Answer:
D. Grams liquid x mol/g x delta Hfreezing
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to reason that the stoichiometry used to calculate energy released when a mass of liquid freezes, involves the grams of the liquid, the molar mass of the liquid, as given in all the group choices, and the enthalpy of freezing because that is the process whereby a liquid goes solid.
In such a way, we infer that the correct factor would be D. Grams liquid x mol/g x delta Hfreezing which sometimes is the negative of the enthalpy of fusion as they are contrary processes.
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B is the answer I think better sure
Answer:
pV= nRT
Explanation:
(p1 × V1)/ T1/ (p2 × V2)/ T2
Answer:
stay the same.
Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):
Na: 1s2 2s2 2p6 3s1
Ar: s2 2s2 2p6 3s2 3p6
As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:
Na: 3s1
Ar: 3s2, 3p6
All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.
Answer:
the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.
Explanation:
assuming ideal gas behaviour:
PV=nRT
therefore
P= 109 Kpa= 1.07575 atm
V= 67 m3/hr = 18.6111 L/s
T= 215 °C = 488 K
R = 0.082 atm L /mol K
n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)
n= 0.5 mol/s
since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:
Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW
