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4 years ago

How is desalination different from water reclamation

Chemistry

2 answers:

Zarrin [17]4 years ago

5 0

Answer: Desalination provides water suitable for drinking.

Explanation:

The reclamation of effluents for soaking and obscure freshwater practices is swiftly emerging as an option to seawater desalination. Two membrane-based choices ready to use residue for water reuse, tertiary filtration (TF) of the effluent from a traditional activated sludge (CAS) method and a multicultural membrane bioreactor (MBR) are searched; in both instances, the level of approach can be complemented by reverse osmosis (RO). These choices are connected to desalination of seawater with both traditional multi-media layer filtration.

Sonbull [250]4 years ago

4 0

Desalination is different from water reclamation because: Desalination provides water suitable for drinking.

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A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium,

Scorpion4ik [409]

Answer: The value of $K_c$ is coming out to be 0.412

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $Sb_2S_3$

Given mass of $Sb_2S_3$ = 1.00 kg = 1000 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $Sb_2S_3$ = 339.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol$

For hydrogen gas:

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol$

For hydrogen sulfide:

Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol$

The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:

$Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)$

Initial: 2.944 5

At eqllm: 2.944-x 5-3x 2x 3x

We are given:

Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

$\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712$

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

The expression of $K_c$ for above equation, we get:

$K_c=\frac{[H_2S]^3}{[H_2]^3}$

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

$K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412$

Hence, the value of $K_c$ is coming out to be 0.412

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Sonja [21]

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3 years ago

The element that make up a ---------------- are chemically combined

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Compound is your answer

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3 years ago

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Which molecule is solid at Room temperature? NH3 HCI C4H10 MgCl2

Rufina [12.5K]

NH3 because the molecule in the room is more solid than and room temperature

6 0

3 years ago

9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1

WITCHER [35]

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.

3 0

4 years ago