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Nata [24]
3 years ago
9

Of the following reactions occurring at 25ºC, which one involves the greatest increase in entropy?H2(g) + Cl2(g) = 2 HCl(g)H2O(s

) = H2O(l)Pb2+(aq) + 2 Cl-(aq) = PbCl2(s)CO2(s) = CO2(g)
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer: [tex]CO_2(s)\rightarrow CO_2(g)[/tex]

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) H_2(g)+Cl_2(g)\rightarrow 2HCl(g)

2 molesof gas are converting to 2 moles of another gas , thus \Delta S is zero.

b) H_2O(s)\rightarrow H_2O(l)

1 mole of solid is converting to 1 mole of liquid, the randomness increases and thus \Delta S is positive.

b) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

2 moles of ions are converting to 1 mole of solid, the randomness decreases and thus \Delta S is negative

d) CO_2(s)\rightarrow CO_2(g)

1 mole of solid is converting to 1 mole of gas, the randomness increases drastically and thus \Delta S is highly positive.

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What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of th
zaharov [31]

Answer:

0.35 V

Explanation:

(a) Standard reduction potentials

                            <u> E°/V</u>

Fe²⁺ + 2e- ⇌ Fe; -0.41

Cr³⁺ + 3e⁻ ⇌ Cr; -0.74

(b) Standard cell potential

                                             <u>  E°/V</u>

2Cr³⁺ + 6e⁻ ⇌ 2Cr;               +0.74

<u>3Fe  ⇌ 3Fe²⁺ + 6e-;             </u>  <u>-0.41 </u>

2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33

3. Cell potential

2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr

<u>3Fe  ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e- </u>

2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)

The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation

E = E^{\circ} - \dfrac{RT}{zF}\ln Q

(a) Data

  E° = 0.33 V

   R = 8.314 J·K⁻¹mol⁻¹

   T = 298 K

   z = 6

   F = 96 485 C/mol

(b) Calculations:  

Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}

 

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8 0
3 years ago
Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o
vivado [14]

<u>Answer:</u>

<u>For 1:</u> The amount of potassium iodate that were titrated is 2.5\times 10^{-4} moles

<u>For 2:</u> The amount of sodium thiosulfate required is 1.25\times 10^{-4} moles

<u>Explanation:</u>

  • <u>For 1:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of KIO_3 solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol

Hence, the amount of potassium iodate that were titrated is 2.5\times 10^{-4} moles

  • <u>For 2:</u>

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = \frac{1}{2}\times 0.00025=0.000125mol of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is 1.25\times 10^{-4} moles

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3 years ago
Inertia increases as an object's_______increases.
lilavasa [31]

Answer:

<h2><em><u>MASS</u></em></h2>

Explanation:

Inertia increases as an object's <u>Mass</u> increases.

3 0
3 years ago
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