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Nataliya [291]
2 years ago
8

More active metals will cause the reduction of less active metals. Less active metals will cause no reaction (N.R.) in more acti

ve metals. What is the complete and balanced reaction of solid iron with aqueous zinc(II)? Use the following activity series to arrive at your answer:

Chemistry
1 answer:
beks73 [17]2 years ago
3 0

Activity of metals with most active and less active metals are given below.

Explanation:

1. Activity of metals -Divide metals Based on the activity.

2. The primary difference between metals is the ability with which they undergo chemical reactions. The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive. Lithium, sodium, and potassium all react with water, for example. The rate of this reaction increases as we go down this column, however, because these elements become more active as they become more metallic.

3. Common Metals Divided into Classes on the Basis of Their Activity

  • Class I Metals: The Active Metals -Li, Na, K, Rb, Cs (Group IA)  ,Ca, Sr, Ba (Group IIA)
  • Class II Metals: The Less Active Metals -Mg, Al, Zn, Mn
  • Class III Metals: The Structural Metals -Cr, Fe, Sn, Pb, Cu
  • Class IV Metals: The Coinage Metals -Ag, Au, Pt, Hg

4. The most active metals are so reactive that they readily combine with the O2 and H2O vapor in the atmosphere and are therefore stored under an inert liquid, such as mineral oil. These metals are found exclusively in Groups IA and IIA of the periodic table.

5. Metals in the second class are slightly less active. They don't react with water at room temperature, but they react rapidly with acids.

6.The third class contains metals such as chromium, iron, tin, and lead, which react only with strong acids. It also contains even less active metals such as copper, which only dissolves when treated with acids that can oxidize the metal.

7. Metals in the fourth class are so unreactive they are essentially inert at room temperature. These metals are ideal for making jewelry or coins because they do not react with the vast majority of the substances with which they come into daily contact. As a result, they are often called the "coinage metals."

Fe⁺² (aq)+Zn(s)=>Zn⁺² (aq)+Fe(s)

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2 years ago
Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
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Answer : The value of K_{eq} is, 11.2

The value of \Delta G_{rxn} is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

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

where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

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

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol

Therefore, the value of \Delta G_{rxn} is -9.04 kJ/mol

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2 years ago
Your task is to measure the amount of energy evolved during the combustion of some hydrocarbon. Which of the following would be
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Answer:

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The coffee-cup calorimetry is used to measure the heat of a dissolution reaction and the bomb manometry is used to measure the pressure.

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If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/
riadik2000 [5.3K]

<u>Answer:</u> The energy of the complex is 2.72\times 10^2kJ

<u>Explanation:</u>

To calculate the energy of the complex, we use the equation given by Planck which is:

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where,

\lambda = Wavelength of the complex = 440nm=4.40\times 10^{-7}m    (Conversion factor:  1m=10^9nm )

h = Planck's constant = 6.624\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

N_A = Avogadro's number = 6.022\times 10^{23}

\Delta E = energy of the complex

Putting values in above equation, we get:

\Delta E=\frac{6.022\times 10^{23}\times 6.624\times 10^{-34}\times 3\times 10^8}{4.40\times 10^{-7}}\\\\\Delta E=2.72\times 10^{5}J=2.72\times 10^2kJ

Conversion factor used:  1 kJ = 1000 J

Hence, the energy of the complex is 2.72\times 10^2kJ

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