Products such as antifreeze are composed of organic compounds that are classified as <em>alcohols</em>. (a)
Maybe those other classes of chemicals also lower the freezing temperature of water, just like alcohol does. I don't know. But alcohol is what's used to make anti-freeze. I'm guessing alcohol must be cheaper, less toxic, and less corrosive inside the engines' cooling systems than any of that other stuff is.
Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
<span>Heating food under a heat lamp is an example of heat transfer by
<span>Radiation</span></span>
Answer:
henry moseley was an english physicist, whose contribution to the science of physics was the justification from physical laws of the previous chemical concept of the atomic number. one of his developments were of moseley's law in x-ray spectra.
Explanation:
To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:
Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to
Recall that the speed of light is equivalent to
Then replacing,
Our values are given as
Replacing we have,
Therefore the magnetic field around this circular area is
