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3 years ago

Which type of reaction occurs in the following equation?

Chemistry

2 answers:

aniked [119]3 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]3 years ago

5 0

Answer:

disproportionation reaction

Explanation:

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71.1%. , is a fertilizer.

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2 years ago

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2 years ago

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

Part A: nicotine

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

Part B: caffeine

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

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3 years ago

2) Write a hypothesis for the Observation and question below. Be sure to use the proper “If…, then…” format in your answer. Iden

Anna [14]

Answer:

If one astronaut used more force, then that one would be faster than the other. The independent variable is force. The dependent variable is speed.

Explanation:

Force is what is to be changed. Speed is what is being measured.

IV = Changed/factor DV=Measured/Changes by factor

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3 years ago

The density of water at 400C is 0.992 g/mL What is the volume of 27.0 g of water at this temperature?

pantera1 [17]

Answer:

Volume of water at this temperature is 27.2 mL

Explanation:

We know that $density=\frac{mass}{volume}$

Here density of water is 0.992 g/mL

Here mass of water is 27.0 g

So $volume=\frac{mass}{density}$

= $\frac{27.0g}{0.992g/mL}$

= 27.2 mL

7 0

3 years ago