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docker41 [41]
3 years ago
10

Calculate the cell potential, the equilibrium constant, and the free energy change for

Chemistry
1 answer:
Tresset [83]3 years ago
5 0

Answer:

Explanation:

Ba(s) + Mn²⁺ (aq,1M)  →      Ba²⁺ (aq,1M) + Mn(s)

Ba⁺²(aq) +2e →  Ba(s) ,  E° =  −2.90 V  

Mn⁺²(aq) +2e → Mn(s),  E⁰  =0.80 V

Anode reaction :

Ba(s)  →  Ba⁺²(aq) +2e        E° =  −2.90 V  

Cathode reaction :

Mn⁺²(aq) +2e → Mn(s)          E⁰  =0.80 V

Cell potential = Ecathode  - Eanode

Ecell  = .80 - ( - 2.90 )

Ecell = 3.7 V .

equilibrium constant ( K ) :

Ecell = .059 log K  / n

n = 2

3.7  = .059 log K  / 2

log K = 125.42

K = 2.63 x 10¹²⁵ .

Free energy change :

ΔG = - n F Ecell

= - 2 x 96500 x 3.7

= 714100 J

= 7.141 x 10⁵ J .

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klasskru [66]

Answer: Temperature final = 103 °C

Explanation: To solve for final temperature we use the equation of heat:

Q= mc∆T

Next derive the equation to find final temperature

Q = mc(T final - T initial)

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Transpose T initial and change the sign so that T final will be left.

T final = Q / mc + T initial

Substitute the values:

T final = 305 J / 28.8 g x 0.128 J/(g°C)

= 305 J / 3.6864 J/°C

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3 years ago
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1/λ =R(1/n₁² -1/n₂²), where
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In the Paschen series is an emission spectrum of hydrogen when the energy level is at least n=4. So, this covers n=4 to n=7.

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Answer:

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Answer:

in an ionic bond.

Explanation:

i hope thats right

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