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Setler79 [48]
1 year ago
15

The difference between oxygen and ozone

Chemistry
1 answer:
jeyben [28]1 year ago
4 0

Differences between oxygen and ozone

  • oxygen exists as a diatomic molecule whereas ozone exists as triatomic molecules.
  • oxygen mass is 32 amu on the other hand ozone has 48 amu
  • Oxygen has no odor but  ozone has a very sharp odor.
  • when oxygen atoms  react with oxygen  molecules then ozone forms
  • . The average length of an O=O double bond is 121 pm on the other hand  ozone exists in resonance
  • ozone is a  polar molecule and have stronger intermolecular forces. The contribution of the electron pair from the central oxygen atom leaves a small positive charge on the central O atom. hence ozone molecules become dipole whereas diatomic oxygen is non-polar.

To know more about oxygen visit :

brainly.com/question/13905823

#SPJ9

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Ira Lisetskai [31]

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Explanation:

When potassium chloride reacts with lead acetate in aqueous medium, it gives white precipitate of lead chloride and potassium acetate in aqueous medium as a product.

As shown in chemical reaction given below:

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The element that can act like a metal when it is under tremendous pressure and is probably responsible for Jupiter and Saturn's
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1. Hydrogen

Explanation:

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3 0
3 years ago
Use this equation for the following problems: 2NaN3 --&gt; 2Na+3N2
olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

<em>2NaN₃ → 2Na + 3N₂,</em>

  • It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

<em>Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?​ </em>

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

  • Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

  • We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

  • We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

<em><u>using cross multiplication:</u></em>

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

  • Finally, we can get the grams of NaN₃ needed:

<em>mass = no. of moles x molar mass</em> = (0.517 mol)(65.0 g/mol) =<em> 33.6 g.</em>

<em />

<em>Q2: How many mL of N₂ result if 8.3 g Na are also produced?</em>

  • We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

  • We can get the no. of moles of N₂ produced with 0.36 mol of Na:

<em><u>using cross multiplication:</u></em>

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

  • We can get the mass of 0.54 mol of N₂:

mass = no. of moles  x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

  • Now, we can get the mL of 15.12 g of N₂:

<em><u>using cross multiplication:</u></em>

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

<em>∴ The volume of N₂ result </em>= (1.0 L)(15.12 g)/(0.92 g) = <em>16.434 L = 16434 mL.</em>

4 0
3 years ago
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