Answer: The Kelvin scale is related to the Celsius scale. The difference between the freezing and boiling points of water is 100 degrees in each, so that the kelvin has the same magnitude as the degree Celsius.
Explanation:
Celsius is, or relates to, the Celsius temperature scale (previously known as the centigrade scale). The degree Celsius (symbol: °C) can refer to a specific temperature on the Celsius scale as well as serve as a unit increment to indicate a temperature interval(a difference between two temperatures or an uncertainty). “Celsius” is named after the Swedish astronomer Anders Celsius (1701-1744), who developed a similar temperature scale two years before his death.
K = °C + 273.15
°C = K − 273.15
Until 1954, 0 °C on the Celsius scale was defined as the melting point of ice and 100 °C was defined as the boiling point of water under a pressure of one standard atmosphere; this close equivalence is taught in schools today. However, the unit “degree Celsius” and the Celsius scale are currently, by international agreement, defined by two different points: absolute zero, and the triple point of specially prepared water. This definition also precisely relates the Celsius scale to the Kelvin scale, which is the SI base unit of temperature (symbol: K). Absolute zero—the temperature at which nothing could be colder and no heat energy remains in a substance—is defined as being precisely 0 K and −273.15 °C. The triple point of water is defined as being precisely 273.16 K and 0.01 °C.
Answer:
C-12 or C with a 12 superscripted on Upper left and 6 Subscripted on bottom left
Explanation:
Isotopic notation
Answer:
Phase changes that require a loss in energy are condensation and freezing.
Explanation:
Explanation:
The given data is as follows.
Concentration of solution = 0.5 M
Volume of solution = 1 L
Molar mass of Glycylglycine = 132.119 g/mol
As molarity is the number of moles present in liter of solvent.
Mathematically, Molarity = 
Hence, calculate the number of moles as follows.
No. of moles = Molarity × Volume
= 
= 0.5 mol
Therefore, mass of glycylglycine = mol × molar mass
= 
= 66.06 g
Thus, we can conclude that 66.06 g glycylglycine is required.
