Why do scientists sometimes discard theories?

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

3 years ago

Aryan want to deposit silver on an iron spoon using silver nitrate as electrolyte. Which terminal of the battery he should conne

34kurt

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

<h3>What is electroplating?</h3>

Electroplating is a way of electrolysis, where a thin layer of metal is used to plate a determined object. It is a kinda process to purify a material you want.

The anode contains the metal you want to plate with, in this case, the Ag.

The anode contains the half-reaction of oxidation so:

Ag(s) → Ag ⁺ (aq) + e⁻

In the cathode, you have the spoon, which it takes place the half-reaction of reduction:

Ag ⁺ (aq) + e⁻ → Ag(s)

The electrolytic cell, where the redox reaction takes place, must be filled with a AgNO₃ solution.

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

Learn more about the electroplating here:

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#SPJ1

8 0

2 years ago

PLEASE HELP ASAPAHHH

Yanka [14]

Answer:

if the oil is already 60 c and you heat up the hot plate to the same degree you are not changing anything

hope this helps :)

4 0

2 years ago

Read 2 more answers

How do water particles move in a wave? A. They move forward with the wave. B. They move in a circular motion. C. They move up an

Olin [163]

Sorry if I'm wrong but I think that it is B.

4 0

2 years ago

When 3.93 grams of lactic acid, CHoOs(s), are burned in a bomb

aliya0001 [1]

The heat released in the combustion of lactic acid is absorbed by the

calorimeter and in the decomposition of the lactic acid.

ΔH°f of lactic acid is approximately <u>-716.2 kJ</u>

Reasons:

Known parameters are;

Mass of the lactic acid = 3.93 grams

Heat capacity of the bomb calorimeter = 10.80 kJ·K⁻¹

Change in temperature of the calorimeter, ΔT = 5.34 K

ΔHrxn = ΔErxn

ΔH°f of H₂O(l) = -285.8 kJ·mol⁻¹

ΔH°f of CO₂(g) = -393.5 kJ·mol⁻¹

The chemical equation for the reaction is presented as follows;

C₃H₆O₃ + 2O₂ → 3CO₂ + 3H₂O

The heat of the reaction = 10.80 kJ·K⁻¹ × 5.34 K = 57.672 kJ

Molar mass of C₃H₆O₃ = 90.07 g/mol

Number of moles of C₃H₆O₃ = $\dfrac{3.93 \, g}{90.07 \, g/mol}$ = 0.043633 moles

Number of moles of CO₂ produced = 3 × 0.043633 moles = 0.130899 moles

Heat produced = 0.130899 mole × -285.8 kJ·mol⁻¹ = -37.4109342 kJ

Moles of H₂O produced = 0.130899 moles

Heat produced = 0.130899 mole × -393.5 ≈ -51.51 kJ

Therefore, we have;

Heat absorbed by the lactic acid = ΔH°f of H₂O + ΔH°f of CO₂ + Heat absorbed by the calorimeter

Which gives;

Heat absorbed by lactic acid = -37.4109342 kJ - 51.51 kJ + 57.672 kJ ≈ -31.249 kJ

The heat absorbed by the lactic acid ≈ -31.249 kJ

$\Delta H^{\circ}f \ of \ C_3H_6O_3 = \dfrac{-31.249}{0.043633} \approx -716.2$

ΔH°f of C₃H₆O₃ ≈ -716.2 kJ

Heat of formation of lactic acid ≈ <u>-716.2 kJ</u>.

Learn more here:

brainly.com/question/13185938

5 0

2 years ago