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Anna71 [15]
2 years ago
8

Ferric chloride is being applied for coagulation at a 15.0 MGD surface water treatment plant, at a rate of 195 pounds per day. A

t this application rate, how many pounds of this chemical will be used in a 30-day month
Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
6 0

5850 pounds of this chemical will be used in a 30-day month.

195 pounds per day.

So, 30 days = 195 * 30

= 5850 pounds

Iron(III) chloride is the inorganic compound with the components FeCl 3. additionally known as ferric chloride, it is a commonplace compound of iron inside the +3 oxidation country. The anhydrous compound is crystalline and stable with a melting point of 307.6 °C.

The primary use of ferric chloride is to dispose of impurities in water and for wastewater treatment. Ferric chloride is likewise one of the few water remedy chemical compounds that can sequester odors.

Reasonably poisonous with the aid of ingestion but a strong irritant to the skin and eyes. it's far moderately corrosive and skin touch should be avoided and avoid skin and eye contact.

Learn more about ferric chloride here brainly.com/question/13617260

#SPJ4

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In a study of the decomposition of the compound XX via the reaction
Leya [2.2K]

Answer:

( About ) 0.03232 M

Explanation:

Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / [X_o]"

This formula would be true for the following information -

{ X_o = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }

________

Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6

1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X

1 / [X] = 28.8 + 1 / ( 0.467 ),

1 / [X] = 28.8 + 2.1413...,

1 / [X] = 31,

[X] = 1 / 31 = ( About ) 0.03232 M

Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M

3 0
4 years ago
What are the equilibrium partial pressures of pcl3 , cl2, and pcl5, respectively? express your answers numerically in atmosphere
aliina [53]
The question is incomplete, so I tried to find a similar problem online. It is shown in the attached picture. The reaction is

PCl₃ + Cl₂ ⇆ PCl₅

Then, we use the ICE (Initial-Change-Excess) approach as follows:

        PCl₃ + Cl₂ ⇆ PCl₅
I        0.5      0.5      0.3
C      -x         -x        +x
E    0.5-x    0.5-x    0.3+x

Total pressure: 0.5 - x + 0.5 - x + 0.3 + x = 1.3 

Kp = [PCl₅]/[PCl₃][Cl₂]
0.18 = (0.3+x)/(0.5-x)²
Solving for x,
x= 0.21

Partial pressures would be:
<em>PCl₃ = 0.5 - 0.21 = 0.29 atm</em>
<em>Cl₂ = 0.5 - 0.21 = 0.29 atm</em>
<em>PCl₅ = 0.3+0.21 = 0.51 atm</em>

8 0
4 years ago
Give 3 safety measures in handling acids or basic
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3 years ago
The standard lab thermometer in a high school science lab contains
earnstyle [38]
Distilled water.high school Lab
6 0
3 years ago
Read 2 more answers
Identify the elements given the orbital diagrams listed below.​
Mrrafil [7]

Answer:

See explanation.

Explanation:

1. There are 8 electrons. Elements that end with 2p orbitals are in the 2nd period (aka row) of the periodic table. Elements that have 4 electrons in 2p are in the 16th group (aka column) (column 16 may also be referred to as 6A) of the periodic table. So looking at row 2, column 16, we can see that the first diagram is of O, Oxygen.

2. 8 electrons. This is the same diagram as the one above.

3. 13 electrons. Elements ending with 3p are in period 3. Elements with 1 valence electron in a p orbital are in group 13 (aka group 3A).

4. 7 electrons. We already know 2p is period 2. 3 valence electrons in a p orbital means that it is in group 15/group 5A.

I did not write the answers for #3 and 4 but they can be easily found on a periodic table with the info I gave.

4 0
4 years ago
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