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il63 [147K]
3 years ago
12

DIPOLES FORM AT EACH BOND BETWEEN HYDROGEN AND OXYGEN IN A WATER MOLECULE BECAUSE _

Chemistry
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

Dipole form on each bond between hydrogen and oxygen atom in water molecule because of the electronegativity difference

between hydrogen and oxygen.

Explanation : Electronegativity means attraction of shared pair of electron toward itself in the covalent bond. Those atoms having same electronegativity values, the shared electron remain in the center whereas those atoms having different electronegativity values, the shared pair of electron is attracted by the atom having high value of electronegativity. In water molecule, electronegativity of oxygen is higher so it attracts electron pair towards itself and forms negative pole and hydrogen aquire positive pole.

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The VSEPR model was developed before any xenon compounds had been prepared. Thus, these compounds provided an excellent test of
Natalka [10]

The maximum amount of XeF4 that could be produced is 0.5 moles.

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<h3>What is mole ratio?</h3>

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6 0
2 years ago
2Na(s) + Cl2(g) - 2NaCl(s) + 822 kj
pychu [463]

Answer:

Is there any other part to this question? If not I'm pretty sure the answer is 205.5 kJ

Explanation:

8 0
3 years ago
Read 2 more answers
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
3 years ago
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