All elements in their standard states (oxygen<span> gas, solid carbon in the form of graphite, etc.) have a standard </span>enthalpy of formation<span> of </span>zero<span>, as there is no change involved in their </span>formation<span>.</span>
Answer:
A. 28 years
Explanation:
Applying,
R = R'(2ᵃ/ⁿ).............. Equation 1
Where R = Original sample, R' = Sample left after decay, a = Total time taken to decay, n = half life.
From the question,
Given: R = 12 g, R' = 6 g, a = 28 years.
Substitute into equation 1 and solve for n
12 = 6(2²⁸/ⁿ)
12/6 = 2²⁸/ⁿ
2²⁸/ⁿ = 2
Equation the base,
28/n = 1
n = 28 years.
Hence the half-life is 28 years
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.
Because during combustion reaction, heat energy is released and it's this energy that is converted to work
