What is the smallest packet of electromagnetic energy that can be absorbed or emitted?
2 answers:
Answer:
<em><u>Photons</u></em> is the smallest packet of electromagnetic energy that can be absorbed or emitted.
Explanation:
<u>Photons</u>
The energy packets of the electromagnetic radiation that can be either absorbed or emitted.these are particles which transmit light into space.
There energy of the photon is given by Planck's equation:
where,
E = energy of photon = ?
h = Planck's constant =
c = speed of light =
= frequency of the photon
= wavelength light emitted by photon
You might be interested in
The elements are arranged by their atomic number but are also grouped by what type of element they are.
For example, metals, non metals, metalloids, and noble gasses will be grouped with elements that match them.
Hope this makes sense!
For example, metals, non metals, metalloids, and noble gasses will be grouped with elements that match them.
Hope this makes sense!
Answer:
You must remove
.
Explanation:
There are three heat transfers in this process:
Total heat = cool the vapour + condense the vapour + cool the liquid
q = q₁ + q₂ + q₃
q = nC₁ΔT₁ + nΔHcond + nC₂ΔT₂
Let's calculate these heat transfers separately.
Data:
You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.
C₁ = specific heat capacity of vapour = 90 J·K⁻¹mol⁻¹
C₂ = specific heat capacity of liquid = 115 J·K⁻¹mol⁻¹
ΔHcond = -38.56 kJ·mol⁻¹
Tmax = 300 °C
b.p. = 78.4 °C
Tmin = 25.0 °C
n = 0.782 mol
Calculations:
ΔT₁ = 78.4 - 300 = -221.6 K
q₁ = 0.782 × 90 × (-221.6) = -15 600 J = -15.60 kJ
q₂ = 0.782 × (-38.56) = -30.15 kJ
ΔT = 25.0 - 78.4 = -53.4 K
q₃ = 0.782 × 115 × (-53.4) = -4802 J = 4.802 kJ
q = -15.60 - 53.4 - 4.802 = -50.6 kJ
You must remove of heat to convert the vapour to a gas.