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PSYCHO15rus [73]

3 years ago

14

Please answer the question that is attached to this question. It will be a BIG help for my homework if you give a proper explana

tion. I promise I will eventually give brainly for the best answer.

1 answer:

vichka [17]3 years ago

8 0

Naw man I ain’t the correct answer

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Which factors determine the length of a bond between two elements?

Tanya [424]

Answer:

The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms.

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3 years ago

Write, Balance and Label the type of reaction: Potassium Hydroxide plus phosphoric acid yields potassium phosphate and water.

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In order to balance this, you have to count each element where the elements in the reactants side and the product side should have equal number of molecules. The balanced reaction is as follows:

KOH + H3PO4 = KH2PO4 +H2O

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3 years ago

In order to say that carbon dioxide caused global warming scientists had to A.Come up with a mechanism that made sense and test

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Answer:

Option A.

Explanation:

Option A directly tests the cause and effect. Option B simply argues it argumentatively without any solid evidence to show cause and effect. Option C only shows correlation and Option D only shows correlation as well.

6 0

3 years ago

Is xenon (xe) more or less reactive than bromine (br)? why?

galben [10]

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Explanation:

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7 0

3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago