Answer:
Explanation:
![\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28product%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28reactant_%7Bj%7D%29%5D)
Where 
and 
are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).
is standard heat of formation and 
is standard enthalpy change for reaction at 
So, ![\Delta H_{rxn}=[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%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)
or, ![\Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B3mol%5Ctimes%20-393.509kJ%2Fmol%5D%2B%5B4mol%5Ctimes%20-241.818kJ%2Fmol%5D-%5B1mol%5Ctimes%20-103.8kJ%2Fmol%5D-%5B5mol%5Ctimes%200kJ%2Fmol%5D)
or,
Answer:
Acid solutions contain hydrogen ions. The higher the concentration of hydrogen ions, the lower the pH. Strong acids are fully ionised but weak acids are only partly ionised in solution.
Example:
Hydrochloric acid is a strong acid and ethanoic acid is a weak acid.
The volume of O₂ : 21 L
<h3>Further explanation</h3>
Given
8.7 grams of C₂H₄
Required
Volume O₂
Solution
Reaction
C₂H₄ + 3 O₂ ⇒ 2 CO₂ + 2 H₂O
mol C₂H₄(MW= 28 g/mol) :
= mass : MW
= 8.7 g : 28 g/mol
= 0.311
From the equation, mol O₂ :
= 3/1 x mol C₂H₄
= 3/1 x 0.311
= 0.933
At STP, 1 mol gas=22.4 L, so for 0.933 mol :
= 0.933 x 22.4 L
= 20.899 L ≈ 21 L
There are 4 types of orbitals. The s, p, d and f orbitals. They can contain a maximum of 2, 6, 10 and 14 electrons, respectively. So, for the 1s energy level, you would expect a maximum of 2 electrons. That includes elements 1 and 2, hydrogen and helium. <em>These are the only 2 elements in the first period.</em>
