Answer: 
Explanation: <u>Heats</u> <u>of</u> <u>formation</u> is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.
<u>Standard</u> <u>Enthalpy</u> <u>Change</u> is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.
Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:
For the reaction
2NH₃ + 3N₂O → 4N₂ + 3H₂O
2(-46.2) + 3(82.05) 4(0) + 3(-241.8)
![\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7B0%7D%3D3%28-241.8%29-%5B%202%28-46.2%29%2B3%2882.05%29%5D)
<u>The standard enthalpy change for the reaction is </u><u> kJ</u>
Ozone which is present in the stratospheric region of atmosphere is helpful for preventing harmful UV rays from reaching the surface of earth. Due to human activity, several compounds (specifically chlorofluorocarbons) are released in atmosphere. Due to inherent chemical stability of these compounds, the remain stable in lower region of atmosphere and slowly diffuse into stratosphere. On reaching the stratosphere, these compounds reacts with ozone and thereby depletes the effective concentration of ozone present in atmosphere. Hence, <span>the Montreal Protocol was signed in 1987 by major countries of the world. This aim of this protocol was to protect the stratospheric ozone layer by phasing out the production and consumption of ozone-depleting substances.</span>
Answer:
81.59%
Explanation:
First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:
- 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃
Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:
- 6.32 mol NH₃ *
= 6.32 mol NO
Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:
- 6.32 mol NO * 30 g/mol = 189.60 g NO
Finally we <u>calculate the percent yield</u>:
- 154.70 g / 189.60 g * 100% = 81.59%
