What would the cathode be in a nickel and copper electrolytic cell

Answer:

[tex]T_2=453.05K[/tex]

Explanation:

Hello,

In this case, the temperature-variable Arrhenius equation is written as:

[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]

Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:

[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]

Regards.

Answer:

Kb = 1.77x10⁻⁵

Explanation:

When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

And the dissociation constant, Kb, for this equilibrium is:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

To find Kb you need to find the concentration of each species. The equilibrium concentrations are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate.

You can know [OH⁻] and, therefore, X, with pH of the solution, thus:

pH = -log [H⁺] = 11.612

[H⁺] = 2.4434x10⁻¹²

As 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]

4.0926x10⁻³ = [OH⁻] = X

Replacing, concentrations of the species are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

[NH₃] = 0.9459M

[NH₄⁺] = 4.0926x10⁻³M

[OH⁻] = 4.0926x10⁻³M

Replacing in Kb expression:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]

The branch of science which deals with the chemical bond is called Chemistry.

The correct answer to the question is  [tex]Kb = 1.77*10^{-5[/tex]

Explanation:

When NH₃, is acts as a  weak base it forms an equilibrium with water the reaction occurs is:

[tex]NH_3(aq) + H_2O(l) ---><NH_4^+(aq) + OH^-(aq)[/tex]

The formula we gonna use is as follows:-

[tex]Kb = \frac{[NH_4^+] [OH^-]}{[NH_3]}[/tex]

The data is given in the question is as follows:-

Where X stands for reaction coordinate.

After solving the ph of the compound the value is as follows:-

Putting the value in the formula.

[tex]Kb = \frac{[4.0926*10^{-3}M] [4.0926*10^{-3}M]}{[0.9459M]}[/tex]

After solving the equation the value of Kb is [tex]1.77*10^{-5[/tex]

Hence, the correct answer is [tex]1.77*10^{-5[/tex]

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Answer:

Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹

Explanation:

Argon is a monoatomic gas that behaves as an ideal gas at 298K.

Using the first law of thermodinamics you can obtain:

Work, Q, for constant pressure molar heat capacity,CP:

CP = (5/2)R

For constant-volume molar heat capacity,CV:

CV = (3/2)R

That means:

2CP/5 = 2CV/3

3/5 = CV / CP

As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:

3/5 = CV / CP

3/5 = CV / 20.79 J K ⁻¹mol⁻¹

12.47 J K ⁻¹mol⁻¹ = CV

Answer:

4.2g of gasoline

Explanation:

In the problem, you need to give energy to the cup from the combustion of gasoline. The energy you need to give is:

Qcup + QWater + QCopper

As you need to increase (110ºC - 21ºC = 89º = Increase 89K) 89K, the Qcup is:

Qcup = 89K × (47J/K) = 4183J.

You can find Qwater using its specific heat, C (4.18Jg⁻¹K⁻¹), its mass (450mL = 450g) and the change of temperature, 89K:

QWater = CₓmₓΔT

QWater = 4.184Jg⁻¹K⁻¹ ₓ 450g×89K

QWater = 167569J

And Q of Copper, QCu, could be obtained in the same way (Specific heat Cu: 0.387 J/g⁻¹K⁻¹:

QCu = CₓmₓΔT

QCu = 0.387 J/g⁻¹K⁻¹ₓ850gₓ89K

QCu = 29277J

Thus, total heat you need is:

Q = Qcup + QWater + QCopper

Q = 4183J + 167569J + 29277J

Q = 201029J = 201kJ

The combustion of gasoline (Octane) produce 47.8kJ/g (Its heat of combustion). that means to produce 201kJ of energy you require:

201kJ × (1g / 47.8kJ) =

Answer:

NaClO + CH₃COOH ----> HClO + CH3CO- + Na

Explanation:

This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction

yes yes yes yes

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yes

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yes

Answer:

A.Electrolysis

Explanation:

A.Electrolysis

For example,  electrolysis of solution of NaCl in water gives H2 and O2.

Answer:

please look at the picture below.

Explanation:

Answer:

0.83 g

Explanation:

Step 1: Write the balanced equation

Fe + CuSO₄ ⇒ Cu + FeSO₄

Step 2: Calculate the moles corresponding to 0.75 g of Fe

The molar mass of Fe is 55.85 g/mol.

[tex]0.75g \times \frac{1mol}{55.85g} = 0.013 mol[/tex]

Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe

The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.

Step 4: Calculate the mass corresponding to 0.013 moles of Cu

The molar mass of Cu is 63.55 g/mol.

[tex]0.013mol \times \frac{63.55g}{mol} = 0.83 g[/tex]

Answer:

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Explanation:

You know the following balanced reaction:

Fe + CuSO₄ ⇒ Cu + FeSO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:

Then you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?

[tex]mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}[/tex]

mass of Cu= 0.85 grams

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Answer:

The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.

