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3 years ago

Carbon monoxide coming from your car smells like

Engineering

1 answer:

Lemur [1.5K]3 years ago

3 0

I wouldn’t smell like anything carbon monoxide is an odorless gas

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A water contains 50.40 mg/L as CaCO3 of carbon dioxide, 190.00 mg/L as CaCO3 of Ca2 and 55.00 mg/L as CaCO3 of Mg2 . All of the

Galina-37 [17]

Answer:

Total sludge = 123426kg/d

Explanation:

The reaction is given as;

H2Co3 + Ca(OH)2 ⇆ CaCo3 + 2H20

1 1 1 2 moles

Calculating the concentration of C02, we have

Concentration of C02 = concentration of CaCo3/Molecular weight of Caco3

= 50.4/100.09

= 0.5035mol/L

Sludge of Co2 = Conc. of Co2 * Q * MW of CaCo3 *10^-6

= 0.5035 * 253.6 *10^6 * 100.09 * 10^-6

= 12780kg/d

From the equation Ca2+ + 2HCo3- + Ca(OH)2 ⇄ 2CaCo3 + 2H2O

1 mole of calcium yields 2 moles of CaCo3

Therefore, Concentration of Ca2+ = Conc. of CaCo3/Mw of CaCO3

= 190-30/100.09

=1.599mol/L

Calculating sludge of calcium:

Sludge of Ca = 2 * Conc. of ca * Q * mw of CaCO3 * 10^-6

= 2 * 1.599 *253.6*10^6* 100.09 * 10^-6

= 811742kg/d

From the equation,

Mg2+ +2HCO3- + Ca(OH)2 ⇄ MgCO3 + 2CaCO3 + 2H2O

1 mole of mg yields 2 moles CaCO3 and 1 mole of Mg(OH)2

Concentration of Mg2+ = Conc, of CaCO3 /Mw of CaCo3

= 55- 10/100.09

= 0.4496mol/L

Sludge of Mg = 2 * Conc. of Mg * Q * mw of CaCO3 * 10^-6 +* Conc. of Mg * Q * mw of Mg(OH)2 * 10^-6

= 2 * 0.4496 * 253.5*10^6 * 100.09 * 10^-6 + 0.4996* 253.5*10^6 58.3 * 10^-6

= 29472kg/d

Total Sludge = Sludge of CO2 + Sludge of Ca + Sludge of Mg

12780+ 81174 + 29472

= 123426kg/d

6 0

3 years ago

Read 2 more answers

A city emergency management agency and a construction company have formed a public-private partnership. The construction company

Goryan [66]

Answer:

Explanation:

4 0

3 years ago

The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

slamgirl [31]

Answer:

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2 ⇒X=4 ft

A=8 x 10=80 $ft^2$

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

4 0

4 years ago

3. When starting an automatic transmission

Alexxandr [17]

Answer:

It should be in Park or Neutral.

Explanation:

4 0

4 years ago

A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo

Natalija [7]

Answer:

For $20^{\circ}$ - 5.556 lb/s

For $15^{\circ}$ - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = $20^{\circ}$

Temperature rise, T' = $15^{\circ}$

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, $C_{p} = 0.24 btu/lb ^{\circ}F$

Now, for a rise of $20^{\circ}$ in temeprature:

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s$

Now, for $15^{\circ}$ :

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s$

4 0

3 years ago