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Otrada [13]
3 years ago
9

PLEASE PLEAS HELP Which of the following compounds is insoluble in water?

Chemistry
1 answer:
Alex17521 [72]3 years ago
7 0

Answer:

Your answer is d

Explanation: silver carbonate Ag2CO3 is insoluble in water

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The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.
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What is the concentration of 60 mL of H3PO4 if it is neutralized by 225 mL of 2 M Ba(OH)2?
goldfiish [28.3K]

7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.

Explanation:

Data given:

volume of phosphoric acid, Vacid =60 ml

volume of barium hydroxide, Vbase = 225 ml

molarity of barium hydroxide, Mbase = 2M

Molarity of phosphoric acid, Macid =?

the formula for titration is used as:

Macid x Vacid = Mbase x Vbase

rearranging the equation to get Macid

Macid = \frac{Mbase x Vbase}{Vacid}

Macid =\frac{225 X 2}{60}

Macid = 7.5 M

the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.

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3 years ago
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In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined
faltersainse [42]

<u>Answer:</u>

<u>For A:</u> The standard cell potential of the reaction is 4.4 V

<u>For B:</u> The standard Gibbs free energy of the reaction is -8.50\times 10^5J

<u>For C:</u> The reaction is spontaneous as written.

<u>Explanation:</u>

  • <u>For A:</u>

The given chemical reaction follows:

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

The given half reaction follows:

<u>Oxidation half reaction:</u>  Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V ( × 2)

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

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

E^o_{cell}=1.36-(-3.04)=4.4V

Hence, the standard cell potential of the reaction is 4.4 V

  • <u>For B:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

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

where,

n = number of electrons transferred = 2mol\text{ e}^-

F = Faradays constant = 96500J/V.mol\text{ e}^-

E^o_{cell} = standard cell potential = 4.4 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J

Hence, the standard Gibbs free energy of the reaction is -8.50\times 10^5J

  • <u>For C:</u>

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

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