Explanation:

The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.

A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.

Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.

SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M

Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M

GIven that :

Ka = [tex]1.9 x 10^{-5}[/tex]

Thus; it's pKa = 4.72

[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]

[tex]pH =4.72 + log(1.1376)[/tex]

[tex]pH =4.72 + 0.05598[/tex]

[tex]pH =4.77598[/tex]

pH ≅ 4.80

The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.

pH of the given solution will be used by using the following equation:

pH = pKa + log[conjugate base] / [weak acid]

Given chemical reaction will be represented as:

HN₃ + OH⁻ → N₃⁻ + H₂O

Moles will be calculated as:

n = M×V, where

M = molarity

V = volume

Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375  mol

Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol

From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:

0.00375 - 0.001995 = 0.001755 mol

And 0.001995 mol of N₃⁻ is preduced by the reaction.

Total volume of the solution = 0.025 +  0.0133 = 0.0383  L

To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:

[N₃⁻] = 0.001995 mol / 0.0383  L = 0.0521 M

[HN₃] = 0.001755 mol / 0.0383  L = 0.0458 M

Ka for HN₃ = 1.9 × 10⁻⁵

pKa = -log( 1.9 × 10⁻⁵ ) = 4.72

On putting all these values on the above equation, we get

pH = 4.72 + log (0.0521) / (0.0458)

pH = 4.80

Hence, pH of the solution is 4.80.

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Answer:

To the left.

Explanation:

Step 1: Write the balanced reaction at equilibrium

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

Step 2: Calculate the reaction quotient (Qc)

Qc = [SO₃]² / [SO₂]² × [O₂]

Qc = 0.254² / 0.500² × 0.00855

Qc = 30.2

Step 3: Determine in which direction will proceed the system

Since Qc > Kc, the system will shift to the left to attain the equilibrium.

The correct answer is C. Surface craters

Explanation:

Short-term environmental changes involve temporary changes and effects in the ecosystem, which are mainly minor. In the case of a small asteroid or comet, this will likely lead to surface craters or changes in the surface of the impact zone. This is because the craters and asteroids impact the surface at hight speed. Also, because this is a minor event it might lead to the death of some organisms but not the extinction of these and it is not expected this has major effects such as changes in weather. Thus, the short-term effect that this will most likely cause is "surface craters."

Answer:

surface

Explanation:

Answer:

Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred

Explanation:

Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]

[tex]M=200g/mol[/tex]

Thus the molar mass of the unknown compound is 200 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:

Depression in freezing point = Van't Hoff factor × Freezing point constant × molality

The molality can be defined as the moles of solute per kg solvent

Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]

The depression in freezing point can be given as:

Depression in freezing point = Van't Hoff factor × Freezing point constant ×  [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)

Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]

Van't Hoff factor = 1 (Molecular compound)

Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m

Mass of solute = 26.7 g

Mass of solvent = 0.250 kg

Substituting the values in equation (i):

2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12

[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]

0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]

Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol

Molecular mass of solute = 199.626 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

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Answer:

Coals

Explanation:

The land feature that supports the theory of continental drift is ; ( C ) coal fields

Continental drift is the gradual shift in position of the earth tectonic plates ( i.e. gradual shift in the continents in relation to ocean basins) and this due to the heat from the earths' mantle.

Coal fields supports this theory because the it is an area where coal is found in large quantities and mined for commercial purposes. coal fields areas are found as a result of continental drift.

Hence we can conclude that the land feature that supports the theory of continental drift is coal fields

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Answer:

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

However, since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis of ATP for this reaction is said to be spontaneous

Explanation:

From the question; The equation for this reaction can be represented as :

[tex]ATP_{(aq)} + H_2O_{(l)} \to ADP_{(aq)}+ HPO_4^{2-}} _{(aq)}[/tex]

where:

[tex]\Delta G ^0 _{rxn} =[/tex]-30.5 kJ/mol

= -30.5 kJ/mol × 1000 J/ 1 kJ

= -30.5 × 10 ⁻³ J/mol

Temperature T = 37 ° C

= (37+273)

= 310 K

pH = 7.0

[ATP] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

[ADP] = 0.30 mM

= 0.30 mM × 1M/1000mM

= 0.0003 M

[tex][HPO_4^{2-}}][/tex] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

The objective is to calculate the value for Δ [tex]G_{rxn}[/tex] in the biological cell and to determine if the hydrolysis  of  ATP is spontaneous under these conditions.

