The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.
Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:
p/T is constant ==> p_1 / T_1 = p_2/T_2
From which you obtain:
p_2 = [p_1 / T_1] * T_2
T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K
p_1 = 1014 kPa
p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa
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:
Explanation:
These instrument works on the analysis of the emisson spectral of light received from the star in this way.
Think of a steel knife in your kitchen. Initially, it has this shiny silver colour that typifies it. When the knife is placed on a hot plate, it becomes hotter and begins to go red as the heating continues. If we stop the heating and pour cold water on it, the red dissapears and our knife is back to itself, although the silvery shine would be lost. This is simply how the atomic absorption spectroscopy works. When you see the hot knife you can say a couple of things about it. Different metals have their various melting point. We can compare the temperature at which our knife will melt with a standard melting point scale to know the type of metal it is made of.
In atomic absorption spectroscopy, an atom gains energy and it becomes excited. Every atom is known to have a peculair amount of absorbant energy that cause them to excite. The more the particles in the atom, the more the energy required. When we analyse the absorbent energy of the atom, it differs from other atoms and we truly identify such an atom even if we don't know it. Most times, the energy is given off as light.
<span> Substance 1 was activated by the heat and inactivated by ice, Substance 2 was activated by the change in concentration, and Substance 3 increased the surface area.</span>
