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11Alexandr11
26 days ago
5

The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its

effects on human health if the receiving water is later used as a drinking water source. One way in which chromate can be removed from solution is by its reaction with ferrous ions (Fe2+) to form a mixture of chromic hydroxide and ferric hydroxide solids [Cr(OH)3(s) and Fe(OH)3(s), respectively], which can then be filtered out of the water. The overall reaction can be represented as
CrO42- + 3 Fe2+ + 8 H2O --> Cr(OH)3(s) + 3 Fe(OH)3(s) + 4 H+
How much particulate matter would be generated daily by this process at a facility that treats 60 m3/h of a waste stream containing 4.0 mg/L Cr, if the treatment reduces the Cr concentration enough to meet a discharge limit of 0.1 mg/L?
Chemistry
1 answer:
KiRa [2.7K]26 days ago
7 0

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

The total particulate matter produced daily = 24 × 1905.3 = 45727g.

You might be interested in
What types of compounds are CaCl2, Cu, C2H6, respectively.
KiRa [2770]

Response:

Ionic, metal, organic

Clarification:

For this scenario, we should examine each compound:

-) CaCl_2

In this compound, there is a non-metal atom (Cl) paired with a metal atom (Ca). This leads to a significant difference in electronegativity, indicating that an ionic bond will form. Ions can be generated:

CaCl_2~->~Ca^+^2~+~2Cl^-

The positive ion would be Ca^+^2 while the negative ion is Cl^-. Thus, we have an ionic compound.

-) Cu

Here, we are looking at a single atom. Consulting the periodic table shows that this atom belongs to the transition metals section (central part of the periodic table). Hence, Cu (Copper) is identified as a metal.

-) C_2H_6

Within this molecule, carbon and hydrogen are linked by single bonds. The difference in electronegativity between C and H is insufficient to lead to ion formation. Therefore, we have covalent bonds. This property is typical of organic compounds. (Refer to figure 1)

5 0
17 days ago
he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp
eduard [2583]

1) Drift velocity: 3.32\cdot 10^{-4}m/s

2. 5.6\cdot 10^{13} electrons per individual

Explanation:

1)

In a conducting material with an electric current, the drift velocity of electrons can be calculated using this equation:

v_d=\frac{I}{neA}

where

I stands for current

n represents the density of free electrons

e=1.6\cdot 10^{-19}C indicates the charge of an electron

A signifies the wire's cross-sectional area

The wire's cross-sectional area can be determined as

A=\pi r^2

where r denotes the wire's radius. Thus, the equation transforms to

v_d=\frac{I}{ne\pi r^2}

In this scenario, we have:

I = 8.0 A as the current

8.5\cdot 10^{28} m^{-3} indicates the free electron concentration

d = 1.5 mm is the diameter, making the radius

r = 1.5/2 = 0.75 mm = 0.75\cdot 10^{-3}m

So, the resulting drift velocity is:

v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s

2)

The entire length of the cord is

L = 3.00 m

And the cross-sectional area is

A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2

Consequently, the volume of the cord is

V=AL (1)

The number of electrons per unit volume is n, thus the total electrons in this cord would be

N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}

Overall, the Earth's population rounds to 8 billion individuals, equating to

N'=8\cdot 10^9

Hence, the number of electrons distributed to each person is:

N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}

7 0
1 month ago
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [2611]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

4 0
1 month ago
Read 2 more answers
Bonds between two atoms that are equally electronegative are _____. bonds between two atoms that are equally electronegative are
Alekssandra [2790]
When two atoms with equal electronegativity bond together, they form nonpolar covalent bonds.

Your second statement mirrors the first; the second statement likely reads, "Bonds between two atoms with unequal electronegativity are termed polar covalent bonds."
4 0
1 month ago
The value of delta G at 141.0 degrees celsius for the formation of phosphorous trichloride from its constituent elements,
VMariaS [2759]
The appropriate answer is option E. Gibbs free energy can be expressed using the equation: ΔG = ΔH - TΔS, where ΔH denotes the change in enthalpy of the reaction, T is the reaction temperature, and ΔS signifies entropy change. For our calculations, we have ΔH = -720.5 kJ/mol which converts to -720500 J/mol (given that 1 kJ = 1000 J), ΔS = -263.7 J/K, and T = 141.0°C, which equals 414.15 K. Consequently, the Gibbs free energy for the specified reaction at 141.0°C is calculated as -611.3 kJ/mol.
6 0
12 days ago
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