What compound do the ions Al3+ and O2- form
1 answer:
You might be interested in
Taking into account the reaction stoichiometry, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 AgBr + Na₂S₂O₃ → Ag₂S₂O₃ + 2 NaBr
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- AgBr: 2 moles
- Na₂S₂O₃: 1 mole
- Ag₂S₂O₃: 1 mole
- NaBr: 2 moles
The molar mass of the compounds is:
- AgBr: 187.77 g/mole
- Na₂S₂O₃: 158 g/mole
- Ag₂S₂O₃: 327.74 g/mole
- NaBr: 102.9 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- AgBr: 2 moles ×187.77 g/mole= 375.54 grams
- Na₂S₂O₃: 1 mole ×158 g/mole= 158 grams
- Ag₂S₂O₃: 1 mole ×327.74 g/mole= 327.74 grams
- NaBr: 2 moles ×102.9 g/mole= 205.8 grams
The following rule of three can be applied: if by reaction stoichiometry 375.54 grams of AgBr form 327.74 grams of Ag₂S₂O₃, 125 grams of AgBr form how much mass of Ag₂S₂O₃?
<u><em>mass of Ag₂S₂O₃= 109.09 grams</em></u>
Then, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
Learn more about the reaction stoichiometry:
#SPJ1
Answer:
a) IUPAC Names:
1) (<em>trans</em>)-but-2-ene
2) (<em>cis</em>)-but-2-ene
3) but-1-ene
b) Balance Equation:
C₄H₁₀O + H₃PO₄ → C₄H₈ + H₂O + H₃PO₄
As H₃PO₄ is catalyst and remains unchanged so we can also write as,
C₄H₁₀O → C₄H₈ + H₂O
c) Rule:
When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.
d) C is not Geometrical Isomer:
For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.
