Answer:
True?
Im sorry I have no clue...
Bose Einstein condensation occur at temperature very close to absolute zero i.e 273.15 degrees centigrade.. Under such conditions, large quantity of Boson occupies the lowest quantum state, at which point microscopic quantum phenomena becomes apparent macroscopically. A BEC is formed by cooling a gas of extremely low density about one hundred thousandth the density of normal air, to ultra low temperature.
The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:
CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol
As observed O2 will not have enthalpy of formation as it is a pure substance.
To calculate for the enthalpy of reaction,
enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
= 294.5 kJ/mol
ANSWER: 294.5 kJ/mol
Answer: 0.424 J/g°C
Explanation:
For this problem, we would have to manipulate the equaiton for heat, q=mCT. Specific heat is the C in the equation. Since we are looking for specific heat, we manipulate the equation so that it says C=.
*I didn't know how to type in delta so I just wrote the word delta, but pretend you see a Δ.
Now that we have our equation, we can plug in our values and solve.
*Please ignore the capital A in the equation. It pops up every time I type in the ° sign.