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3 years ago

Toxic Cr(VI) can be precipitated from an aqueous solution by bubbling SO2 through the

Chemistry

1 answer:

gavmur [86]3 years ago

3 0

Answer:

2.48 × 10⁴ mol

Explanation:

Step 1: Write the balanced equation

2 CrO₄²⁻ + 3 SO₂ + 4 H⁺ ⇒ Cr₂(SO₄)₃ + 2 H₂O

Step 2: Calculate the reacting moles of CrO₄²⁻

3.00 × 10⁸ L of 5.50 × 10⁻² mM CrO₄²⁻ react. The reacting moles are:

3.00 × 10⁸ L × 5.50 × 10⁻² × 10⁻³ mol/L = 1.65 × 10⁴ mol

Step 3: Calculate the reacting moles of SO₂

The molar ratio of CrO₄²⁻ to SO₂ is 2:3. The reacting moles of SO₂ are 3/2 × 1.65 × 10⁴ mol = 2.48 × 10⁴ mol.

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3 years ago

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago

A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will b

Angelina_Jolie [31]

Answer:

50.4 % of the water will be vapor

Explanation:

Step 1: Data given

Mass of water = 1.68 grams

volume of the flask = 5.3 L

Temperature = 65°C

Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

Step 2: Calculate moles of H2O

p*V=n*R*T

⇒ p = the pressure of water = 0.2467 atm

⇒ V = the volume of the flask = 5.3 L

⇒ n = moles of water

⇒ R = gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = 65°C = 338 Kelvin

n = (p*V)/(R*T)

n = (0.2467 * 5.3) /(0.08206* 338)

n = 0.047 moles

Step 3: Calculate mass of water

Mass of water = moles of water * molar mass of water

Mass of water = 0.047 moles *18.02 g/mol

Mass of water = 0.84694 grams

Step 4: Calculate the percent of water vaporized

% = (0.84694 grams/1.68 grams) *100%

% = 50.4%

50.4 % of the water will be vapor

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3 years ago