Well there's a couple reasons why. First, the land itself can become “tired" and less fertile<span>. This is because the same type of </span>crop<span> planted repeatedly in the same area keeps draining the land of the same nutrients needed for that plant's growth. Pest can also plant their nest near the crop if the farmer continues to plant the same crop in the same place.
Which is the reason why things such as "crop rotation" exist to keep things fresh, and prevent what I described above from happening.</span>
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)
The heat that is required to raise the temperature of an object is calculated through the equation,
heat = mass x specific heat x (T2 - T1)
Specific heat is therefore calculated through the equation below,
specific heat = heat / (mass x (T2 - T1))
Substituting,
specific heat = 645 J / ((28.4 g)(15.5 - - 11.6))
The value of specific heat from above equation is 0.838 J/g°C.
Answer: The strange liquid would float to the top of a cup of water.
Explanation:
Density = Mass/Volume
Strange Liquid Density = 70g/84mL
Strange Liquid Density = 0.833g/mL
Density of water in g/mL = 1 g/mL
Strange Liquid Density < Water Density
A substance with a lower density would be suspended above a substance with a higher density.
Since the density of the strange liquid is less than that of water, it would float to the top of a cup of water.
Answer:
2H2(g]+O2(g]→2H2O(l]]. Notice that the reaction requires 2 moles of hydrogen gas and 1
Explanation:
