<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
Answer:
7.16x10⁻⁸M = [Ag+]
Explanation:
Using the equation:
E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)
<em>Where E</em>⁰<em>= 0.4249V</em>
<em>E(Cell) = -(-0.0019V) -Measured value-</em>
<em>[Cu2+] = 1M</em>
<em />
Replacing:
0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)
-0.423V = - 0.0296 • log (1M/[Ag+]²)
14.29 = log (1M/[Ag+]²)
1.95x10¹⁴ = 1M / [Ag+]²
[Ag+]² = 5.12x10⁻¹⁵M
7.16x10⁻⁸M = [Ag+]
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