Answer:
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Explanation:
Here, 
H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation
H(diene) depends on stability of diene.
More stable a diene, lesser will be it's H(diene) value (more neagtive).
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Hence, 
is higher (less negative) for trans-1,3-pentadiene
Explanation:
Let the mass of isoamyl acetate be 100g.
Moles of Carbon = 60.58/12 = 5.048mol
Moles of Hydrogen = 7.07/1 = 7.07mol
Moles of Oxygen = 32.28/16 = 2.018mol
Mole Ratio of C : H : O
= 5.048 : 7.07 : 2.018
= 5 : 7 : 2.
Hence the empirical formula of isoamyl acetate is C5H7O2.
Answer:
This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol
Explanation:
The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.
Answer:
48%
Explanation:
Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:
P1T2 = P2T1
<em>Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:</em>
P1 = 10.0atm
T2 = 1420K
P2 = ?
T1 = 730K
P2 = 10.0atm*1420K / 730K
P2 = 19.45 atm
The CO2 reacts as follows:
2CO2 → 2CO+ O2
Where 2 moles of gas react producing 3 moles of gas
Assuming the 100% of CO2 react, the pressure will be:
19.45atm * (3mol / 2mol) = 29.175atm
As the pressure rises just to 24.1atm the moles that react are:
24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present
The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:
0.48mol / 1mol * 100 = 48%
Answer:
involuntary, attached to the eyeball, nonstriated.
Explanation:
