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3 years ago

Please help me and explanation would be really awesome thank you!

Chemistry

1 answer:

valentina_108 [34]3 years ago

4 0

Answer:

So the answer would be 10 moles

Explanation:

1) Start with the molecular formula for water: $H_{2} O!$

2) If there are 10 moles of water use a mole ratio to calculate the moles of oxygen it would produce.

(This question is... interesting... since they chose an element that is diatomic in free state so It could TECHNICALLY be two answers, moles of O or moles of $O_{2}$ )

The mole ratio is 1 moles of $H_{2}O$ to 1 moles of O. This is because the coefficient for oxygen in water is simple 1, so the ratio is 1:1.

3) that means if 10 moles of water decompose, they decompose into 10 moles of $H_{2}$ and 10 moles of O.

Extra:

About what I was saying before about the question being slightly interesting:

10 moles of pure oxygen is produced but free state oxygen exists as $O_{2}$ so it could possibly be 10 OR 5! However, notice it says elements. This leads me to believe the answer is 10 (monatomic oxygen) instead of 5 (free state/diatomic oxygen).

I hope this helps!

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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

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

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$\tau$ = 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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