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3 years ago

Using the appropriate conversion factor from the first part, convert 0.600 mil potassium chlorate to grams.

Chemistry

1 answer:

Eduardwww [97]3 years ago

3 0

Answer:

0.600 mol potassium chlorate is 73.52969999999999 grams or rounded, 73.53 g Potassium Chlorate.

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A 75 um thick polysulphone microporous membrane has an average porosity of E 0.35. Pure water flux through the membrane is 35 m'

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Explanation:

The given data is as follows.

Water flux, $J_{w}$ = 25 $m^{3}/m^{2}h$

= $\frac{25}{3600} m^{3}/m^{2}h$

So, let velocity (u) = $\frac{25}{3600}$ m/s = $6.9 \times 10^{-3}$

$\rho$ = 998 $kg/m^{3}$

Pore size, d = $0.8 \times 10^{-6} m$

$\mu$ = 0.9 cP = $9 \times 10^{-4}$ Pa.s

Hence, calculate the reynold number as follows.

$R_{e} = \frac{\rho \times u \times d}{\mu}$

= $\frac{998 kg/m^{3} \times 6.9 \times 10^{-3} \times 0.8 \times 10^{-6} m}{9 \times 10^{-4} Pa.s}$

= $612.1 \times 10^{-5}$

= 0.006

This means that the flow is laminar.

Now, we use Hagen-Poiseuille equation as follows.

$J_{w} = \frac{\varepsilon \times d^{2}}{32 \times \mu \times \tau} \times \frac{\Delta P}{L_{m}}$

where, $\varepsilon$ = membrane porosity = 0.35

d = $0.8 \times 10^{-6}$ m

$\Delta P$ = $2 \times 10^{5} Pa$

$\mu$ = $9 \times 10^{-4}$

$\tau$ = tortuosity

$L_{m}$ = membrane thickness = $75 \times 10^{-6} m$

$\frac{25}{3600} = \frac{0.35 \times (0.8 \times 10^{-6})}{32 \times (9 \times 10^{-4}) \times \tau}$

$\tau$ = 3.73

Hence, the tortuosity factor of the pores is 3.73.

As flow resistance = $R_{m}$

$J_{w} = \frac{\Delta P}{r \times R_{m}}$

$R_{m} = 3.2 \times 10^{10} m^{-1}$

Water permeability is represented by $L_{p}$ .

$J_{w} = L_{p} \times \Delta P$

$6.9 \times 10^{-3}$ = $L_{p} \times 2 \times 10^{5} Pa$

$L_{p} = 3.45 \times 10^{-8} m^{3}/m^{2}s Pa$

Therefore, the resistance to flow is $3.2 \times 10^{10} m^{-1}$ and its water permeability is $3.45 \times 10^{-8} m^{3}/m^{2}s Pa$ .

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