The standard heats of combustion (δh∘) per mole of 1,3-butadiene, c4h6(g); butane, c4h10(g); and h2(g) are −2540.2,−2877.6, and −285.8kj, respectively. use these data to calculate the heat of hydrogenation of 1,3-butadiene to butane. c4h6(g)+2h2(g)→c4h10(g)

solution:

Hydration is the addition of water; hydrogenation is the addition of hydrogen.  

desire rxn: _C4H6(g) + 2 H2(g)> C4H10(g)___dHhy = ??  

knowns:  

C4H6 + 11/2 O2 > 4CO2 + 3H2OdHox = -2540.2 kJ/mole  

4CO2 + 5H2O > C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole  

2(1/2 O2 + H2 > H2O)2*dHox = 2(-285.8 kJ/mole)  

Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem

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

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."