Answer:
V = 177.4 L.
Explanation:
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In this case, since this gas can be assumed as ideal due to the given data, we can use the following equation:
Thus, by solving for volume we obtain:
So we can plug in the temperature in Kelvins (537 K), the pressure in atmospheres (0.404 atm) and the molar mass (54 g/mol) to obtain:
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Answer:
The temperature change was 10.04 °C.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system. In this way, there is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case you know:
Replacing:
7,529 cal= 1 *750 g* ΔT
Solving you get:
ΔT=
ΔT= 10.04 °C
So, <u><em>the temperature change was 10.04 °C.</em></u>
The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.
