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.
Regards!
Answer:
Atom has the tendency to attain stable electronic configuration of the nearest noble gas in accordance with the duplet and octet rule.hence,when an atom gains electron it becomes negatively charged and that is its negative charge making it an anion
Answer:
c. 2.16 × 10^8 kJ
Explanation:
In the given question, 2 C-12 nuclei were used for the reaction and the mass of C-12 is 12.0000 amu. Therefore, for 2 C-12 nuclei, the mass is 2*12.0000 = 24.0000 amu.
In addition, a Na-23 and a H-1 were formed in the process. The combined mass of the products is 22.989767+1.007825 = 23.997592 amu
The mass of the reactant is different from the mass of the products. The difference = 24.0000 amu - 23.997592 amu = 0.002408 amu.
Theoretically, 1 amu = 1.66054*10^-27 kg
Thus, 0.002408 amu = 0.002408*1.66054*10^-27 kg = 3.99858*10^-30 kg
This mass difference is converted to energy and its value can be calculated using:
E = mc^2 = 3.99858*10^-30 *(299792458)^2 = 3.59374*10^-13 J
Furthermore, 1 mole of hydrogen nuclei contains 6.022*10^23 particles. Thus, we have:
E = 3.59374*10^-13 * 6.022*10^23 = 2.164*10^11 J = 2.164*10^8 kJ
Answer:
38.4 atm
Explanation:
Data obtained from the question include:
V1 (initial volume) = 3200 L
P1 (initial pressure) = 3.00 atm
V2 (final volume) = 250.0 L
P2 (final pressure) = ?
Using Boyle's law equation P1V1 = P2V2, the final pressure can be obtained as follow:
P1V1 = P2V2
3 x 3200 = P2 x 250
Divide both side by 250
P2 = 3 x 3200/250
P2 = 38.4 atm
Therefore, the pressure of the gas if ethylene is supplied by a 250.0 L tank is 38.4 atm
