Answer:
- The graphs that come with this question are in the picture attached.
- The answer is graph identified with the letter A.
Explanation:
The normal boiling point of water is 100°C. That is the temperature at which water boils when the atmospheric pressure is 1 atm, i.e. at sea level.
The liquids boil when its vapor pressure equals the atmospheric pressure; so the higher the atmospheric pressure the higher the boiling point, and the lower the atmospheric pressure the lower the boiling point.
Since, it is stated that the altitude of Denver, Colorado is 1,600 m, the atmosperic pressure (ther pressure exerted by the columnn of air above the place) is lower than 1 atm (atmospheric pressure at sea level).
Hence, water boiling point in Denver is lower than 100°C.
The graphs shown represent the temperature (T °C) as water is heated. Since when liquids boil their temperature remains constant during all the phase change, the flat portion of the graph represents the time during which the substance is boiling.
In the graph A, the flat portion is below 100°C; in the graph B, the flat portion is at 100 °C; in the graph C the flat part is above 100ªC, and, in graph D, there is not flat part. Then, the only graph that can illustrate water boiling in Denver, Colorado is the graph A.
Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:

Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:

The expression for equilibrium constant is:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
Now put all the given values in this expression, we get:

x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M
To solve this we assume that the gas is an ideal
gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At
a constant temperature and number of moles of the gas the product of PV is
equal to some constant. At another set of condition of temperature, the
constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 x V1 / P2
<span>V2 = 153 x 4 / 203</span>
V2 = 3 L