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2 years ago

Which type of reaction occurs in the following equation?

Chemistry

2 answers:

aniked [119]2 years ago

6 0

Answer:

The right choice is → a reduction reaction

Explanation:

From the balanced equation

3CIO⁻₍aq₎ → CIO⁻ ₍aq₎ + 2Cl⁻ ₍aq₎

we can calculate the change in oxidation number of Cl in reactant and product

in reactant ClO⁻

let x= oxidation number of Cl

and oxidation number of oxygen in its compound is (-2)

and the total charge on the ClO⁻ ion = -1

∴x + (-2) = -1 →→→ x = +1

in product Cl⁻

let x= oxidation number of Cl

and the total charge on the Cl⁻ ion = -1

∴x = -1

so the oxidation number of Cl decreases from +1 in reactant to -1 in product so it is reduction reaction.

But, disproportionation reaction is defined as the chemical reaction in which a single substance gets reduced as well as oxidized.

For a disproportionated substance , it should contain at least three oxidation states. which isn't the case here.

So, the right choice is:

→ a reduction reaction

Crazy boy [7]2 years ago

5 0

Answer:

disproportionation reaction

Explanation:

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Answer:

Tug of War

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1 year ago

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Diamond is forever" is one of the most successful advertising slogans of all time. but is it true? for the reaction shown below,

Zielflug [23.3K]

The given reaction is

$C _{(Diamond)}\rightarrow C_{(Graphite)}$

An element can exist in 2 or more different forms which have totally different chemical and physical properties. They are known an allotropes.

Diamond and Graphite are allotropic forms of carbon. In the given reaction, diamond is changing to graphite and we have to find out the standard free energy of this reaction. This will help us to find out whether this reaction is spontaneous at 298 K or not.

The following data is needed for calculations which is taken from standard reference table.

$H_{f}^{0}(diamond)= 1.895 kJ/mol$

$H_{f}^{0}(graphite)= 0$

$S^{0}(diamond)=2.337 J/mol-K$

$S^{0}(graphite)=5.740 J/mol-K$

Step 1: Find ΔH⁰ rxn for the given reaction.

The formula to calculate ΔH⁰ rxn is given below.

$\bigtriangleup H^{0}_{rxn}= H_{f}(product)- H_{f}( reactant)$

We have graphite on product side and diamond on reactant side.

Therefore, $\bigtriangleup H^{0}_{rxn}= H_{f}(graphite)- H_{f}(diamond)$

Let us plug in the values given above.

$\bigtriangleup H^{0}_{rxn}= 0 - 1.895 kJ/mol$

$\bigtriangleup H^{0}_{rxn}= - 1.895 kJ/mol$

ΔH⁰ rxn for the given reaction is -1.895 kJ/mol

Step 2 : Find ΔS⁰ rxn for the given reaction.

The formula to calculate ΔS⁰ rxn is

$\bigtriangleup S^{0}_{rxn}= S^{0}(product)- S^{0}( reactant)$

$\bigtriangleup S^{0}_{rxn}= S^{0}(graphite)- S^{0}(diamond)$

$\bigtriangleup S^{0}_{rxn}= (5.740J/mol.K) - (2.337 J/mol.K)$

$\bigtriangleup S^{0}_{rxn}= 3.403 J/mol.K$

Let us convert this to kJ.

$\frac{3.403J}{mol.K}\times \frac{1 kJ}{1000J}= 3.403\times 10^{-3}kJ/mol.K$

ΔS⁰ rxn for the given reaction is 3.403 x 10⁻³ kJ/mol.K

Step 3: Find standard free energy ΔG⁰ rxn.

ΔG⁰ rxn for the given reaction is calculated as

$\bigtriangleup G^{0}_{rxn}= \bigtriangleup H^{0}_{rxn}- T\times \bigtriangleup S^{0}_{rxn}$

We have T = 298 K. Let us plug in the calculated values of ΔH⁰ rxn and ΔS⁰ rxn.

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [(298K)\times 3.403\times 10^{-3}kJ/mol.K]$

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [1.014 kJ/mol]$

$\bigtriangleup G^{0}_{rxn}= - 2.909 kJ/mol$

The standard free energy change for the given reaction is -2.909 kJ/mol

The negative value of delta G⁰ suggests that the given reaction is spontaneous at room temperature. That means diamond will slowly convert to graphite. The speed of this reaction is extremely slow, but yet the reaction is taking place. So over a period of time diamond will become graphite.

Therefore "Diamond is forever" is not true as it is going to get converted to graphite.

7 0

3 years ago

The concentration from analysis question 5 represents the concentration in the 10.00 mL sample that was prepared in the volumetr

jolli1 [7]

Answer:

0.048 M

Explanation:

Note that the Concentration from Question 5= .00096 M

To Prepere of commercial aspirin solution, take the following steps:

Mix 1 aspirin tablet and 10 mL of 1 M NaOH in a 125 mL Erlenheyemer flask, heat to boil.

Transfer solution to 100 mL volumetric flask and fill to 100 mL mark with deonized water. Cover and mix solution thoroughly.

Using pipette transfer a 0.200 mL of solution to a 10 mL volumetric flask and dilute it with the 0.02 M buffered iron (III) chloride solution. Transfer it to test tube.

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3 years ago

A balloon was partially filled with helium gas at room temperature. It occupied 4.0 liters of volume at 700.0 mmHg atmospheric p

Tatiana [17]

Answer:

28.28 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and T are constant, and have two different values of V and P:

P₁V₁ = P₂V₂

P₁ = 700.0 mm Hg, V₁ = 4.0 L.

at burst: P₂ = 99.0 mm Hg, V₂ = ??? L.

∴ V₂ = P₁V₁/P₂ = (700.0 mm Hg)(4.0 L)/(99.0 mm Hg) = 28.28 L.

3 0

3 years ago

How did Rutherford discredit Thomson's plum pudding model of an atom?

Dafna1 [17]

Answer:

He conducted an experiment using gold foil and alpha particles.

Explanation:

Ernest Rutherford in 1911 performed the gold foil experiment which provided a better outlook to the structure of the atom. In his experiment, he bombarded a thin gold foil with alpha particles. Most of the alpha particles passed through the gold foil and just a few was deflected back.

This observation led Rutherford to propose the nuclear model of the atom in which an atom has a small positively charged centre and electrons moving round it.

5 0

2 years ago

Read 2 more answers