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yarga [219]
3 years ago
6

A sample of oxygen has a volume of 128 milliliter at a pressure of 500.0 mmhg calculate the volume this sample would occupy at s

tandard pessure
Chemistry
1 answer:
leva [86]3 years ago
4 0

Answer:

84.2mL

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 128mL

Initial pressure (P1) = 500mmHg

Final pressure (P2) = stp = 760mmHg

Final volume (V2) =..?

Thus, the new volume of the gas can be obtained by using the Boyle's law equation as follow:

P1V1 = P2V2

500 x 128 = 760 x V2

Divide both side by 760

V2 = (500 x 128)/760

V2 = 84.2mL

Therefore, the new volume of the gas is 84.2mL

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1. Using the balanced equation, answer the following questions:
nalin [4]

Answer:

                     a)  2.53 × 10²³ molecules of O₂

                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: <u>Find out  moles of O₂ produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

                     X =  0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules  =  Moles × 6.022 × 10²³

No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: <u>Find out  moles of KCl produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

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3 years ago
Atomic mass of nitrogen
erma4kov [3.2K]

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8 0
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Solve the problem for the moles of oxygen.<br> mol O2<br> DONE
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Answer: The answer is 0.245

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Inga [223]

Answer:

90.3 L

Explanation:

Given data:

Volume of water produced = 77.4 L

Volume of oxygen required = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

It is known that,

1 mole = 22.414 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

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Now we will compare:

                               H₂O           :              O₂    

                               134.5         :              156.9

                                 77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

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