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In-s [12.5K]
2 years ago
10

Complete the following single replacement reaction. If they don’t react, just write “NR”

Chemistry
1 answer:
Kipish [7]2 years ago
4 0

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq)  + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

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A dehydration reaction starting with 3.8 g cyclohexanol produces 2.6 g cyclohexene. Calculate the theoretical yield for this rea
diamong [38]

Answer:

Theoretical yield of C6H10 = 3.2 g.

Explanation:

Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.

Equation of the reaction

C6H11OH --> C6H10 + H2O

Moles of C6H11OH:

Molar mass of C6H110H = (12*6) + (1*12) + 16

= 100 g/mol

Mass of C6H10 = 3.8 g

number of moles = mass/molar mass

=3.8/100

= 0.038 mol.

Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.

Therefore, 0.038 moles of C6H10 was produced.

Mass of C6H10 = molar mass * number of moles

Molar mass of C6H10 = (12*6) + (1*10)

= 82 g/mol.

Mass = 82 * 0.038

= 3.116 g of C6H10.

Theoretical yield of C6H10 = 3.2 g

4 0
3 years ago
Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO2 and 4.96g of H2O. What is the empirical formula for the hydr
vlabodo [156]

Let empirical formula for hydrocarbon is CxHy

it will undergo combustion as

CxHy + (x + y/4) O2  ---> xCO2 + (y/2 )H2O

Given that mass of CO2 produced = 9.69 g

So moles of CO2 produced = 9.69 / 44 = 0.22 moles

So moles of carbon present = 0.22 moles

mass of H2O produced = 4.96 g

Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles

So moles of H present = 2 X 0.28 = 0.56 moles

Let us divided the moles of each with lowest value of moles

Moles of Carbon = 0.22 / 0.22 = 1 moles

moles of H = 0.56 / 0.22 = 2.55

Multiplying with two to get whole number

the ratio of carbon and hydrogen will be : C:H = 2:5

empirical formula : C2H5


4 0
3 years ago
Read 2 more answers
A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

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