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Effectus [21]
1 year ago
6

Consider the following voltaic cell. Identify the anode and cathode, and indicate the direction of Na ion and NO3- ion flow from

the salt bridge. A. Sn is the anode and Fe is the cathode; Na ions flow into half-cell compartment (A) and NO3- ions flow into half-cell compartment (B). B. Fe is the anode and Sn is the cathode; NO3- ions flow into half-cell compartment (A) and Na ions flow into half-cell compartment (B). C. Sn is the anode and Fe is the cathode; NO3- ions flow into half-cell compartment (A) and Na ions flow into half-cell compartment (B). D. Fe is the anode and Sn is the cathode; Na ions flow into half-cell compartment (A) and NO3- ions flow into half-cell compartment (B).
Chemistry
1 answer:
levacccp [35]1 year ago
4 0

Answer:

Fe is the anode and Sn is the cathode; NO3- ions flow into half-cell compartment (A) and Na ions flow into half-cell compartment (B).

Explanation:

The purpose of a salt bridge in a voltaic cell is not to move electrons from the electrolyte, it's main function is the maintenance of charge balance between the half cells .The electrons flow from the anode to the cathode.

The iron half cell which has a higher reduction potential serves as anode. Iron atoms give up electrons to form positive ions according to the equation;

Fe(s)-----> Fe^2+(aq) + 2e

Hence there will be an excess of positive ions in the anode (compartment A). NO3^- ions from the salt bridge will migrate towards this compartment to ensure charge balance.

Similarly, in compartment B(cathode) Sn^2+ ions will accept two electrons according to the reaction;

Sn^2+(aq) + 2e ------> Sn(s)

This reaction causes a depletion of positive ions in compartment B, hence Na^+ from the salt bridge move towards this compartment to ensure charge balance.

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Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0
ExtremeBDS [4]

Answer:

The freezing point will be 2.9^{0}C

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

Depressioninfreezing point=K_{f}Xmolality

Where

Kf= 20\frac{^{0}C}{m}

the molality is calculated as:

molality=\frac{moles_{solute}}{mass_{solvent}}

moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol

massofcyclohexane=25g=0.025Kg

molarity=\frac{0.0045}{0.025}=0.18m

Depression in freezing point = 20X0.18=3.6^{0}C

The new freezing point = 6.5^{0}C-3.6^{0}C=2.9^{0}C

5 0
2 years ago
The temperature at the boiling point remained constant despite the continued addition of heat by the bunsen burner. What was the
Paul [167]

Concept:

<em><u>Latent Heat of Vaporization</u></em>: It is defined as the amount of heat required to change the state of mater without changing of its temperature.

From the given question, the temperature at the boiling point remained constant despite the continued addition of heat by the Bunsen burner. <em>Actually,</em> this amount of heat is used by water to break the intermolecular bonds between the water molecules in the form of latent heat that converts the liquid state of water into vapor state of water.

Hence, the correct option will be d.<u>The energy was used to break the intermolecular bonds between the water molecules. </u>

3 0
2 years ago
At 10 K Cp,m(Hg(s)) = 4.64 J K−1 mol−1. Between 10 K and the melting point of Hg(s), 234.3 K, heat capacity measurements indicat
umka21 [38]

Answer:

S°m,298K = 85.184 J/Kmol

Explanation:

∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol

∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol

∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol

∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol

⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)

⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol

⇒ S°m,298K = 85.184 J/Kmol

4 0
2 years ago
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K
Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0
1 year ago
A group of students measured the average monthly temperature of 1000 cities around the world and plotted the cumulative frequenc
mr Goodwill [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

The deviations will be added then.

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

The deviations will be added then.

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

8 0
1 year ago
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