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katrin [286]
2 years ago
14

The ozone "hole" over Antarctica formed as the result of a series of chemical reactions. The loss occurs in part through three c

onsecutive steps in which ozone is broken down: Step 1: Chlorine atoms react with ozone (O3) to form chlorine monoxide and molecular oxygen. Step 2: Chlorine monoxide molecules combine to form ClOOCl gas Step 3: ClOOCl absorbs sunlight and breaks into chlorine atoms and molecular oxygen. a) Write a balanced equation for each step. b) Write an overall balanced equation for the sequence
Chemistry
1 answer:
mixer [17]2 years ago
7 0

Answer:

A)

Cl + O3 --> ClO + O2

2ClO --> ClOOCl

ClOOCl --> 2 Cl + O2

B)

Cl2 + 4O3 -> 2 ClO + 5O2

Explanation:

Step 1: Chlorine atoms react with ozone (O3) to form chlorine monoxide and molecular oxygen

Cl + O3 --> ClO + O2

Step 2: Chlorine monoxide molecules combine to form ClOOCl gas

2ClO --> ClOOCl

Step 3: ClOOCl absorbs sunlight and breaks into chlorine atoms and molecular oxygen

ClOOCl --> 2 Cl + O2

B) Overall Balanced equation

Cl2 + 4O3 -> 2 ClO + 5O2

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Ilia_Sergeevich [38]

Explanation:

Moles=mass/molar mass

moles × molar mass = mass

0.206 x 119= mass

Mass= 24. 51grams

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3 years ago
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Define electronegativity. Where on the periodic table is the highest? Explain why this is the case.
geniusboy [140]

Answer:

Explanation:

Electronegativity is a measure of the ability of an atom to attract the electrons when the atom is part of a compound. Electronegativity values generally increase from left to right across the periodic table. The highest electronegativity value is for fluorine.

5 0
3 years ago
Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T
Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 730.5 mmHg  

V = Volume of the gas = 9.49 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol

According to the reaction shown below as:-

2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when \frac{2}{3}\times 0.3794 moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 735.5 mmHg  

V = Volume of the gas = 9.99 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0
3 years ago
Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.
Ronch [10]

Answer : The limiting reagent is O_2

Solution : Given,

Moles of methane = 2.8 moles

Moles of O_2 = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 5 moles of O_2 react with \frac{5}{2}=2.5 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is O_2

8 0
3 years ago
Calculate the osmotic pressure (in torr) of 6.00 L of an aqueous 0.958 M solution at 30.°C, if the solute concerned is totally i
cluponka [151]

<u>Answer:</u> The osmotic pressure is 54307.94 Torr.

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iCRT

where,

\pi = osmotic pressure of the solution = ?

i = Van't hoff factor = 3

C = concentration of solute = 0.958 M

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = temperature of the solution = 30^oC=[30+273]K=303K

Putting values in above equation, we get:

\pi=3\times 0.958mol/L\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 303K\\\\\pi=54307.94Torr

Hence, the osmotic pressure is 54307.94 Torr.

7 0
3 years ago
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