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AURORKA [14]
3 years ago
11

if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro

m 27?
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0
<h3>Answers:</h3>

             1) 2 Units of Ozone

             2)  3 Units of Ozone

              3)  9 Units of Ozone

<h3>Solution:</h3>

1)  From 6 Oxygen Particles;

As given,

                          3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      6 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

                           3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      9 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  3 Units of Ozone

3)  From 27 Oxygen Particles;

As given,

                             3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      27 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  9 Units of Ozone

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Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuri
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Answer:

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b) 0.05625 mol

Explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g)         Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g)                        Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

   CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______  46.0684 g

x                            ______   50.0 g

x = 1.085 mol  CH₃CH₂OH

1 mol  CH₃CH₂OCH₂CH₃ ______  74.12 g g

y                           ______   35.9 g

y = 0.48 mol   CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w                _____  0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

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