Answer:
The heat capacity and the specific heat are related by C=cm or c=C/m. The mass m, specific heat c, change in temperature ΔT, and heat added (or subtracted) Q are related by the equation: Q=mcΔT. Values of specific heat are dependent on the properties and phase of a given substance.
Explanation:
This is the full question: what is the strongest intermolecular force in a liquid containing molecules with nonpolar bonds?
A. Covalent Bonds
B. Dispersion Forces
C. Hydrogen Bonds
D. none of these
This is the answer: B. Dispersion forces
Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
Answer: it's part of the immune system that focuses on specific foreign particles. Rather than generically attack any antigens, T cells circulate until they encounter their specific antigen. As such, T cells play a critical part in immunity to foreign substances !
Explanation:
Answer:
exothermic entropy is increased
Explanation:
An exothermic process is one whose rate increases when the temperature is decreased. Hence if a decrease in temperature favours the dissolution of more solute at equilibrium, then the process is exothermic.
Similarly, the dissolution of a solute in a solvent increases the disorderliness (entropy) of the system because of the increase in the number of particles present. Hence once a solute in dissolved, the entropy of the system increases.
