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natita [175]
3 years ago
13

6. Identify each element as a metal, metalloid, or nonmetal

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
7 0
A. Metal
B. Metalloid
C. Nonmetal
D. Metal
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Please write a balanced equation for this reaction!! WILL GIVE BRAINLIEST!! 50 points!!
Alex73 [517]

Answer:

Cu+2AgNO3--->2Ag+Cu(NO3)2

Explanation:

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3 years ago
Which explanations did you include in your response? Check all that apply. The oxidation number of carbon changes from -1 to +4.
Kaylis [27]

Answer:

B, E

Explanation:

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5 0
2 years ago
Assuming equal concentrations, arrange these solutions by pH.
LenKa [72]

Answer:

Highest pH(most basic)

Sr(OH)2(aq)

KOH (aq)  

NH3(aq)

HF (aq)  

HClO4(aq)

Lowest pH(most acidic)

Explanation:

The concentration of H+ ion will determine the pH of a solution. The pH actually reflects the ratio of H+ ion and OH- since both of them can combine into water. Solution with more H+ ion will have a lower pH and called acidic, while more OH- will have high pH and be called basic. Strong acid/base will be ionized more than weak acid/base.

Sr(OH)2(aq) = strong base, release 2 OH- ion per mole

KOH (aq) = Strong base, release 1 OH- per mole

NH3(aq) = weak base, release less than 1 OH- per mole

HF (aq) =strong acid, release 1 H+ per mole

HClO4(aq) = stronger acid, release 1 H+ per mole

8 0
3 years ago
Unscramble dreor to get a science answer
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Order. 
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2 years ago
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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
2 years ago
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