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:
When waves act together, you talk about "interference".
When they reinforce each other, it is "constructive interference".
When they cancel each other, it is "destructive interference".
Answer: -
The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.
Explanation: -
Temperature of the hydrogen gas first sample = 10 °C.
Temperature in kelvin scale of the first sample = 10 + 273 = 283 K
For the second sample, the temperature is 350 K.
Thus we see the second sample of the hydrogen gas more temperature than the first sample.
We know from the kinetic theory of gases that
The kinetic energy of gas molecules increases with the increase in temperature of the gas. The speed of the movement of gas molecules also increase with the increase in kinetic energy.
So higher the temperature of a gas, more is the kinetic energy and more is the movement speed of the gas molecules.
Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.
The molecule have TETRAHEDRAL HYBRID ORBITAL.
SP3 hybridization involves the mixing of one orbital of S sub level and three orbitals of P sub level of the valence shell. All the orbitals possess equivalent energies and shapes. The SP3 orbital has 25% S character and 75% P character. S and P refers to the s and p sub shells.<span />
