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Luda
8 days ago
9

Charlize wants to measure the depth of an empty well. She drops a ball from a height of 3 feet into the well

Mathematics
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day and night kennel charges $20 per day plus a food fee of $15 to board a pet. home away from home kennel charges $30 per day p
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Let n denote the number of days and C the total cost; Charges by Day and Night kennel, C=15+20n Charges by Home Away from Home; C=5 + 30n The resulting system of equations would be; C=15+20n C=5+30n
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The volume of water remaining in a hot tub when it is being drained satisfies the differential equation dV/dt = −3(V)^1/2 , wher
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The provided equation is a differential equation that allows for variable separation. You should group similar terms, integrate, use the correct limits, and present V as a function of t. This is achieved in the following manner: dV/dt = -3(V)^1/2, which rearranges to dV/-3V^1/2 = dt. Initially, when V equals 225, after integration, we arrive at -2/3(√V - √225) = t, which can be further detailed as -2/3(√V - 15) = t. This represents the function for V at a specific time t. I trust this information is helpful, have a pleasant day.
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The solutions to the inequality y ≤ −x + 1 are shaded on the graph. Which point is a solution?
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1 month ago
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What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re
lawyer [12517]

Answer:

4.0921 reflects the logarithm of the equilibrium constant.

Step-by-step explanation:

Fe^{2+} (aq) +2e^{-}\rightarrow Fe(s); ​E​° = - 0.41 V

Ag^+(aq) + e^-\rightarrow Ag(s); E° = 0.80 V

Iron has a negative reduction potential, indicating its tendency to lose electrons and undergo oxidation, and thus it will be at the anode.

E^{o}_{cell}=Reduction potential of cathode - Reduction potential of anode

E^{o}_{cell}=E^{o}_c-E^{o}_a

=0.80 V-(-0.41 V)=1.21 V

Fe^{2+} (aq) + 2e^{-}\rightarrow Fe(s); ​E​° = - 0.41 V

2Ag^+(aq) + 2e^-\rightarrow 2Ag(s); E° = 0.80 V

Net reaction: Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)

n = 2

To determine the equilibrium constant, we utilize the correlation with Gibbs free energy, as follows:

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

and,

\Delta G^o=-RT\ln K_{eq}

Aligning these two equations yields:

nfE^o_{cell}=RT\ln K_{eq}

where,

n = electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 1.21 V

R = gas constant = 8.314 J/K.mol

T = reaction temperature = 25^oC=[273+25]=298K

Substituting values into the equation, we arrive at:

2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}

\ln K_{eq}=9.3478

\log K_{eq}=\frac{9.3478}{2.303}=4.0921

4.0921 represents the logarithm of the equilibrium constant.

7 0
2 months ago
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