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Olegator [25]
3 years ago
6

The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/ml. the density of mercury is 13.6g/ml. part a what

do you predict for the height of a barometer column based on 1-iodododecane, when the atmospheric pressure is 751 torr ?
Chemistry
1 answer:
valina [46]3 years ago
3 0

As the atmospheric pressure is, P = dgh

Here d is the density of the mercury,

g is gravitation = 9.8 m/s²

h is height of the column, P = 751 torr = (751 torr × 1 atm / 760 torr) (101325 Pa) (1 N/m² / 1 Pa) = 100125 N/m²

Where, 1 N = 1 Kg / ms²

Thus, P = 100125 Kg / m³. s²

Therefore, height of the mercury column, when the atmospheric pressure is 751 torr,

h = P / gd

= (100125 kg / m³. s²) / (9.8 m/s²) (13.6 × 10³ kg / m³) = 0.751 m

As, d₁h₁ = d₂h₂

Here, d₁ is the density of the non-volatile liquid = 1.20 g/ml

d₂ is the density of the mercury = 13.6 g/ml

h₂ = 0.751 m

Thus, putting the values we get,

h₁ = d₂h₂ /d₁ = 13.6 g/ml × 0.751 m / 1.20 g/ml

= 8.5 m


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Mᵣ:       239.27   32.00        207.2

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\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

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