Answer:
man, i have the same problem, i don't know hpw to solve this
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>
Water Solution
Because you are testing the water solutions pH
Answer:
Less
Explanation:
The hydronium from the HCl is used to neutralize the bicarbonate in the baking soda. The hydronium is the acid and the bicarbonate ion is the base while the sodium and chloride ions are pH-neutral. Since theres a 1:1 mole ratio of hydronium to HCl and bicarbonate to sodium bicarbonate, it would require less HCl to neutralize a less concentration baking soda solution.
Answer:
Explained below.
Explanation:
A tennis player with two sneakers wouldn't bond to any other tennis player because he is already stable and complete with the 2 and doesn't need another players assistance to make him stand well.
However, helium atom with two electrons wouldn't bond to any other atoms because it is stable. This stability arises from the fact that it has two protons and 2 electrons, of which the 2 electrons completely fill its valence shell/outer most shell to make it neutral.