Answer:
t = 7.58 * 10¹⁹ seconds
Explanation:
First order rate constant is given as,
k = (2.303
/t) log [A₀]
/[Aₙ]
where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>
[A₀] = 615 calories;
[Aₙ] = 615 - 480 = 135 calories
k = 2.00 * 10⁻²⁰ sec⁻¹
substituting the values in the equation of the rate constant;
2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)
(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)
t = 2.303 / 3.037 * 10⁻²⁰
t = 7.58 * 10¹⁹ seconds
Answer:
The correct answer is because the molecular structure.
Explanation:
The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.
In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.
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Answer:
30.33L
Explanation:
Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant and Charles law states that the volume of a given mass of gas is directly proportional to the temperature provided the pressure remains constant
P1V1/T1 = P2V2/ T2
P1 = 78atm, V1 = 21L , T1 = 900K
P2 = 45atm, V2 = ? , T2 =750K
78× 21 / 900 = 45×V2 / 750
1638/900 = 45 V2 / 750
1638×750 = 900×45V2
1228500 = 40500V2
Divide both sides by 40500
1228500÷40500= V2
V2 = 30.33L
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answer:
This allows us to tell alkenes apart from alkanes using a simple chemical test. Bromine water is an orange solution of bromine. It becomes colourless when it is shaken with an alkene. Alkenes can decolourise bromine water, but alkanes cannot.
explanation:
