Answer:
Pp O2 = 82.944 KPa
Explanation:
heliox tank:
∴ %wt He = 32%
∴ %wt O2 = 68%
∴ Pt = 395 KPa
⇒ Pp O2 = ?
assuming a mix of ideal gases at the temperature and volumen of the mix:
∴ Pi = RTni/V
∴ Pt = RTnt/V
⇒ Pi/Pt = ni/nt = Xi
⇒ Pi = (Xi)*(Pt)
∴ Xi: molar fraction (ni/nt)
⇒ 0.68 = mass O2/mass mix
assuming mass mix = 100 g
⇒ mass O2 = 68 g
∴ molar mass O2 = 32 g/mol
⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2
⇒ mass He = 32 g
∴ molar mass He = 4.0026 g/mol
⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He
⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles
molar fraction O2:
⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099
⇒ Pp O2 = (X O2)(Pt)
⇒ Pp O2 = (0.2099)(395 KPa)
⇒ Pp O2 = 82.944 KPa
The value of Kc for the equilibrium is 0.150 mole² / litre ²
<u>Explanation:</u>
<u>Given:</u>
An equilibrium mixture in an 1.00 L vessel contains 5.30 moles of
Mg(OH )₂ 0.800 moles of Mg²⁺ and 0.0010 moles OH₋
We have to find the value of Kc
- Step 1: Find the equilibrium Concentration.
- Step 2: Substitute the values in the equation.
- Step 3: Find the value of Kc.
- I have attached the document for the detailed explanation
The value of Kc for the equilibrium is 0.150 mole² / litre ²
Boyle's law gives the relationship between pressure and volume of gases.
it states that for a fixed amount of gas at constant temperature, pressure of gas is inversely proportional to volume.
PV = k
where P - pressure , V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting this equation
P x 0.5650 L = 715.1 bar x 1.204 L
P = 1 524 bar
