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vesna_86 [32]
3 years ago
9

Calculate the standard potential, e∘, for this reaction from its equilibrium constant at 298 k. x(s)+y4+(aq)↽−−⇀x4+(aq)+y(s)k=4.

98×10−5
Chemistry
1 answer:
laila [671]3 years ago
6 0
First, we need to get the number of moles:

from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+  

∴ the number of moles n = 4

we are going to use this formula:

㏑K = n *F *E/RT

when K is the equilibrium constant = 4.98 x 10^-5

and F is Faraday's constant = 96500

and the constant R = 8.314

and T is the temperature in Kelvin = 298 K

and n is number of moles of electrons = 4 

so, by substitution:

㏑4.98 x 10^-5 = 4*96500*E / 8.314*298

∴E = -0.064 V
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Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

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Thus we have

ΔH = m·c·(T₂ - T₁) where  ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C  and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c  or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

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3 years ago
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Answer:

2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}

Explanation:

Hello there!

In this case, according to the rules for the oxidation states in chemical reactions, it is possible to realize that lone elements have 0 and since magnesium is in group 2A, it forms the cation Mg⁺² as it loses electrons and oxygen is in group 6A so it forms the anion O⁻²; therefore resulting oxidation numbers are:

2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}

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