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dalvyx [7]
3 years ago
11

A solution containing less than the equilibrium amount is called ________. a solution containing less than the equilibrium amoun

t is called ________. a dilute solution an unsaturated solution a saturated solution a supersaturated solution a concentrated solution
Chemistry
2 answers:
IrinaK [193]3 years ago
8 0

Answer: unsaturated solution.

Explanation:

Saturation can be defined as the the degree in which solute is dissolved. There are three main types of saturation, they are; supersaturated solution, saturated solution and unsaturated solution.

Let us take a quick look at the three types below.

(1). Supersaturated solution: a supersaturated solution is that solution containing more undissolved solids than the saturated solution. In other words, supersaturated solution is a solution containing containing more than the equilibrium amount in a solute more than the saturated solution. Leaving undissolved solute at the buttom.

(2). Saturated solution: saturated solution is that solution that dissolves solute to its maximum level where it can dissolve no more. Here, the undissolved solute at the buttom is not as much as the one in the Supersaturated solution.

(3). Unsaturated solution: is a solution A solution containing less than the equilibrium amount of solute.

Alex787 [66]3 years ago
6 0
<span>I believe it is an unsaturated solution.</span>
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The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
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Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol  

k_2=2.3\times 10^8

T_2=280\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (280 + 273.15) K = 553.15 K  

T_2=553.15\ K  

k_1=4.8\times 10^8  

T_1=376\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (376 + 273.15) K = 649.15 K  

T_1=649.15\ K  

So,  

\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=22.87\ kJ/mol

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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