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Reil [10]
3 years ago
7

Ammonium Phosphate (NH4)3PO4 is a strong electrolyte. What will be the concentration of all the ions in a 0.9 M solution of ammo

nium phosphate?
Chemistry
1 answer:
katrin2010 [14]3 years ago
3 0
There are 3 moles of

NH_4^+

<span>per 1 mole of salt and 1 mole of

PO_{4}^{3-}

</span>per mole of salt, the total ionic concentrations must be

2.7 M of NH_4^{+}, and 0.9 M of PO_4^{3-}
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Question 2 answer is A
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What best describes the bonding in a carbon dioxide molecule?
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A is the answer just did it
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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
3 years ago
How many grams of methionine (MW = 149.21) are needed to make 20 mL of a 150 mM solution?
Leni [432]

Answer:

0.45 g

Explanation:

Step 1: Given data

  • Molar mass of methionine (M): 149.21 g/mol
  • Volume of the solution (V): 20 mL
  • Concentration of the solution (C): 150 mM

Step 2: Calculate the moles of methionine (n)

We will use the following expression.

n = C × V

n = 150 × 10⁻³ mol/L × 20 × 10⁻³ L

n = 3.0 × 10⁻³ mol

Step 3: Calculate the mass of methionine (m)

We will use the following expression.

m = n × M

m = 3.0 × 10⁻³ mol × 149.21 g/mol

m = 0.45 g

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3 years ago
How many variables are you allowed to change a purpose in a experiment
evablogger [386]
Only one. variable to get a reliable results you only have to change one.
4 0
3 years ago
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