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:
Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.
Explanation:
-Air becomes increasingly thinner with increasing altitudes.
-As such, oxygen becomes limited at higher altitudes and makes it difficult or almost impossible to breath a condition called hypoxia.
-Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.
A proton is a positively charged particle found within the nucleus of an atom, a neutron carries no charge and is also found in the nucleus. An electron is a very small negatively charged particle found in the outer "shells" or orbitals of the atom.
<h2>
Answer: 131.9 g</h2>
<h3>
Explanation:</h3>
<u>Write a Balanced Equation for the decomposition</u>
CaCO₃ → CaO + CO₂
<u></u>
<u>Find Moles of CO₂ Produced</u>
Since the mole ratio of CaCO₃ to CO₂ is 1 to 1,
the moles of CaCO₃ = moles of CO₂
moles of CaCO₃ = mass ÷ molar mass
= 300 g ÷ 100.087 g/mol
= 2.997 moles
∴ moles of CO₂ = 2.997 moles
<u>Determine Mass of CO₂</u>
Mass = moles × molar mass
= 2.997 mol × 44.01 g/mol
= 131.9 g
<u></u>
<h3>∴ when 300 g of calcium carbonate is decomposed, it produces 131.9 g of carbon dioxide.</h3>
<h3><u>Answer;</u></h3>
10 ft
<h3><u>Explanation;</u></h3>
The wave base of a wave is the depth at which there is no more perturbation of the medium caused by the wave. The wave base of a wave is determined to be half the wavelength of the wave itself.
Therefore;
Wave base = wavelength/2
= 20 ft/2
<u> = 10 ft</u>
