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Anna11 [10]
3 years ago
5

Which type of bond will form between two chlorine atoms?

Chemistry
2 answers:
beks73 [17]3 years ago
6 0
The link between the same atom is always non-polar Covalent.Is Covalent because there is no possibility of an atom donating electrons to the other, since they are equal and need the same number of electrons, and is why the formal charge non-polar (electronegativity) of two atoms is equal. When there is no difference of formal charges (dipole moment) between the atoms, the connection is always apolar.

hope this helps!
malfutka [58]3 years ago
3 0
Maybe (nonpolar) covalent bond.
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Formula of preparation of sodium hydroxide​
Umnica [9.8K]

The chemical formula of sodium hydroxide is NaOH, and its molar mass is 40.01 g/mol. It is the alkali salt of sodium, and its structure is shown below:

It is an ionic compound consisting of sodium cation (Na+) and hydroxide (OH-) anion.

7 0
2 years ago
He reaction produces 1.6 g of gas in 30 seconds.Calculate the mean rate of the reaction in the first 30 seconds.Use the equation
devlian [24]

Answer:

Mean rate of reaction produced = 0.533 g/sec (approx.)

Explanation:

Given:

Reaction produced = 1.6 gram

Time taken = 30 sec

Find:

Mean rate of reaction produced

Computation:

Mean rate of reaction produced = Reaction produced / Time taken

Mean rate of reaction produced = 1.6 / 30

Mean rate of reaction produced = 0.533 g/sec (approx.)

5 0
2 years ago
How many moles are in 25.0 grams of KMnO4?
ElenaW [278]
The answer is: 0.158 mol
You find this by doing:
number of moles (n) = mass (m) / molar mass (M)
n=158.034/25.0
6 0
2 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
2 years ago
Mathias Schleiden, one of the writers of the cell theory, studied _________ tissue and noticed that every sample he looked at wa
JulijaS [17]
You need to put there “poetry “
8 0
3 years ago
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