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3 years ago

Does anyone know this ?Please I need help

Chemistry

1 answer:

Marrrta [24]3 years ago

8 0

Explanation:

Entropy refers to the degree of disorderliness of system. Basically it refers to how random a system is.

Gong through the options.

A. The phase change is from solid to liquid. Liquid has a greater degree of disorderliness compared to solid hence entropy is increasing.

B. The phase change is from solid to gas. Gas particles has a greater degree of disorderliness compared to solid hence entropy is increasing.

C. The phase change is from aqueous to liquid. Aqueous solutions has a greater degree of disorderliness compared to liquid hence entropy is decreasing.

D. The reactant side has; solid + gas. In the product side, there is only gas. The product side has a greater degree of disorderliness hence the entropy is increasing.

E. The entropy of the product is greater than the entropy of the reactants hence entropy is increasing.

F. The phase change is from aqueous to solid. Aqueous solutions has a greater degree of disorderliness compared to solid hence entropy is decreasing.

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The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

<u>Where:</u>

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

4 0

3 years ago

A dextrose/saline aqueaout solution doctors commonly use to replace fluids in the body contains 1.75g/L NaCl and 40.0g/L dextros

qwelly [4]

<em>Mixed molarity = 0.25 M</em>

<em>Osmotic pressure solution: 6.15 atm</em>

<h3><em>Further explanation</em></h3>

Osmotic pressure is the minimum pressure given to the solution so that there is no osmotic displacement from a more dilute solution to a more concentrated solution.

General formula:

$\large{\boxed {\bold {\pi \: = \: M \: x \: R \: x \: T}}}$

π = osmotic pressure (atm)

M = solution concentration (mol / l)

R = constant = 0.08205 L atm mol-1 K-1

T = Temperature (Kelvin)

Mixing solution

To find the molarity of the mixed solution we can use the following formula:

Vc. Mc = V1.M1 + V2.M2

where

Mc =Mixed molarity

Vc = mixed volume

Mr. NaCl = 58.5

Mr. Dextrose = C₆H₁₂O₆ = 180

mole NaCl = gram / Mr

mole NaCl = 1.75 / 58.5 = 0.03

Dextrose mole = 40/180 = 0.22

Assuming 1 liter of solution

M mixture = mole NaCl + mole dextrose / 1 liter

M mixture = 0.03 + 0.22 / 1 L

M mix = 0.25 M

T = 25 + 273 = 298 K

The osmosis pressure of the solution becomes:

π = 0.25. 0.0825. 298 = 6.15 atm

<h3><em>Learn more</em></h3>

The osmotic pressure brainly.com/question/8195553

<h3><em>Answer details </em></h3>

Grade: Senior High School

Subject: Chemistry

Chapter: Colligative property

Keywords: osmotic pressure, molarity, mole,dextrose, NaCl

7 0

3 years ago

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densk [106]

Answer:a

Explanation:

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Answer:

Explanation:

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I believe it is A compound

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