Answer:
The heat of combustion is -25 kJ/g = -2700 kJ/mol.
Explanation:
According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.
Qcomb + Qcal = 0
Qcomb = - Qcal
The heat absorbed by the calorimeter can be calculated with the following expression.
Qcal = C × ΔT
where,
C is the heat capacity of the calorimeter
ΔT is the change in temperature
Then,
Qcomb = - Qcal
Qcomb = - C × ΔT
Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ
Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:
The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:
It is false, hope this helps!
C=10⁻⁶ mol/L
pH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg10⁻⁶=14-6=8
B. pH = 8
Answer:
Plants use photosynthesis to make glucose. Glucose is also know as sugar. You can tell it is present if the plant receives sunlight as well as water.
Explanation:
MARK BRAINLIEST
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
