The bonds that are broken when water vaporizes are
2 answers:
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Answer:
2.90 x 10⁻¹¹moldm⁻³
Explanation:
Given parameters:
[H⁺] = 3.5 x 10⁻⁴mol/dm³
Unknown
[OH⁻] = ?
Solution;
The ionic product of water can be used to solve this problem. It has been experimentally determined to be 1 x 10⁻¹⁴mol² dm⁻⁶
[H⁺] [OH⁻] = 1 x 10⁻¹⁴
Therefore;
[OH⁻] = =
= 0.29 x 10⁻¹⁰moldm⁻³
= 2.90 x 10⁻¹¹moldm⁻³
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.
