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Stels [109]
3 years ago
13

The average flowrate of hydrochloric acid solution from a plant is 11,900 Lt/hr. The density is the same as that of water (1kg/L

t). The concentration of HCl in the solution is 0.015gmol/Lt.
Calculate the following:
(i) Flowrate in gallons per minute
(ii) Mass flowrate in lbm / hr
(iii) The number of gram mol of HCl flowing per minute
(iv) The mass fraction of HCl in the solution
(v) The number of lb mols of HCl in 1m3 of solution
Chemistry
1 answer:
skelet666 [1.2K]3 years ago
8 0

Answer:

See explanation.

Explanation:

Hello,

Here, considering the given flowrate, we perform the following units conversions for each case:

(i)

11,900 \frac{L}{h} *\frac{0.264172 gal}{1L} *\frac{1h}{60min} =52.39\frac{gal}{min}

(ii)

11,900 \frac{L}{h} *\frac{1kg}{1L}*\frac{2.20462lbm}{1kg}=26,234\frac{lbm}{h}

(iii)

11,900 \frac{L}{h} *\frac{0.015gmolHCl}{1L}*\frac{1h}{60min}=2.975\frac{gmolHCl}{min}

(iv) We find the kg of HCl per kg of acid solution:

w_{HCl}=0.015\frac{gmolHCl}{Lsln}*\frac{36.45gHCl}{1gmolHCl} *\frac{1kgHCl}{1000gHCl}*\frac{1Lsln}{1kgsln} =5.4675x10^{-4}

(v)

n_{HCl}=0.015\frac{gmolHCl}{L}*\frac{1000L}{1m^3}  *1m^3=15gmolHCl

Best regards.

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3 years ago
How many total electrons can be contained in the 4d sublevel? <br> 2<br> 6<br> 10<br> 14
qaws [65]
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So, having this in mind, 10 electrons in total can be contained in the 4d sublevel.
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5 0
4 years ago
Read 2 more answers
To what volume (in mL) would you need to dilute 25.0mL of a 1.45 m solution of Kcl to make a 0.0245m solution of KCl
dezoksy [38]

Answer:

The answer is

<h2>1479.60 mL</h2>

Explanation:

In order to calculate the volume needed we use the formula

V_2 =  \frac{C_1V_1}{C_2}

where

C1 is the concentration of the stock solution

V1 is the volume of the stock solution

C2 is the concentration of the diluted solution

V2 is the volume of thevdiluted solution

From the question

C1 = 1.45 M

V1 = 25 mL

C2 = 0.0245 M

So we have

V_2 =  \frac{1.45 \times 25}{0.0245}   =  \frac{36.25}{0.0245}  \\  = 1479.59...

We have the final answer as

<h3>1479.60 mL</h3>

Hope this helps you

4 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
Identify each definition that applies to the compound in red. Check all that apply.
andre [41]

Answer:

Arrhenius acid & Bronsted-Lowry acid

Explanation:

4 0
4 years ago
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