55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
O2 gas, where there are two Oxygen atoms which are covalently bonded together
Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
Answer:
Explanation:
Using freezing point depression formula,
ΔTemp.f = Kf * b * i
Where,
ΔTemp.f = temp.f(pure solvent) - temp.f(solution)
b = molality
i = van't Hoff factor
Kf = cryoscopic constant
= 1.86°C/m for water
= (0 - (-5.58))/1.86
= 3.00 mol/kg
Assume 1 kg of water(solvent)
= (3.00 x 1)
= 3.00 mol.
A. deposition
this means that it was moved from one place to another and can be transported by ice, water, gravity, or wind.
