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Tanzania [10]
3 years ago
10

The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his

original periodic table because he suspected the atomic weights for these elements were inaccurate. What was Mendeleev's reason for this suspicion?
Chemistry
1 answer:
QveST [7]3 years ago
3 0

Explanation :

As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.

He placed the elements with similar nature in the same group.

According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.

However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.

As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127  and tellurium isotopes are tellurium-128 and tellurium-130.

Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.

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The water-gas shift reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is used industrially to produce hydrogen. The reaction enthalpy is ΔH∘=−4
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Answer:

The answers are in the explanation

Explanation:

A. For the reaction:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g);  ΔH°=−41kJ.

As the reaction is exothermic ( ΔH°<0), you need to use low temperature to increase the equilibrium yield of hydrogen -LeChatelier's principle-.

We would use <em>low </em>temperature. For an <em>exothermic </em>reaction such as this, <em>decreasing </em>temperature increases the value of K and the amount of products at equilibrium.

B.

c. No. We cannot increase the equilibrium yield of hydrogen by controlling the pressure of this reaction.

It is possible to increase the equilibrium yield of reaction by controlling the amount of reactants added. As reactants and products are gases, the pressure of the reaction will not change the amount of reactants or products in the equilibrium.

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8 0
3 years ago
An unknown compound is processed using elemental analysis and found to contain 117.4g of platinum 28.91 carbon and 33.71g nitrog
dlinn [17]

Answer:

1 mole of platinum

Explanation:

To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.

The empirical formula for the compound can be obtained as follow:

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Nitrogen (N) = 33.71 g

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Pt = 117.4 / 195 = 0.602

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Pt = 0.602 / 0.602 = 1

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The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

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8 0
3 years ago
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.

7 0
4 years ago
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