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Kay [80]
3 years ago
8

The oxidation of copper(I) oxide, Cu 2 O ( s ) , to copper(II) oxide, CuO ( s ) , is an exothermic process. 2 Cu 2 O ( s ) + O 2

( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol Calculate the energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure.
Chemistry
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

The energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure is -10.142 kJ

Explanation:

Here we have

2Cu₂O ( s ) + O₂ ( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol

In the above reaction, 2 Moles of Cu₂O (copper (I) oxide) react with one mole of O₂ to produce 4 moles of CuO, with the release of − 292.0 kJ/mol of energy

Therefore,

1 Moles of Cu₂O (copper (I) oxide) react with 0.5 mole of O₂ to produce 2 moles of CuO, with the release of − 146.0 kJ  of energy

We have 9.94 g of Cu₂O with molar mass given as 143.09 g/mol

Hence the number of moles in 9.94 g of Cu₂O is given as

9.94/143.09 = 6.95 × 10⁻² moles of Cu₂O

6.95 × 10⁻² moles of Cu₂O will therefore produce 6.95 × 10⁻² ×  − 146.0 kJ mol  or -10.142 kJ.

You might be interested in
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
Degger [83]
a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

       = 9.2 x 10^-5 M

when PH = -㏒[H+]

   PH = -㏒(9.2 x 10^-5)

        = 4  

b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]


first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

                                        =0.07 M

and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

                                    = 0.07 M

and when Ka = 4 x 10^-8 

∴Pka =-㏒Ka

         = -㏒(4 x 10^-8)

         = 7.4 

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4 

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume 

                                                        = 0.21 M * 0.05L

                                                        = 0.0105 moles

then moles KOH = molarity * volume
                            =  0.22 M* 0.035 L 

                            =0.0077 moles 

∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume 

                                      = 8 x 10^-5 / 0.085

                                      = 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

                                          = 0.0077M / 0.085L

                                          = 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

                                   = 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

       = 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO] 

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

           = 3.8
∴PH = 14- POH

        =14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

 when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M 

V2 is the excess volume added  of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

           = -㏒0.02 

           = 1.7

∴PH = 14- POH

       = 14- 1.7 

      = 12.3 
8 0
3 years ago
Read 2 more answers
PLZ HELP ILL GIVE BRAINLIST!
Schach [20]

Answer:

The energy of a wave is inversely proportional to the wavelength of the wave.

As wavelength increases, the energy of the wave decreases.

As wavelength decreases, the energy of the wave increases.

Explanation:

The energy of a wave is directly proportional to the wave's frequency. As frequency increases, so does the energy of the wave.

E\propto f (energy E is proportional to frequency f)

<u>How is this related to wavelength?</u>

Frequency is inversely proportional to wavelength. That means that as frequency increases, wavelength decreases and as frequency decreases, wavelength increases.

f\propto \frac{1}{\lambda} (frequency f is inversely proportional to wavelength \lambda)

Therefore, as wavelength increases, the energy of a wave decreases and as wavelength decreases, the energy of a wave increases.

E\propto f\propto \frac{1}{\lambda}

6 0
3 years ago
How many oxygen atoms are in 10 formula units of AI2(SO4)3
Alisiya [41]

<em>Answer:</em>

  • 120 Oxygen atoms

<em>Data given</em>

  • Formula unit of AL2(SO4)3 =10
  • 1 formula unit of AL2(SO4)3  contain oxygen atoms = 12
  • 10 formula unit of AL2(SO4)3  contain oxygen atoms = 12 × 10
  • 1 formula unit of AL2(SO4)3  contain oxygen atoms = 120
3 0
4 years ago
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The following is the chemical formula for Acetic Acid. Which answer lists the correct amount of atoms for each Element? A. 2 Car
leonid [27]

Answer:

c

Explanation:

6 0
4 years ago
what is the empirical formula of a compound that contains 15.77% aluminum, 28.11% sulfur and 56.12% oxygen
Marizza181 [45]
<h3>Answer:</h3>

Al₂(SO₄)₃

<h3>Explanation:</h3>

We are given percentage composition of elements in a compound;

  • Aluminium is 15.77%
  • Sulfur is 28.11 %
  • Oxygen is 56.12%

We are required to calculate the empirical formula of the compound.

  • Assuming the mass of the compound is 100 g then the masses of the elements is;

Aluminium = 15.77 g

Sulfur = 28.11

Oxygen = 56.12

We can determine the number of moles of each;

Moles of Aluminium = 15.77 g ÷ 26.98 g/mol

                                 = 0.585 moles

Moles of sulfur = 28.11 g ÷ 32.07 g/mol

                         = 0.877 moles

Moles of Oxygen = 56.12 g ÷ 16.0 g/mol

                            = 3.5075 moles

  • But, the empirical formula is the simplest whole number ratio of elements in a compound.
  • Therefore; we need to get the ratio of moles of the above elements;

Aluminium : Sulfur : Oxygen

0.585 mol  :  0.877 mol : 3.5075 mol

0.585/0.585 : 0.877/0.585 : 3.5075/0.585

    1 : 1.5 : 6

But, we need whole number ratios, therefore;

= (1 : 1.5 : 6 ) × 2

= 2 : 3 : 12

Therefore; the formula of the compound is Al₂S₃O₁₂

The compound is written as Al₂(SO₄)₃

Thus, the empirical formula of the compound is Al₂(SO₄)₃

3 0
3 years ago
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