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lisabon 2012 [21]

2 years ago

7

A solid substance is an excellent conductor of electricity. the chemical bonds in this substances are most likely

2 answers:

PSYCHO15rus [73]2 years ago

7 0

The answer is (4) metallic, because the valence electrons are mobile

marishachu [46]2 years ago

6 0

Answer;

-4. metallic, because the valance electrons are mobile

Explanation;

-Electrical conductivity in metals is a result of the movement of electrically charged particles.The atoms of metal elements are characterized by the presence of valence electrons (electrons in the outer shell of an atom) that are free to move about.

-Therefore; Metallic elements such as sodium and potassium conducts electricity in solid form due to the presence of delocalized valence electrons. These electrons can move freely within the structure of a metal when an electric current is applied.

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Balance kclo3-kcl+02

jeka94

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2 years ago

Calculate the equilibrium constant for the reaction: 2 Cr + 3 Pb2+ ----> 3 Pb + 2 Cr3+ at 25oC. Eocell = 0.61 V

sattari [20]

Answer:

The value is $K = 8*10^{61}$

Explanation:

From the question we are told that

The equation is $2 Cr + 3Pb^{2+} \to 3Pb + 2Cr^{3+}$

The temperature is $T = 25^oC = 298 K [room \ temperature ]$

The emf at standard condition is $E^o_{cell} = 0.61 \ V$

Generally at the cathode

$3Pb^{2+}(aq) + 6 e- --> 3Pb(s)$

At the anode

$2Cr^{3+} + 6e^- \to 2Cr$

Generally for an electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as

$G = n* F * E^o_{cell}$

Here n is the no of electron with value n = 6

F is the Faraday's constant with value 96487 J/V

=> $G = 6 * 96487 * 0.61$

=> $G = 3.5 *10^{5} \ J$

This Gibbs free energy can also be represented mathematically as

$G = RTlogK$

Here R is the cell constant with value 8.314J/K

K is the equilibrium constant

From above

=> $K = antilog^{\frac{G}{ RT} }$

Generally antilog = 2.718

=> $K = 2.718^{\frac{3.5 *10^5}{ 8.314* 298} }$

=> $K = 8*10^{61}$

6 0

2 years ago

How many moles of ammonium ions are in 125 mL of 1.40 M NH4NO3 solution? ________ moles (give answer with correct sig figs in un

Sholpan [36]

The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:

Volume = 125 mL = 125 / 1000 = 0.125 L

Molarity = 1.40 M

<h3>Mole of NH₄NO₃ =? </h3>

Mole = Molarity x Volume

Mole of NH₄NO₃ = 1.40 × 0.125

<h3>Mole of NH₄NO₃ = 0.175 mole</h3>

Finally, we shall determine the number of mole of ammonium ion, NH₄⁺ in the solution. This can be obtained as follow:

NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)

From the balanced equation above,

1 mole of NH₄NO₃ contains 1 mole of NH₄⁺

Therefore,

0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺

Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

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