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Semmy [17]
3 years ago
12

HELP 50 POINTS WILL BE AWARDED I need 3-5 sentences that explains the importance of reaction types (combustion decomposition syn

thesis single/double replacement) in determining products
Chemistry
1 answer:
Fofino [41]3 years ago
6 0

Explanation:

Combustion reaction: Type of a chemical reaction in which compound and an oxidant reacts together to form a new product with release of an energy in the form of heat.

Decomposition reaction: Type of a chemical reaction in which compound breaks down into two or more elements or into new compounds.

ABC →  A + B + C

Synthesis reaction: Type of a chemical reaction in which two or more than two reactant reacts together to form a single product.

A + B + C  →  ABC

Single replacement reaction: Type of a chemical reaction in which element reacts with a compound and takes the place of another element in the compound.

A + BC → BA + C

Double replacement reaction: Type of chemical reaction in which two compounds reacts is such a way that the cations and anions of the compounds exchange their places to form new products.

AB + CD  →  AD + CB

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A balloon contains 2.0 L of air at 101.5 kPa. You squeeze the balloon to a volume of .75 L. What is the pressure of air on the i
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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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