Explanation:
The experiment is performed at a constant atmospheric pressure. The experiment proceeds by placing an empty flask in a boiling water bath. As the temperature increases, the air inside the flask expands. Afterwards, the gas is cooled in a water bath by maintaining the amount of the air in the flask constant.
This is called adaptation. An example would be a polar bear with white fur. They have this to be more efficent hunters, as they use it as a type of camouflage in order to sneak up on prey. Brown bear use this in the same way but their environment requires them to be brown in order to blend in beter with trees. So basicly adaptation
Answer:
Specific heat of metal = 0.26 j/g.°C
Explanation:
Given data:
Mass of sample = 80.0 g
Initial temperature = 55.5 °C
Final temperature = 81.75 °C
Amount of heat absorbed = 540 j
Specific heat of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 81.75 °C - 55.5 °C
ΔT = 26.25 °C
540 j = 80 g × c × 26.25 °C
540 j = 2100 g.°C× c
540 j / 2100 g.°C = c
c = 0.26 j/g.°C
Answer: 
Explanation:
Lattice energy : It is defined energy released when ions combine together in a gaseous phase to form a compound. it is energy possessed by the crystal lattice of a compound. Denoted by symbol 
.
, energy is absorbed while forming of the lattice
, energy is release while forming of the lattice
One mole of sodium ion when combines with one mole chloride ion release 786 kJ of energy.
Answer:
The root mean square speed of O2 gas molecules is
<u>519.01 m/s</u>
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Explanation:
The root mean square velocity :
Molar mass , M
For He = 4 g/mol
For O2 = 2 x 16 = 32 g/mol
O2 = 32/1000 = 0.032 Kg/mol
First calculate the temperature at which the K.E of He is 4310J/mol
K.E of He =
K.E of He = 4310 J/mol
<u>Now , Use Vrms to calculate the velocity of O2</u>
