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3 years ago

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

1 answer:

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

How many days does it take the earth to revolve around the Sun? this is for science class plz help.

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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

KNO3

8 0

4 years ago

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

san4es73 [151]

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