PH is the test of acidity or basicity of a solution. it follows the formula:
pH = pKa + log [salt] / [acid] where NaF is the salt and HF is the acid in this case.
By literature, Ka of HF is 3.5*10^-4
<span>pKa= -log(Ka)=</span><span> 3.46 </span>
<span>pH = pKa + log [NaF / [HF] </span>
4.05 = 3.46 + log [NaF / [HF]
log [NaF / [HF]<span> = 0.59
</span>
[NaF / [HF] = 3.89
The formula to calculate osmotic pressure is
Osmotic Pressure = M R T
M = Molarity
R = Ideal Gas Constant
T = Temperature in Kelvin
So,
24.6/.2254kg=109.139g /kg >>>>> Molarity
109.139 x mols/92 g = 1.186 mols kg^-1
1.186 x 0.08134 x 298 K = 28.755 atm
<span>1.06852 x 0.08134 x 298K= 26.5 atm
The answer is 26.5</span>
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Answer:
Mass = 2.77 g
Explanation:
Given data:
Mass of HCl = 2 g
Mass of CaCl₂ produced = ?
Solution:
Chemical equation:
2HCl + Ca → CaCl₂ + H₂
Number of moles of HCl:
Number of moles = mass / molar mass
Number of moles = 2 g/ 36.5 g/mol
Number of moles = 0.05 mol
now we will compare the moles of HCl with CaCl₂.
HCl : CaCl₂
2 : 1
0.05 : 1/2×0.05 = 0.025 mol
Mass of CaCl₂:
Mass = number of moles × molar mass
Mass = 0.025 mol × 110.98 g/mol
Mass = 2.77 g
Answer:
8510 mol
Explanation:you must divide both of the molar masses into each other
