Do anyone know how to do question B
1 answer:
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.
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Answer:
The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case, you know:
- Q=?
- c= 4.186
- m= 12 g
- ΔT=Tfinal - Tinitial= 21 °C - 16°C= 5 °C
Replacing:
Q= 4.186 *12 g *5 °C
Solving:
Q=251.16 J
Since 1 J is equal to 0.2388 cal, then the following rule of three can be applied: if 1 J is equal to 0.2388 cal, then 251.16 J to how many cal are?
cal= 59.98 ≅ 60
<u><em>The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.</em></u>