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2 years ago

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To what pH should you adjust a standard hydrogen electrode to get an electrode potential of -0.128 V ? (Assume that the partial

pressure of hydrogen gas remains at 1 atm.) Express your answer using two decimal places.\

1 answer:

8090 [49]2 years ago

8 0

The pH of the standard hydrogen electrode that has electrode potential of -0.128 V is 4.3.

The equation of the hydrogen electrode is;

2H^+(aq) + 2e ⇄ H2(g)

The standard electrode potential of hydrogen is 0.00 V

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Now;

E°cell = 0.00 V

n = 2

Q = 1/[H^+]

-0.128 = 0.00 - 0.0592/2 log 1/[H^+]

-0.128 = 0.00 - 0.0296 log 1/[H^+]

-0.128 = - 0.0296 log 1/[H^+]

-0.128/ - 0.0296 = log 1/[H^+]

1/[H^+] = Antilog (4.32)

[H^+] = 4.79 × 10^-5

Now;

pH = -log[H^+]

pH = -log (4.79 × 10^-5)

pH = 4.3

Learn more: brainly.com/question/11155928

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pishuonlain [190]

Answer:

A. K

Step-by-step explanation:

Remember the trends in the Periodic Table:

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At a certain temperature the vapor pressure of pure heptane C7H16 is measured to be 454.mmHg. Suppose a solution is prepared by

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Answer:

Mass of heptane = 102g

Vapor pressure of heptane = 454mmHg

Molar mass of heptane = 100.21

No of mole of heptane = mass/molar mass = 102/100.21

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Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane

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3 years ago

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

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Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

<u>E = 2.18 x 10^-18 J</u>

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3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

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Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

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