<span>11.3 kPa
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant (8.3144598 L*kPa/(K*mol) )
T = Absolute temperature
We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon.
Atomic weight argon = 39.948
Atomic weight neon = 20.1797
Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol
Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol
Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol
Now take the ideal gas equation and solve for P, then substitute known values and solve.
PV = nRT
P = nRT/V
P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L
P = 113.8892033 L*kPa / 5.00 L
P = 22.77784066 kPa
Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles.
0.024777375 mol / 0.049809918 mol = 0.497438592
Now multiply by the pressure
0.497438592 * 22.77784066 kPa = 11.33057699 kPa
Round the result to 3 significant figures, giving 11.3 kPa</span>
Answer:
D.Lowering the temperature is the best option.
Explanation:
The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.
In the reaction below for example;
A + B <==>C+D
If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?
Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?
If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.
For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
