🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
Chemistry
1
answer
0
watching
431
views
11 Nov 2019

The raw water supply for a community contains 18 mg/L total particulate matter and is to be treated by addition of 60 mg alum (Al2(SO4)3·18 H2O) per liter of water treated. What will the pH be if the treated solution is bubbled with air, so that it reaches equilibrium with atmospheric CO2?

Watch
1
answer
0
watching
431
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Hubert Koch
Hubert KochLv2
1 Mar 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share
discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Basic Chemistry
5 Edition,
Timberlake
ISBN: 9780134138046
Principles of Chemistry Molecular Approach
4th Edition,
Tro
ISBN: 9780134112831
Principles of Chemistry Molecular Approach
3rd Edition, 2014
Tro
ISBN: 9780321971944
Chemistry: Structure and Properties
2nd Edition,
Tro
ISBN: 9780134293936
Chemistry: A Molecular Approach
3rd Edition,
Tro
ISBN: 9780321809247
Chemistry: A Molecular Approach
5th Edition,
Tro
ISBN: 9780134874371
Principles of Chemistry: A Molecular Approach
4th Edition,
Tro
ISBN: 9780134895741
Chemistry: The Central Science
14th Edition, 2017
Brown
ISBN: 9780134414232

Related questions

A water treatment plant with an average flow of 40 million gallons/day treats its water with 30 mg/L alum (Al2(SO4)3-18H2O). Alum coagulation is used to remove particulate matter, reduce the concentration of organic matter, and reduce the alkalinity of the water (the reduction of alkalinity is shown below):

Al2(SO4)3-18H2O + 6HCO3- ↔ 2Al(OH)3(s) + 6CO2 +18 H2O +3SO42-
If the organic matter concentration is reduced from 12 mg/L to 5 mg/L, determine the total mass of alkalinity (i.e. HCO3-) consumed and the total mass of dry solids produced per day. Give your answers in kg/day.

The reaction of divalent manganese with oxygen in aqueous solution is given as follows:

Mn2+ + 1/2O2(aq) + H2O = MnO2(s) + 2H+

The equilibrium constant (K) for this reaction is 23.7. It has been found that a lake water sample that contains no oxygen at 25 C, pH = 8.5, originally contained 0.6 mg/L of Mn2+. The sample was aerated (atmospheric conditions of the dissolved oxyen concentration is 9.2 mg/L) and after 10 days of contact with atmospheric oxygen, the Mn2+ concentration was 0.4 mg/L. The molecular weight of Mn is 55, O is 16 and H is 1. The change in free energy of formation at standard conditions (kcal/mole) are as follows: Mn2+ = -54.4, O2(aq) = +3.93, H2O = -56.69, MnO2(s) = -111.1, H+ = 0.

(a) Assuming that the pH remains constant during aeration, will the precipitate continue to form after the measurement on the tenth day?

(b) What should the Mn2+ concentration be (in moles/L) at equilibrium, assuming that pH and presence of dissolved oxygen are the same as in part "a". Assume ideal conditions

1. A sludge containing 4% solid (by weight) needs to be thickened by centrifugation to ≥15% solid for disposal. The centrifuge has a high constant speed and produces a dewatered sludge containing 20% solid. To increase the processing capacity while still meeting the disposal requirement, you decide to bypass (i.e., not dewater) some of the feed sludge and blend the untreated sludge with the dewatered (20% solid) sludge, so as to produce a final sludge with a 15% solid content. Assume all sludge streams have the density of water.

(a) Give the centrate production rate (in L/min) and the percent solid (of the total coming in) discharged via the centrate?

(b) What % of the feed sludge is bypassed (i.e., not centrifuged)?

2. Eutrophication is a natural aging process all lakes undergo as nutrients and organic matter accumulate from their drainage basins. Over time the accumulated nutrients enable increased aquatic growth and stimulate biological activities. The increase in phytoplankton causes the lake to become murky, while decaying organic matter depletes the available dissolved O2. Further accumulation of silt, nutrients, and organic matter causes the eutrophic ("well-fed") lake to become shallower and warmer with more plants growing in the shallow edges. This process transforms lakes into bogs and marshes over time. Natural eutrophication is usually quite slow, often measured in thousands of years before major changes occur. What human activity has done is rapidly accelerate this process (called "cultural eutrophication") by introducing large quantities of nutrients. Sources of nutrients include municipal wastewater, industrial wastes, and runoff from fertilized agricultural land. The introduction of superfluous nutrients disturbs the natural balance of the system, causing excessive algae growth and water quality deterioration. Algal blooms are the unsightly and often malodorous clumps of rotting debris along the shoreline and thick mats of dead organic matter in the lake. A nutrient of major importance that enter fresh water bodies such as lakes from the aforementioned sources is phosphorous. C = 20% C = 15% Q = 100 L/min C = 4% bypass centrifuge C = 0.1% ("centrate") 2 Consider a phosphorus-limited lake with a surface area equal to 8 x 107 m 2 that is being fed by a stream with a flow rate of 15 m3 /s and a phosphorus concentration of 0.01 mg/L. An outfall from a wastewater treatment plant also adds 1 g/s of phosphorus to the lake, but its volumetric flow rate is small and can be ignored. The outflow rate from the lake is the same as the inflow (15 m3 /s). Because phosphorus can attach to suspended particles in lake water, the phosphorus in the lake can also be removed through particle settling (to sediment), which occurs at a rate of 10 m/yr. For the questions below, assume the lake is a cylindrical CSTR.

(a) Draw a diagram to represent the processes involved in phosphorus mass balance. List all sources and sinks. Estimate the steady-state phosphorus concentration in the lake.

(b) An acceptable level of phosphorus in the lake is 0.01 mg/L, above which eutrophication is considered likely. By what percent must the wastewater treatment plant reduce its phosphorous discharge into the lake to achieve this concentration?

3. A textile finishing process involves drying fabric that has been treated with a volatile solvent. The drying process involves evaporation of solvent and removal of solvent vapor by air. The wet fabric entering the dryer contains 50% solvent by weight. Air enters the dryer at a rate of 8 kg per kg of solvent-free fabric. The drying process is 92% efficient; i.e., 92% of entering solvent is removed from fabric and carried out by air. What is the weight % of solvent in the dried fabric? What is the mole % of solvent vapor in the exhaust? The molecular weight of the solvent is 46 g/mol.(Note: Dried fabric still contains some solvent; i.e., dried fabric is not solvent-free).

4. A process heater burns 4.8 kg of natural gas per minute, and the air supply rate is 20% higher than the stoichiometric rate. Exhaust from the heater is sent to a heat exchanger, where water vapor is condensed and removed from the exhaust. Assume natural gas is all methane (CH4), air contains 21% O2 and 79% N2, and the combustion of CH4 is complete, what is the mole % of CO2 in the cooled exhaust?

Weekly leaderboard

Home
Homework Help 3,900,000
Chemistry 830,000
Start filling in the gaps now
Log in