A large volume of material will have a small amount of mass when the
material in question is gaseous or porous, and therefore has a lot of
space within it despite taking up a large amount of space overall.
Answer:
C(graphite) → C(diamond), ΔH = - 0.45 kcal
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal
Explanation:
C(graphite) → C(diamond), ΔH = - 0.45 kcal
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal
These reactions are exothermic reaction because heat is evolved.
The energy changes occur during the bonds formation and bonds breaking.
There are two types of reaction endothermic and exothermic reaction.
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
The reaction described above is the formation of an acetal. The initial starting material has a central carbonyl and two terminal alcohol functional groups. In the presence of acid, the carbonyl will become protonated, making the carbon of the carbonyl susceptible to nucleophilic attack from one of the alcohols. The alcohol substitutes onto the carbon of the carbonyl to provide us with the intermediate shown.
The intermediate will continue to react in the presence of acid and the -OH that was once the carbonyl will become protonated, turning it into a good leaving group. The protonated alcohol leaves and is substituted by the other terminal alcohol to give the final acetal product. The end result of the overall reaction is the loss of water from the original molecule to give the spiroacetal shown in the image provided.
Answer:
20.25 W
Explanation:
Applying,
P = I²R.................... Equation 1
Where P = Power, I = current drawn by the test light, R = Resistance of the test light
From the question,
Given: I = 1.5 A, R = 9.00 Ω
Substitute these values into equation 1
P = (1.5²)(9)
P = 2.25×9
P = 20.25 W