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BigorU [14]
3 years ago
8

Hydrogen peroxide is used to clean small wounds. It has a limited shelf life, however, because the following decomposition react

ion occurs:
2 H₂O₂ (aq) → 2H₂O (l) + O₂ (g)

This is why you often hear a gas release when you open a bottle of old hydrogen peroxide. If 35 mL of a 1.20 M solution of hydrogen peroxide completely decomposes, how many grams of oxygen are produced?
Chemistry
1 answer:
g100num [7]3 years ago
4 0

Answer:

                      0.672 g of O₂

Explanation:

                   The balance chemical equation for the decomposition of hydrogen peroxide is as follow;

                                   2 H₂O₂ → 2 H₂O + O₂

Step 1: <u>Calculate Moles of H₂O₂ as;</u>

                     Molarity  =  Moles / Volume of Solution

Solving for Moles,

                     Moles  =  Molarity × Volume of Solution

Putting values,

                     Moles  =  1.20 mol.L⁻¹ × 0.035 L         ∴ 1 mL  =  0.001 L

                     Moles  =  0.042 moles of H₂O₂

Step 2: <u>Calculate Moles of O₂ as;</u>

According to equation,

                        2 moles of H₂O₂ produced  =  1 mole of O₂

So,

                  0.042 moles of H₂O₂ will produce  =  X moles of O₂

Solving for X,

                     X =  0.042 mol × 1 mol / 2 mol

                     X =  0.021 moles of O₂

Step 3: <u>Calculate Mass of O₂ as;</u>

                     Mass  =  Moles × M.Mass

                     Mass  =  0.021 mol × 32 g/mol

                     Mass  = 0.672 g of O₂

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6.02 x 10²³ atoms

Explanation:

Given parameters:

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Unknown:

Number of Ca atoms in the given compound = ?

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The given compound is:

          CaSO₄

     1 mole of CaSO₄  is made up of 1 mole of Ca atoms

Now;

   1 mole of any substance contains 6.02 x 10²³ atoms

  1 mole of Ca atoms will also contain 6.02 x 10²³ atoms

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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
4 years ago
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