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Nina [5.8K]
1 year ago
13

Consider the reaction in the lead-acid cell pb(s) pbo2(s) 2 h2so4(aq) 2 pbso4(aq) 2 h2o(l) for which e°cell = 2. 04 v at 298 k.

δg° for this reaction is:________
Chemistry
1 answer:
aleksley [76]1 year ago
8 0

δg° for the reaction having e° cell = 2.04v. is 393.658kJ.

The lead (Pb) in the lead storage battery act as an anode

<h3>What is Voltaic Cell?</h3>

A voltaic cell or galvanic cell is an electrochemical cell that uses chemical reactions to produce electrical energy.

<h3>Parts of voltaic cell</h3>

Anode where oxidation occurs Cathode where reduction occurs.

Therefore, Pb acts as anode and PbO2 act as a cathode.

Given,

E° cell = 2.04v

Faraday constant= 96485 C/mol

Moles of electron transferred n= 2

∆G = -nFE°

= -2 × 96485 × 2.04

= 393.658kJ

Thus we concluded that the δg° for the reaction having e° cell = 2.04v. is 393.658kJ.

learn more about voltaic cell :

brainly.com/question/1370699

#SPJ4

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lianna [129]

Answer:

21.6 g

Explanation:

The reaction that takes place is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

  • 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
  • 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:

  • 0.6 mol CH₄ * \frac{2molH_2O}{1molCH_4} = 1.2 mol H₂O

Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:

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4 0
2 years ago
The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r
son4ous [18]

Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

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R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

5 0
3 years ago
explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity
grin007 [14]
The answer is… Ions are highly charged species that can readily conduct an electrical current. But hydrogen chloride the gas does not exist as ions since there is no solvent medium for them to dissolve into. Hope this helps!
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