Answer:
the energy of matter implies the motion of masses and the energy of the wave has no moving masses
The energy of the particles (matter) is the ability to do some work, therefore the energy can be of motion in the form of kinetic energy or in a configuration of the system called potential energy, the sum of these two energy is constant .
The wave is formed by a disturbance of the medium by matter, therefore for the formation of the wave matter supply energy, for example: in the form of movement, in the form of fluctuation of electric or magnetic field, etc.
The waves after being formed can move away from the matter that formed them, transporting the energy that alternately has kinetic and potential energy, but the total energy is constant.
Therefore the energy in matter is due to the movement of the same and the energy in the wave does not require the movement of matter, so it is a more efficient way of doing work.
In conclusion, the energy of matter implies the movement of masses and the energy of the wave has no moving masses.
Answer:
Here's what I get.
Explanation:
The name tells me the compound is a lactone (a cyclic ester).
1. IR spectrum
1770 cm⁻¹
Esters and unstrained lactones normally absorb at 1740 cm⁻¹.
This peak is shifted to a higher frequency by ring strain.
A five-membered lactone absorbs at 1765 cm⁻¹, and a four-membered lactone at 1840 cm⁻¹.
The compound is probably a five-membered lactone.
2. NMR spectrum
2.28 m (2H)
2.48 t (2H)
4.35 t (2H)
This indicates three CH₂ groups arranged as X-CH₂-CH₂-CH₂-Y.
The X-CH₂- and -CH₂-Y signals would each be triplets, being split by the central -CH₂- group.
The central -CH₂- signal would be a multiplet, split by the non-equivalent hydrogens on either side.
The peak at 4.35 ppm indicates that the group is adjacent to an oxygen atom ( -CH₂- = 1.3; -CH₂-O- = 3.3 - 4.5).
The peak at 2.42 ppm indicates that the group is adjacent to a carbonyl group (-CH₂-C=O = 1.8 - 2.5.
The only way to fit these pieces together is if γ-butyrolactone has the structure shown below.
Confirmation:
(a) The IR spectrum shows a carbonyl peak at 1770 cm⁻¹.
(b) The NMR spectrum matches that given in the problem.
M(O) = m(CaCO₃) - m(Ca) - m(C).
m(O) = 36.41 g - 14.58 g - 4.36 g.
m(O) = 17.47 g.
ω(Ca) = m(Ca) ÷ m(CaCO₃) · 100%.
ω(Ca) = 14.58 g ÷ 36.41 g · 100%.
ω(Ca) = 40 %; mass percent of calcium.
ω(C) = m(C) ÷ m(CaCO₃) · 100%.
ω(C) = 4.36 g ÷ 36.41 g · 100%
ω(C) = 12%; mass percent of carbon.
ω(O) = 100% - ω(Ca) - ω(C).
ω(O) = 100% - 40% - 12%.
ω(O) = 48%; mass percent of oxygen.
Answer:
At constant Pressure V /T = constant. sooooo
49 / 79 = 24.5 / T
T = 24.5 × 79/49 = 39.5°C
