<span>
<span> 250g H2SO4 x 1moleH2SO4/ 98g mole H2SO4 x 4moles Oxygen/1moleH2SO4 x 1moleOxgen/16g Oxygen
= 163.27g Oxygen
my figures aren't exact....so it's 160g </span>
</span>
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
Answer: b} The exact time when an individual atom will decay can be accurately predicted.
c} After each half-life, the amount of radioactive material is reduced by half.
Explanation:
All radioactive decay follows first order kinetics.
Rate law expression for first order kinetics is given by:
where,
k = rate constant
t = time taken for decay process
a = initial amount of the reactant
a - x = amount left after decay process
Expression for calculating half life, which is the time taken by the half of the reactants to decompose is:
