Answer:
The answer to your question is: V = 60.3 ml
Explanation:
Data
V1 = 500 ml
T1 = 27°C = 300°K
P1 = 75 mmHg = 0.098 atm
V2 = ?
T2 = - 33°C = 240°K
P2 = 0.65 atm
Formula




V2 = 60.3 ml
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.
The formula of aspartame is <span>C14H18N2O5.
From the periodic table:
molecular mass of hydrogen = 1 grams
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of oxygen = 16 grams
This means that:
molar mass of aspartame = 14(12)+18(1)+2(14)+5(16) = 294 grams
Therefore, each 294 grams of aspartame contains 5(16) = 80 grams of oxygen. To know number of grams of oxygen in 23.6 grams of aspartame, simply use cross multiplication as follows:
mass of oxygen = (23.6x80) / 294 = 6.42 grams</span>