What do S-shadows tell us about the interior of the earth?

(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe

Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration $C_o$ = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

$time (t) = \dfrac{V}{Q}$

$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

$\mathbf{C_e = 102.98 \ mg/l}$

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

3 0

2 years ago

Law of Conservation of Mass in terms of atoms

VARVARA [1.3K]

No atoms are lost or made during the chemical reaction so the total mass of the products is equal to the total mass of the reactants. In an atom, protons and neutrons contribute to the mass and since the number of them doesn’t change, the mass doesn’t either.

4 0

2 years ago

Can someone please help and list what each type of thing they are? (Element, Compound, Mixture of elements, Mixture of compounds

enot [183]

1. B
2. C
3. E
4. D
5. A
6. B
7. A
8. E
9. A

Hope this helps!

5 0

3 years ago

Calculate the concentration in mol/L, M, of an aqueous sugar solution with a concentration of 23.5% (w/w) and density of 1.005 g

Temka [501]

Answer:

The concentration in mol/L is 0.683M

Explanation:

23.5% (w/w)

This data means that in 100 g of solution, we have 23.5 grams of solute.

From this point, we can calculate the moles of sugar.

Moles = Mass / Molar mass

Moles = 23.5 g /342.30 g/m

Moles = 0.068 moles

Density data make us know, the volume of our solution.

solution δ = solution mass / solution volume

δ = 1.005 g/mL = 100 g / solution volume

solution volume = 100g / 1.005 g/ml

solution volume = 99.5 mL

In conclusion, 0.068 moles are in 99.5 mL

Molarity (M) is mol/L

Let's convert 99.5 mL in L

99.5 mL / 1000 = 0.0995 L

0.068 m / 0.0995L = 0.683

If we convert moles in mmoles, we can also get Molarity (M)

mmoles / mL = M

0.068 moles . 1000 = 68 mmoles

68 mmmoles / 99.5mL = 0.683

8 0

2 years ago

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of hydrosulfuric acid (H2S) and sodium hy

vovangra [49]

Answer: 2NH4Br(aq)+Pb(C2H302(aq)-------------------->

2NH4C2H3o2(aq) + PbBr2(s)

Explanation:

The net equation is :Pb2+ (aq)2Br (aq)---------------------->PbBr2(s)

the spectator ions NH4 +C2H3O2 are canceled

6 0

3 years ago