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miss Akunina [59]
3 years ago
6

Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re

ference (pages 861 - 865) section 19.5 while completing this problem.'
Chemistry
1 answer:
lapo4ka [179]3 years ago
8 0
a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K 

when ΔG° is Gibbs free energy 

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 =  298 K 

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

      = -7739 J  = -7.7 KJ


b) then, to get E
° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

              = - 2.7 x 10^-2 V
  
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Determine the molecular formula for each of the following compounds from the
bulgar [2K]

The molecular formula for each of the following compound given the data is

  • C₂H₆
  • C₂H₂
  • C₄H₈Cl₂

<h3>A. How to determine the molecular formula </h3>

We'll begin by calculating the empirical formula. This can be obtained as follow:

  • C = 80%
  • H = 20%
  • Empirical formula =?

Divide by their molar mass

C = 80 / 12 = 6.67

H = 20 / 1 = 20

Divide by the smallest

C = 6.67 / 6.67 = 1

H = 20 / 6.67 = 3

Thus, the empirical formula of the compound CH₃

Thus, the molecular formula of the compound can be obtained as follow

  • Empirical formula = CH₃
  • Molar mass of compound = 30 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Thus, the molecular formula of the compound is C₂H₆

<h3>B. How to determine the molecular formula </h3>

We'll begin by calculating the empirical formula. This can be obtained as follow:

  • C = 92.3%
  • H = 7.7%
  • Empirical formula =?

Divide by their molar mass

C = 92.3 / 12 = 7.7

H = 7.7 / 1 = 7.7

Divide by the smallest

C = 7.7 / 7.7 = 1

H = 7.7 / 7.7 = 1

Thus, the empirical formula of the compound CH

Thus, the molecular formula of the compound can be obtained as follow

  • Empirical formula = CH
  • Molar mass of compound = 26 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH]n = 26

[12 + 1]n = 26

13n = 26

Divide both side by 13

n = 26 / 13

n = 2

Molecular formula = [CH]n

Molecular formula = [CH]₂

Molecular formula = C₂H₂

Thus, the molecular formula of the compound is C₂H₂

<h3>C. How to determine the molecular formula </h3>

We'll begin by calculating the empirical formula. This can be obtained as follow:

  • C = 37.5%
  • H = 6.3%
  • Cl = 55.8%
  • Empirical formula =?

Divide by their molar mass

C = 37.5 / 12 = 3.125

H = 6.3 / 1 = 6.3

Cl = 55.8 / 35.5 = 1.572

Divide by the smallest

C = 3.125 / 1.572 = 2

H = 6.3 / 1.572 = 4

Cl = 1.572 / 1.572 = 1

Thus, the empirical formula of the compound C₂H₄Cl

Thus, the molecular formula of the compound can be obtained as follow

  • Empirical formula = C₂H₄Cl
  • Molar mass of compound = 127 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[C₂H₄Cl]n = 127

[(2×12) + (4×1) + 35.5]n = 127

63.5n = 127

Divide both side by 63.5

n = 127 / 63.5

n = 2

Molecular formula = [C₂H₄Cl]n

Molecular formula = [C₂H₄Cl]₂

Molecular formula = C₄H₈Cl₂

Thus, the molecular formula of the compound is C₄H₈Cl₂

Learn more about empirical formula:

brainly.com/question/24297883

#SPJ1

7 0
2 years ago
Consider the following reaction where Kc = 154 at 298 K.2NO(g) + Br2(g) 2NOBr(g)A reaction mixture was found to contain 4.64×10-
IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = \frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M

Concentration of Br_{2} = \frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M

Concentration of NOBr = \frac{0.102}{1}M=0.102M

Reaction quotient,Q_{c} , for this reaction = \frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}

species inside third bracket represents concentrations at the given interval.

So, Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106

So, the reaction is not at equilibrium.

As Q_{c}< K_{c} therefore reaction must run in forward direction to increase Q_{c} and make it equal to K_{c}.

4 0
2 years ago
An inhibitor is added to an enzyme-catalyzed reaction at a concentration of 26.7 μM. The Vmax remains constant at 50.0 μM/s, but
fomenos

Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

<h3>What is the Ki for the inhibitor?</h3>

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

  • Kma = Km/(1 + [I]/Ki)

Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

where:

  • Kma is the apparent Km due to inhibitor
  • Km is the Km of the enzyme-catalyzed reaction
  • [I] is the concentration of the inhibitor

Solving for Ki:

where

[I] = 26.7 μM

Km = 1.0

Kma = (150% × 1 ) + 1 = 2.5

Ki = 26.7 μM/{(2.5/1) - 1)

Ki = 53.4 μM

Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

Learn more about enzyme inhibition at: brainly.com/question/13618533

5 0
2 years ago
...................................................................
neonofarm [45]

Answer:

Imao ;/

Explanation:

5 0
3 years ago
Read 2 more answers
What is the molality of an aqueous solution that contains 29.5 g of glucose (C6H12O6) dissolved in 950 g of water (H2O)?
Gnesinka [82]
Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg
7 0
3 years ago
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