The bond formed from electrons that are transferred is a(n) _____.
1 answer:
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a. mol O₂=0.5
b. volume O₂ = 25 cm³
c. i. the total volume of the two reactants = 75 cm³
c. ii. the volume of nitrogen dioxide formed = 50 cm³
<h3>Further explanation</h3>Reaction
2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)
a.
mol NO = 1
From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :
b.
From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
Because mol ratio NO : O₂ = 2 : 1, so volume O₂ :
c.
i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³
ii. the volume of nitrogen dioxide formed :
mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.
The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:
The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.
Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.
The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus,
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
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