HELP ITS DUE SOON PLEASEE Identify the type of chemical reaction for each of the following:

SnO2(aq) + 2H2(g) → Sn(s) + 2H2O(g)

AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)

2Fe(s) + 3Br2(l) → 2FeBr3(s)

3H2SO4(aq) + 2Fe(s) → Fe2(SO4)3(aq) + 3H2(g)

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

4Al(s) + 3O2(g) → 2Al2O3(s)

2KI(l) → 2K(s) + I2(s)  

3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)

The options are:
Single replacement
Double replacement
Decomposition
Formation
Hydrocarbon combustion

SnO2(aq) + 2H2(g) → Sn(s) + 2H2O(g) - Redox reaction

AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq) - double replacement reaction

2Fe(s) + 3Br2(l) → 2FeBr3(s) - Formation

3H2SO4(aq) + 2Fe(s) → Fe2(SO4)3(aq) + 3H2(g) - Single replacement

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g) - Hydrocarbon combustion

4Al(s) + 3O2(g) → 2Al2O3(s) - Formation

2KI(l) → 2K(s) + I2(s) - Decomposition

3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq) - Double replacement

Explanation:

In an oxidation - reduction reaction, the oxidation number of the species involved in the reaction changes from left to right. One specie is oxidized while the other is reduced. E.g in SnO2(aq) + 2H2(g) → Sn(s) + 2H2O(g) - Sn^2+ was reduced while H2 was oxidized.

For a single replacement reaction, one specie is replaced in the reaction e.g H^+ is replaced by Fe^2+ in the reaction, 3H2SO4(aq) + 2Fe(s) → Fe2(SO4)3(aq) + 3H2(g) .

In a double replacement reaction, the ions on the reactant side exchange partners at the product side e.g 3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)

In a formation reaction, two reactants combine to form a product. E.g 2Fe(s) + 3Br2(l) → 2FeBr3(s)

In a decomposition reaction, a compound breaks down to yield its components e.g 2KI(l) → 2K(s) + I2(s)

Hydrocarbon combustion refers to the reaction of a hydrocarbon compound with oxygen e.g 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

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