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sergiy2304 [10]

2 years ago

15

Metals present in municipal wastewater may still be present in treated sewage sludge; ______

1 answer:

Marrrta [24]2 years ago

5 0

Metals present in municipal waste water may still be present in treated sewage sludge IN CONCENTRATIONS THAT MAY AFFECT THE PUBLIC HEALTH. Sewage sludge is an end product of municipal waste water treatment and it contains many of the pollutant that are removed from the waste water.

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Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi

ivann1987 [24]

Answer:The formulas of ionic compounds are:

a) $CuBr$

b) $Mn_2O_3$

c) $Hg_2I_2$

d) $Mg_3(PO_4)_2$

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains $Cu^+$ ion.

$Cu^++Br^-\rightarrow CuBr$

(b) Manganese oxide : Given that it contains $Mn^{3+}$ ion.

$Mn^{3+}+O^{2-}\rightarrow Mn_2O_3$

(c)Mercury iodide :Given that it contains $Hg_2^{2+}$

$Hg_2^{2+}+I^-\rightarrow Hg_2I_2$

(d) Magnesium phosphate :Given that it contains $PO_4^{3-}$

$Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2$

4 0

2 years ago

A sample of CO2 gas has a volume of 2.7 L at 78.5 kPa. At what pressure would this sample of gas have a volume of 4.0L? Temperat

lisabon 2012 [21]

Answer:

52.99 kPa

Explanation:

Initial volume V1 = 2.7 L

Initial Pressure P1 = 78.5 kPa

Final Volume V2 = 4.0L

Final Pressure P2 = ?

Temperature is constant

The relationship between these quantities is given by the mathematical expression of Boyles law. This is given as;

V1P1 = V2P2

P2 = V1P1 / V2

P2 = 2.7 * 78.5 / 4.0

P2 = 52.99 kPa

7 0

3 years ago

When converting from moles to molecules OR molecules to moles you use what quantitative value?

miv72 [106K]

24 because it amen r amend

7 0

2 years ago

Need it in 2 minutes ASAP

RUDIKE [14]

Answer: read your book

Explanation: if you read it should be in there

3 0

3 years ago

An isotope of cesium-137 has a half-life of 30 years. If 5.0g of cesium-137 decays over 60 years. How many grams will remain.

LenKa [72]

Answer:

1.25 gram of cesium-137 will remain.

Explanation:

Given data:

Half life of cesium-137 = 30 year

Mass of cesium-137 = 5.0 g

Mass remain after 60 years = ?

Solution:

Number of half lives passed = Time elapsed / half life

Number of half lives passed = 60 year / 30 year

Number of half lives passed = 2

At time zero = 5.0 g

At first half life = 5.0 g/2 = 2.5 g

At 2nd half life = 2.5 g/ 2 = 1.25 g

Thus. 1.25 gram of cesium-137 will remain.

6 0

2 years ago