This is a Charles' Law problem: V1/T1 = V2/T2. As the temperature of a fixed mass of gas decreases at a constant pressure, the volume of the gas should also decrease proportionally.
To use Charles' Law, the temperature must be in Kelvin (x °C = x + 273.15 K). We want to solve Charles' Law for V2, which we can obtain by rearranging the equation into V2 = V1T2/T1. Given V1 = 25 L, T1 = 1200 °C (1473.15 K), and T2 = 25 °C (298.15 K):
V2 = (25 L)(298.15 K)/(1473.15 K) = 5.1 L.
Hydrocarbons
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B. 3.77 L is the new volume occupied by the gas.
<u>Explanation:</u>
As per Avogadro's law, which states that if the pressure and temperature held constant, then an equal volume of the gases will occupy an equal number of molecules. It can be written as,
Here, V1, volume of the helium gas = 2.9 L
V2, volume of the additional helium gas in the balloon = ?
n1, moles of helium gas = 0.150 mol
n2, number of moles of additional helium gas = 0.150 + 0.0450 = 0.195 mol
We have to rearrange the equation for V2 as,
V2 = 
Now Plugin the values as,
V2 = 
= 3.77 L
So the new volume of the balloon is 3.77 L.
Answer:
K2 = 9.701 x 10^-10
Explanation:
K1 = 4.0 x 10^-5
K2 = ?
T1 = -10.8 °C +273 = 262.2K
T2 = -16 °C + 273 = 257K
ΔHrxn = 120. kJ/mol = 120000 J/mol
The formular relating all these parameters is given as;
ln( K2 / K1) = −ΔHrxn / R * (1 / T2 − 1 / T1)
ln (K2 / 4.0 x 10^-5) = - 120000 / 8.314 (1 / 257 - 1 / 262.2)
ln (K2 / 4.0 x 10^-5) = 1.1138
ln K2 - ln4.0 x 10^-5 = 1.1138
ln K2 = 1.1138 + ln4.0 x 10^-5
ln K2 = 1.1138 - 10.1266
ln K2 = -9.0128
K2 = 9.701 x 10^-10
Answer:
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