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Ipatiy [6.2K]
3 years ago
12

An atom of the element iron has an atomic number of 26 and an atomic mass number of 56. if it is neutral, how many protons, neut

rons, and electrons does it have?
Chemistry
2 answers:
allsm [11]3 years ago
6 0
<span> 26 protons, 30 neutrons, 26 electrons</span>
asambeis [7]3 years ago
6 0

Answer:

26 protons, 26 electrons, 30 neutrons

Explanation:

Let P = protons, N = neutrons, E = electrons

The mass number of an atom is the sum of protons and neutrons in the nucleus of the atom.

The number of neutrons is the difference between the mass and atomic numbers of the atom

Therefore, 56 = P + N

where N = 56 - 26

N = 30

That means the number of neutrons is 30

To find P, 56 = P + N, N being 30

56 is now P + 30

56 = P + 30

56 - 30 = P

P = 26

Now, for an atom in its neutral state, the number of electrons is equal to the number of protons

Therefore, E is also 26

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0.08 because 4mol NO divided by 50g gives you your answer
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If you produced 76.10 grams of potassium chloride in the reaction how many grams of potassium were required?
11111nata11111 [884]

Answer:39.8375

Explanation:

The mole for the equation is 1:1

Then the molar mass of KCl is 74.5g

Molar mass of k is 39

74.5g of KCl gives 39g of k

76.10g of KCl gives xg of k

X= 76.10×39/74.5

X= 2967.9/7

X= 39.8375

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3 years ago
In the reaction 4Al+3O2→_Al2O3 , what coefficient should be placed in front of the Al2 to balance the reaction?
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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
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