Answer:
The Dehydrohaogenation of 1-bromo propane with alcoholic KOH gives propene which on again hydrohalogenation with HBr gives 2-bromo propane due to Markonikove's rule for addition.
Explanation:
Answer:
Option A.
Lower air pressure results in a lower boiling point
Explanation:
This is because in an open system, the lower the pressure the lesser the energy that will be required for boiling point. The is little or no collision of air molecules with the surface of the liquid
But if there is increase in pressure, more energy will be required to get to boiling point because there will be strong collision between air molecules and surface of the liquid.
The most appropriate and most commonly used metric when describing a person's mass is D.) kilograms
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Answer:
ΔH3 = 1/2 (629) - ΔH^0
Explanation:
Given data:
Bond energy of H2 = ΔH1 = 436 Kj/mol
Bond energy of Br2 = ΔH2 = 193 Kj/mol
To find:
Let bond energy of HBr = ΔH3 = ?
Equation:
H2 + Br2 → 2HBr
enthalpy of formation of HBr = ΔH1 + ΔH3 - 2(ΔH3)
ΔH^0 = 436 + 193 - 2(ΔH3)
(436 + 193) - ΔH^0 = 2(ΔH3)
ΔH3 = 1/2 (629) - ΔH^0
It is a radio wave
Explanation:
The energy of an electromagnetic waves is related to its frequency by the equation:
where
h is the Planck constant
f is the frequency of the wave
We notice that the energy of an electromagnetic wave is proportional to its frequency. This means that waves with higher frequencies, such as gamma rays, x-rays, are more energetic, while waves with lower frequencies, such as microwave and radio waves, are the less energetic.
In this case, we want to find the frequency of a wave with energy
E = 0.000001 eV
Converting into Joules,
Solving for the frequency,
Which falls in the range of frequency of radio waves.
Learn more about electromagnetic waves:
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