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3 years ago

State boyl"s law and Charle's law

Chemistry

1 answer:

zubka84 [21]3 years ago

6 0

Answer:

hey mateee

Boyle's law :- the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature.

Boyle's law can also be formularized as P1V1 = P2V2

Charle's law :- the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

Charle's law can also be formularized as V1/T1 = V2/T2

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Katarina [22]

Answer:

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3 years ago

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VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

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

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

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

T = temperature = 298 K

K = equilibrium constant = ?

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

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

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Which atom is smaller? (1 point) A. Hydrogen B. Oxygen

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3 years ago

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aleksley [76]

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Explanation:

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3 years ago

State boyl"s law and Charle's law​

State boyl"s law and Charle's law