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Bad White [126]
3 years ago
5

The compound chromium(II) chloride is a strong electrolyte. Write the transformation that occurs when solid chromium(II) chlorid

e dissolves in water. Be sure to specify states such as (aq) or (s).
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
5 0

Answer:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Explanation:

Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

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4.13 moles of Fe

Explanation:

Step 1: Write the balanced equation

3 CO + Fe₂O₃ ⇒ 2 Fe + 3 CO₂

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of CO to Fe is 3:2.

Step 3: Calculate the moles of Fe formed from 6.20 moles of CO

We will use the previously established molar ratio.

6.20 mol CO × 2 mol Fe/3 mol CO = 4.13 mol Fe

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Heat never travels from a cooler object to a warmer object.
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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

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3 years ago
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