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mihalych1998 [28]
3 years ago
9

Use the following steps to balance the redox reaction using half-reactions. Zn(s) + Fe3+ Zn2+ + Fe(s) (unbalanced) a. Write bala

nced oxidation and reduction half-reactions. Indicate which is the oxidation reaction and which is the reduction reaction. (6 points) b. Balance the charge in the two half-reactions by multiplying the half-reactions by the correct multiplication factors. (4 points) c. Add the equations and simplify to get the balanced equation.
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
6 0

Answer:

Explanation:

a )

Zn = Zn²⁺ + 2e   ( oxidation reaction )

Fe³⁺ + 3e = Fe( s ) ( reduction reaction )

b )

[ Zn = Zn²⁺ + 2e ] x 3

[ Fe³⁺ + 3e = Fe( s ) ] x 2

-------------------------------------------------------------------

3 Zn + 2 Fe³⁺ + 6e = 3 Zn²⁺ + 2 Fe ( s ) + 6e

3 Zn + 2 Fe³⁺  = 3 Zn²⁺ + 2 Fe ( s )

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A student dissolves 10.7 g of lithium chloride (LiCl) in 300. g of water in a well-insulated open cup. He then observes the temp
Novosadov [1.4K]

Answer:

1) Exothermic.

2) Q_{rxn}=-8580J

3) \Delta _rH=-121.0kJ/mol

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature increases the reaction is exothermic because it is releasing heat to solution, that is why the temperature goes from 22.0 °C to 28.6 °C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is absorbed by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

Q_{rxn}=-m_{Total}C(T_2-T_1)\\\\Q_{rxn}=-(300g+10.7g)*4.184 \frac{J}{g\°C} (28.6\°C-22.0\°C)\\\\Q_{rxn}=-8580J

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case LiCl, we proceed as follows:

\Delta _rH=\frac{Q_{rxn}}{n_{LiCl}} \\\\\Delta _rH=\frac{-8580J}{10.7g*\frac{1mol}{150.91g} }*\frac{1kJ}{1000J}  \\\\\Delta _rH=-121.0kJ/mol

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8 0
3 years ago
True or false : rocks have different properties such as hardness and texture
harkovskaia [24]

Answer:

true

Explanation:

Rocks have different properties such as hardness and texture

Metamorphic rocks form when heat and pressure cause sedimentary or igneous rocks to recrystallize into new rock.  

Soil is made up of dirt with decomposed organic material

6 0
4 years ago
Read 2 more answers
What is the name of the binary molecule H2O2
skad [1K]

Explanation:

H2O2: hydrogen peroxide.

hope it will help u ✌️

3 0
4 years ago
Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+
DIA [1.3K]

These are four questons and four answers:

Answers:

  • 1)  7⁺
  • 2) 1⁻
  • 3) 3⁺
  • 4) 5⁺

Explanation:

<u><em>Question 1) </em></u><u><em>Cl₂O₇:</em></u>

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

  • 2×X + 7 (-2) = 0
  • 2X - 14 = 0
  • 2X = +14
  • X = +14 /2 = + 7

<em>Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.</em>

<u><em>Question 2) </em></u><u><em>AlCl₄⁻</em></u>

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

  • 1 (+3) + 4X = - 1
  • +3 + 4X = - 1
  • 4X = - 1 - 3
  • 4X = - 4
  • X = - 1

<em>Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.</em>

<em><u>Question 3)</u></em><em><u> Ba(ClO₂)₂</u></em>

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

  • +2 + 2X + 4 (-2) = 0
  • 2X +2 - 8 = 0
  • 2X - 6 = 0
  • 2X = +6
  • X = + 3

<em>Conclusion: the oxidation number of Cl in Ba(ClO₂)₂  is 3⁺.</em>

<u><em>Question 4)</em></u><u><em> CIF₄⁺</em></u>

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

  • X + 4(-1) = +1
  • X - 4 = +1
  • X = +1 + 4
  • X = + 5

<em>Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.</em>

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3 years ago
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3 years ago
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