ANSWER
EXPLANATION
Given that
The energy released by the system is 12.4J
Work done on the surrounding is 4.2J
Follow the steps below to find the change in energy
In the given data, energy is said to be released to the surroundings
Recall, that exothermic reaction is a type of reaction in which heat is released to the surroundings. Hence, change in enthalpy is negative
Step 1; Write the formula for calculating change in energy
Since heat is released to the surrounding, then q = -12J
Recall, that work done by the system on the surroundings is always negative
Hence, w = -4.2J
Step 2; Substitute the given data into the formula in step 1
Therefore, the change i
I would say physical, because a physical change is affecting the form of a chemical substance, but not it's chemical makeup.
<em>Thermal energy</em> is the sum of the kinetic and potential energies of all the particles in an object.
Assume that you have 250 gL of water and 1 kg of water at the same temperature.
Then, each water molecule has the same kinetic energy.
The larger sample contains four times as many molecules, so it contains four times as much thermal energy.
Thus, thermal energy is directly proportional to mass.
In symbols, <em>KE </em>∝ <em>m</em> or <em>KE = km</em>.
The graph of a direct proportion is a <em>straight line passing trough the origin</em>.
It should look something like the graph below.
True, because diatomic elements (H2, O2, F2, Br2, I2, N2, Cl2) consist of only one element but are molecules with covalent bonds.
Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
