[tex]c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}[/tex]

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

                        Ni²⁺             +             6NH₃       ⇌       Ni(NH₃)₆²⁺

I/mol·L⁻¹:    6.106×10⁻³                     0.1462                       0

C/mol·L⁻¹:  -6.106×10⁻³         0.1462-6×6.106×10⁻³             0

E/mol·L⁻¹:           0                              0.1095                6.106×10⁻³

Then we approach equilibrium from the right.

                            Ni²⁺   +   6NH₃       ⇌       Ni(NH₃)₆²⁺

I/mol·L⁻¹:              0           0.1095                6.106×10⁻³

C/mol·L⁻¹:            +x            +6x                           -x

E/mol·L⁻¹:             x         0.1095+6x            6.106×10⁻³-x

[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}[/tex]

Kf is large, so x ≪ 6.106×10⁻³. Then

[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}[/tex]

Answer:

4.96 mol/dm³

Explanation:

From the question,

Mass of NaCl that dissolved in 0.5L of water = 500-346.8 = 153.2 g.

Therefore, 145.2(1/0.5)g of NaCl will dissolve in 1 L of water

mass of NaCl that will dissolve in 1 L of water = 290.4 g/dm³

Molar mass of NaCl = 58.5 g/mol.

Solubility is the amount of substance in mol that will dissolve in 1 L or 1 dm³ Solution.

 solubility in (mol/dm³) = solubility in (g/dm³)/molar mass.

solubility in (mol/dm³) = 290.4/58.5

solubility in (mol/dm³) = 4.96 mol/dm³

Answer:

3-bromobenzoic acid

Explanation:

In this case, we have to remember that the [tex]Br_2/FeBr_3[/tex]  is a reaction in which we add Br into the molecule an electrophilic aromatic substitution. Additionally, we have a carboxylic acid in the benzene. This carboxylic acid is an ortho director because is a deactivating group (it removes electrons from the benzene ring). With this in mind, a "Br" atom would be added in an ortho position respect to the COOH group and we will obtain 3-bromobenzoic acid.

See figure 1.

I hope it helps!

To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group according to electrophilic aromatic substitution.

Electrophilic aromatic substitution is a type of organic reaction in which an atom or group in an aromatic ring is substituted with an electrophile. It is a fundamental reaction in aromatic chemistry that happens due to the aromatic system's high electron density.

It is an electrophilic aromatic substitution process in which Br is incorporated into the molecule. In addition, the benzene contains a carboxylic acid. Because it removes electrons from the benzene ring, this carboxylic acid functions as an ortho director. To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group. The product is seen in the photographs below.

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Answer:

3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O

Explanation:

Let's consider the double-replacement reaction between rubidium hydroxide and phosphoric acid to form rubidium phosphate and water. The cation rubidium replaces the cation hydrogen and the anion hydroxyl replaces the anion phosphate. The balanced chemical reaction is:

3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O

Answer:

200g

Explanation:

n = CV

n = mass/molar mass

mass/molar mass = CV

mass/40 = 2 x 2.5

mass/40 = 5

mass = 5x 40

mass = 200g

Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,  

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.  

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,  

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

Answer:

[tex]3.65~M[/tex]

Explanation:

We have to remember the molarity equation:

[tex]M=\frac{mol}{L}[/tex]

So, we have to calculate "mol" and "L". The total volume is 100 mL. So, we can do the conversion:

[tex]100~mL\frac{1~L}{1000~mL}=~0.1~L[/tex]

Now we can calculate the moles. For this we have to calculate the molar mass:

O: 16 g/mol

H: 1 g/mol

C: 12 g/mol

[tex](16*1)+(1*4)+(12*1)=32~g/mol[/tex]

With the molar mass value we can calculate the number of moles:

[tex]1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH[/tex]

Finally, we can calculate the molarity:

[tex]M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M[/tex]

I hope it helps!

Mixtures can be classified as homogeneous or heterogeneous . Mixtures are composed of substances that are not chemically combined.

Homogeneous mixtures are solutions. The components of a solution are evenly distributed throughout, so that every part of the solution is the same. The components that make up a solution include one or more solutes dissolved in a solvent. Solutes can be solids, liquids, or gases, and solvents can also be solids, liquids or gases.

Brass is an example of a solid/solid solution, saline solution is an example of a solid/liquid solution, diluted ethanol is an example of a liquid/liquid solution. There are many examples of solutions. The components of a solution can be separated by physical means, such as distillation, evaporation, and chromatography, among others.

Answer:

Option B is correct.

Only this option has the first underlined element with a lower oxidation number than the second amongst the options.

Explanation:

Complete Question

In which pair of substances does the first underlined atom have a lower oxidation number than the second?

A. NH₃OH⁺, NH₄⁻ (N is underlined)

B. H₂O, H₂O₂ (O is underlined)

C. SO₃, SO₄²⁻ (S is underlined)

D. HCHO, C (C is underlined)

Solution

In determination of the oxidation number of an atom in a compound, we first name the unknown oxidation number x.

Then, the total oxidation number of the atoms in the compound is equal to the charge of on the compound (or radical).

So, elements in their neutral state have no charge and no oxidation number.

A. NH₃OH⁺, NH₄⁻ (N is underlined)

N in NH₃OH⁺

Oxidation number of N = x

Oxidation number of H = +1

Oxidation number of O = -2

x + (3×+1) + (-2) + (+1) = +1

x - 3 - 2 + 1 = 1

x = +5

N in NH₄⁻

Oxidation number of N = x

Oxidation number of H = +1

x + (4×1) = -1

x + 4 = -1

x = -1 - 4 = -5

First underlined element has a greater oxidation number than the second. So, this doesn't qualify.

B. H₂O, H₂O₂ (O is underlined)

O in H₂O

Oxidation number of H = +1

Oxidation number of O = x

(2×1) + x = 0

2 + x = 0

x = -2

H₂O₂

Oxidation number of H = +1

Oxidation number of O = x

(2×1) + (2×x) = 0

2 + 2x = 0

2x = -2

x = (-2/2) = -1.

First underlined element has a lower oxidation number than the second. So, this qualifies.

C. SO₃, SO₄²⁻ (S is underlined)

S in SO₃

Oxidation number of S = x

Oxidation number of O = -2

x + (3×-2) = 0

x - 6 = 0

x = +6

SO₄²⁻

Oxidation number of S = x

Oxidation number of O = -2

x + (4×-2) = -2

x - 8 = -2

x = 8 - 2 = +6

First underlined element has the same oxidation number as the second. So, this doesn't qualify.

D. HCHO, C (C is underlined)

C in HCHO

Oxidation number of H = +1

Oxidation number of C = x

Oxidation number of O = -2

+1 + x + 1 - 2 = 0

x = 0

C in C

Oxidation number of C = x

x = 0

First underlined element has the same oxidation number as the second. So, this doesn't qualify.

Hope this Helps!!!