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zimovet [89]
3 years ago
15

Determine the equilibrium constant, keq, at 25°c for the reaction 2br– (aq) + i2(s) br2(l) + 2i– (aq).

Chemistry
1 answer:
Travka [436]3 years ago
5 0
Electrochemical cell representation for above reaction is,

Br-/Br2//I2/I-

Reaction at Anode: Br2 + 2e-   →    2Br-               (1)
Reaction at Cathode: 2I-            →  I2   + 2e-          (2)

Standard reduction potential for Reaction 1 =  Ered(anode) = 1.066 v
Standard reduction potential for Reaction 2 = Ered(cathode) = 0.535 v

Eo cell = Ered(cathode)  -  Ered(anode)
            = 0.535 - 1.066
            = -0.531v

Now, we know that ΔGo = -nF (Eo cell)        ..............(3)
Also, ΔGo = RTln(K)         ..........(4)

Equation 3 and 4 we get,

ln (K)   = nF (Eo cell)  / RT
           = 2 X 96500 X (-0.531)/ (8.314 X 298)

∴ K = 1.085 X 10^-18.


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3 years ago
Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa
kumpel [21]

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

P_{50} = 22 mm Hg

h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

Hence, the fractional saturation for hemoglobin is 0.86.

4 0
3 years ago
A compound contains 38.7% K, 13.9% N, and 47.4% O by mass. What is the empirical formula of the compound?
inysia [295]
K ---> 38.7 g / 39.1 g/mol = 0.99
N ---> 13.9 g / 14.0 g/mol = 0.99
O ---> 47.4 g / 16.0 g/mol = 2.96

Divide by smallest:

K ---> 0.99 / 0.99 = 1
N ---> 0.99 / 0.99 = 1
O ---> 2.96 / 0.99 = 3

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4 years ago
Assume that some protein molecule, in its folded native state, has one favored conformation. But when it is denatured, it become
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Answer:

Gibbs Free Energy is given by the next equation: ΔG = ΔH - TΔS, where

ΔH - <u>change</u> in enthalpy

T - temperature in Kelvin

ΔS - <u>change</u> in entropy

ΔG can be:

  • ΔG>0 positive(not spontaneous reaction)
  • ΔG<0 negative(spontaneous reaction)
  • ΔG=0(in equilibrium).

→ When the proteins are denatured, the entropy of the system increases. Therefore, the ΔS also increases and becomes more positive.  

→ From the equation we see that positive value of ΔS contributes to negative value of ΔG.

The proteins are stable when there is no spontaneous reaction, thus, the ΔG>0 is required. Which means that ΔH has to be large and positive to counteract the influence of ΔS.

3 0
4 years ago
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