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
<span>Calculate the mass of 1 L of solution. Mass of solution=1000mL soln ×1.19 g soln1mL soln =1190 g soln (3 significant figures + 1 guard digit)Calculate the mass of HCl . Mass of HCl=1190g soln ×37.7g HCl100g soln =448.6 g HCl.Calculate the moles of HCl . ...Calculate the molarity of the HCl.</span>
Using the pH=-log[ hydrogen ions], assuming the pH of water is 7 then using the concept that kw=[ OH] [H+], then the [H+]= 10 power minus 7,since kw=10 to the power minus 14
<span>2 * 22.4 = 44.8 liters (if using pre 1982 standard)
2 * 22.7 = 45.4 liters (if using 1982 and later standard)
First, let's determine how many moles of F2 we have.
Atomic weight fluorine = 18.998403
Molar mass F2 = 2 * 18.998403 = 37.996806 g/mol
Moles F2 = 76 g / 37.996806 g/mol = 2.000168119 mol
Now we have a minor problem. What definition of STP are you using?
Up until 1982,
STP was defined as 0°C and 1 atmosphere (101.325 kPa)
From 1982 and later,
STP was defined as 0°C and 100 kPa
Because of the difference in pressure between the two different definitions of STP, the molar volume of a gas is 22.414 liter/mol using the pre-1982 definition and 22.711 liter/mol using the 1982 and later definition. So you get to choose which of the following 2 answers.
2 * 22.4 = 44.8 liters (if using pre 1982 standard)
2 * 22.7 = 45.4 liters (if using 1982 and later standard)
Unfortunately there's still a large number of text books in use using the standard that should have been obsolete 35 years ago.</span>
