How many liters are present in 10 grams of H2O ?

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

My sister is a peach" is a metaphor.<br><br><br> TrueFalse

dolphi86 [110]

Yes that is a metaphor

6 0

3 years ago

HgS + O2 → HgO + SO2

Igoryamba

Answer:

2HgS + 3O2 → 2HgO + 2SO2

The coefficients are: 2, 3, 2, 2

Explanation:

HgS + O2 → HgO + SO2

The equation can be balance as follow:

Put 3 in front of O2 as shown below:

HgS + 3O2 → HgO + SO2

Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:

HgS + 3O2 → 2HgO + 2SO2

Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:

2HgS + 3O2 → 2HgO + 2SO2

Now the equation is balanced.

The coefficients are: 2, 3, 2, 2

The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product

5 0

3 years ago

If the visible light spectrum is from 400 to 700 nm, would light with an energy of 2.79 x 10^-19 J be visible with the naked eye

sladkih [1.3K]

Answer:

713 nm. It is not visible with the naked eye.

Explanation:

Step 1: Given data

Energy of light (E): 2.79 × 10⁻¹⁹ J

Planck's constant (h): 6.63 × 10⁻³⁴ J.s

Speed of light (c): 3.00 × 10⁸ m/s

Wavelength (λ): ?

Step 2: Calculate the wavelength of the light

We will use the Planck-Einstein equation.

E = h × c / λ

λ = h × c / E

λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 2.79 × 10⁻¹⁹ J

λ = 7.13 × 10⁻⁷ m

Step 3: Convert "λ" to nm

We will use the relationship 1 m = 10⁹ nm.

7.13 × 10⁻⁷ m × (10⁹ nm/1 m) = 713 nm

This light is not in the 400-700 nm interval so it is not visible with the naked eye.

5 0

3 years ago

The Ka for acetic acid (HC2H3O2) is 1.80 x 10-5 . Determine the pH of a 0.0500mol/L acetic acid solution.

TiliK225 [7]

Answer:

pH = 3.02

Explanation:

Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:

HOAc ⇄ H⁺ + OAc⁻.

In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.

By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02

______________________________________________________

*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.

8 0

3 years ago