Now;

From the equation given; the equilibrium constant [tex]K_{eq}[/tex] can be expressed as:

[tex]K_{eq} = \dfrac{[ADP][ HPO_4^{2-}]} {[ATP]}[/tex]

[tex]K_{eq} = \dfrac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}[/tex]

[tex]K_{eq} = 3*10^{-4}[/tex]

The  Δ [tex]G_{rxn}[/tex] in the biological cell can now be calculated as:

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^{-4})[/tex]

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (-20906.68126)[/tex]

Δ [tex]G_{rxn}[/tex] = −51406.68 J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 × 10³ J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

Thus since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis for this reaction is said to be spontaneous

The given question is incomplete, the complete question is:

Calculating an equilibrium constant from a partial equilibrium... Steam reforming of methane (CH) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 25.0L tank with 8.0 mol of methane gas and 1.9 mol of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 1.5 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer:

The correct answer is 2.47.

Explanation:

Based on the given information, the equation for the synthesis gas is,  

CH₄ (g) + H₂O (g) ⇔ CO (g) + 3H₂ (g)

Based on the given information, 25.0 L is the volume of the tank, the concentration of CH₄ is 8.0 mol, the concentration of water vapor is 1.9 mol, and the concentration of CO gas is 1.5 mol.  

Therefore, 25 L of the solution comprise 8.0 mole of CH₄. So, 1 L of the solution will comprise 8.0 / 25 mole CH₄,  

= 0.32 mole of CH₄

Thus, the concentration of CH₄ or [CH₄] will be 0.32 mole/L or 0.32 M.  

Similarly, the concentration of H₂O or [H₂O] will be 1.9/25 = 0.076 M

and [CO] is 1.5/25 = 0.06 M

The concentration equilibrium constant for the steam will be,  

Kc = [CO] pH₂ / [CH₄] [H₂O] (Here pH₂ is the partial pressure of H₂)

Now lets us assume that the reaction has taken place in a constant atmospheric pressure, therefore, pH₂ will be equal to 1.  

= 0.06 M/0.32 M × 0.076 M  

= 2.47  

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

Answer:

  0.0480 M

Explanation:

The reaction is ...

  H₂SO₄ + 2NaOH ⇒ Na₂SO₄ +2H₂O

That is, 2 moles of NaOH react with each mole of H₂SO₄. Then the molarity of the H₂SO₄ is ...

  moles/liter = (0.186 M/2)(12.9 mL)/(25.0 mL) ≈ 0.0480 M

Answer:

Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.

The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.

To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.

Answer:

CaF2 + CO3- ----> CaCO3 + 2 F-

Explanation:

The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.

Calcium carbonate is a quaternary salt

Answer:

[tex]Q=2582J=2.58kJ[/tex]

Explanation:

Hello,

In this case, for us to compute the absorbed heat, we apply the following equation:

[tex]Q=m_{Al}Cp_{Al}(T_2-T_1)[/tex]

Whereas we use the mass, specific heat and temperature change for the piece of aluminium, thus, we obtain:

[tex]Q=63.1g*0.930\frac{J}{g*\°C}*(67.0-23.0)\°C\\ \\Q=2582J=2.58kJ[/tex]

It is positive as the heat is entering, therefore the temperature raises.

Best regards.

Answer:

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Explanation:

According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺. Let's consider the molecular equation showing that benzoic acid is a Brönsted-Lowry acid.

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The complete ionic equation includes all the ions and molecular species.

C₆H₅COO⁻(aq) + H⁺(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

The sand that grained will be quickly chemically weathered away should be option b. ferromagnesian silicate minerals.

It should be considered as the Silicate minerals where cations of iron and the form of magnesium should be important for the chemical components. It is used for covering up the minerals. Also, calcite should be normal weather via the solution process so it required a lot of water that contains a high amount of carbonic acid.

Hence, the correct option is b.

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Answer:

18.308 KJ

Explanation:

From the given above, we obtained the following:

Activation energy for the catalyzed reaction = 4.6 kJ.

Activation energy for the uncatalyzed reaction =..?

Now, a careful observation of the question revealed that the activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction.

With this vital information, we can thus, calculate the activation energy of the uncatalyzed reaction as follow:

Activation energy for the uncatalyzed reaction = 3.98 times that of the catalyzed reaction.

Activation energy for the uncatalyzed reaction = 3.98 x 4.6 kJ = 18.308 KJ

Therefore, the activation energy of the uncatalyzed reaction is 18.308 KJ.

Answer:

The answer is Neutralization reaction

It occurs when an acid and a base are present in the same solution and react to form salt and water only

Hope this helps you

Answer:

a change in temperature would be observed(ΔH is -ve)

Explanation:

Hydrochloric acid react with sodium hydroxide to give salt(sodium chloride) and water

HCl(aq) + NaOH(aq) =====> NaCl(aq) + H2O(l)

There would be no notable change since sodium chloride dissolved in water but there would be a change in temperature.

Since neutralization is exothermic(heat is evolved), therefore ΔH is negative

Answer:

Explanation:

1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂

In this reaction , oxidation number of Cl decreases  from 0 to -1  so it is reduced  and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .

2 )

Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄

In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.

3 )

CaCO₃ = CaO + CO₂

In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.

So only first reaction is oxidation - reduction reaction.

2nd option is correct.