The experimental absolute zero value is less when compared to the accepted value of absolute zero.
<h3>What is absolute zero?</h3>
Absolute zero is defined as the temperature in which the lowest energy possible is attained in a thermodynamic system.
Absolute zero temperature has an accepted values of 0 Kelvin or -273.15 degrees Celsius.
At absolute zero, it is assumed that the volume of an ideal gas becomes zero. However, it has not been possible to cool any gas to absolute zero.
Based on the graph of temperature against volume of gases, the experimental absolute zero extrapolated from the graph where volume of the gases becomes zero is -285 degrees Celsius.
Therefore, the experimental absolute zero value is less when compared to the accepted value.
Learn more about absolute zero at: brainly.com/question/1191114
#SPJ1
ANSWER:
What is the measured component of the orbital magnetic dipole moment of an electron with the values
(a) ml=3
(b )
ml= −4
a) -278 x 
J/T
b) 3.71 x J/T
STEP-BY-STEP EXPLANATION:
a) ml= 3
Цorb,z = ml Цв = - (3) * (9.27e - 24) = -278 x J/T
b) ml= 3
Цorb,z = ml Цв = - (-4) * (9.27e - 24) = 3.71 x J/T
Answer:
both spheres have a positive charge
Change in velocity over time is acceleration.
You're finding acceleration .
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
