Delta enthalpy = 2x386-3x1x432-3x942=-3350kJ/mol
there is more air near sea level
Answer:
14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.
Explanation:
The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.
During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.
So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.
In this case, being:
- L= 84

and replacing in the expression Q = m*L you get:
Q=172 g*84 
Q=14,448 J
<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Could I hear the options..