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icang [17]
3 years ago
15

The diagram below shows what happens to an atom of uranium-235 when bombarded by a neutron. Which isotope is represented by X in

the diagram?

Chemistry
1 answer:
Margaret [11]3 years ago
8 0

Answer:

X represents barium - 141 nucleus.

Written as: (141, 56) Ba

Explanation:

In nuclear fission reaction, due to the fact that the process produces more neutrons, a nuclear chain reaction can possibly occurr. Now, a chain reaction is a one whereby the material that begins the reaction will also be one of the products and it can start another reaction.

Now, for the nuclear chain reaction for the fission of uranium-235, its nucleus absorbs a neutron after which it splits into a krypton - 92 nucleus and a barium - 141 nucleus. Thereafter it releases three more neutrons.

Thus, X is the image represents barium - 141 nucleus.

(141,56) Ba

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The phenomenon that causes air masses and water to bend or curve is the ______________.
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The Coriolis effect

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Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
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}

5 0
3 years ago
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