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nlexa [21]
3 years ago
14

Two students prepared aqueous solutions of LiCl and measured the properties, as shown in the table above. Both students observed

that the solid LiCl readily dissolved in H2O. The students drew particle diagrams to explain the changes in the enthalpy and entropy of dissolution for LiCl based on their results and observations. Based on this information, the better particle diagram was drawn by which student, and why is that diagram more accurate?
A) The better particle diagram was drawn by Student 1 because when LiCl dissolves in water, it dissociates into Li+ and Cl− ions causing an increase in entropy.

B) The better particle diagram was drawn by Student 1 because when LiCl dissolves in water, it dissociates into Li+ and Cl− ions indicating that the dissolution of LiCl is exothermic.

C) The better particle diagram was drawn by Student 2 because it shows that LiCl does not dissociate, resulting in a decrease in entropy for the dissolution of LiCl.

D) The better particle diagram was drawn by Student 2 because it shows that LiCl does not dissociate, resulting in an endothermic process for the dissolution of LiCl.
Chemistry
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer:

c

Explanation:

it makes sense and the diagram is build like that with high tech

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A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The speci
Molodets [167]

Answer:

35.578g or 36g if you round

Explanation:

Q=mc ∆∅ where ∅ is temperature difference

1160= m x 1.716 x (42-23)

m = 1160/ 1.716 x19

m=35.578g

m = 36g to nearest whole number

6 0
3 years ago
What is the pH of a 0.640 M solution of C₅H₅NHBr (Kb of C₅H₅N is 1.7 × 10⁻⁹)?
Elden [556K]

The pH of a 0.64 M solution of pyridine (C₅H₅N) is 9.52.  

<h3>What is pH ?</h3>

A figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The equation for the protonation of the base pyridine is the following:

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻   (1)

Kb = 1.7 × 10⁻⁹ (Given)

To calculate the pH of the solution we need to use the following equation:

pH + pOH = 14

<em>pH = 14 - pOH</em>

     =14 - [-log[OH⁻]]

    = 14 + log[OH⁻]

Now, we need to find the concentration of the OH⁻ ions. Since pyridine is a weak base, at the equilibrium we have (eq 1):

C₅H₅N  +  H₂O  ⇄  C₅H₅NH⁺  +  OH⁻

0.64 - x                          x              x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

1.7 × 10⁻⁹  =[C₅H₅NH⁺]  [OH⁻]  /  [C₅H₅N]

                = x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

  • x₁ = -3.32 x 10⁻⁵
  • x₂ = 3.32 x 10⁻⁵

Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

<em />

<em>pH = </em>14 + log[OH⁻]

     = 14 + log (3.32 x 10⁻⁵)

 

     = 9.52

Therefore, the pH of the solution of pyridine is 9.52.

Find more about pH here:

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3 0
2 years ago
Consider the bonds in iron(III) oxide. Are the bonds more ionic or more covalent?
kvv77 [185]
<span>the bonds in iron(III) oxide are more ionic</span>
4 0
3 years ago
Read 2 more answers
What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium c
LuckyWell [14K]

Answer:

[A g + ]  =  3.12 *10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^− 12

Step 2: The balanced equation

2 A g + ( a q )  +  C rO4^2-  ( a q )  →Ag2CrO4 (  s)

Step 3:

[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 -0.088 = 0.123 M

1.2 * 10^− 12  = ( 2 x ) ²*( 0.123M + x )

[ C O 3^ −2] f  >>  x

1.2 * 10^− 12  = ( 2 x ) ²* 0.123M

x = 1.56 * 10^-6

[A g + ] f  = 2x = 3.12 *10^-6 M

,

4 0
3 years ago
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zvonat [6]

Explanation:

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