The balanced equation for the ionization of the weak base pyridine,C5H5N in water, H2O
C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)
<h3>What is the balanced equation for the ionization?</h3>
Generally, Pyridine is characterized by a ring structure, in this characteristic ring structure N is sp2 hybridized, hence creating a lone pair present on N so s - character is more, as well as lone pair, is present.
Therefore, Considering The following functions of the equation:weak base pyridine,C5H5N in water, H2O
We write the balanced equation for the ionization as
C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)
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brainly.com/question/11231920
Answer:materials that impede the free flow of electrons from atom to atom and molecule to molecule.
An air mass is a large volume of air in the atmosphere thats mostly uniform in temperature and moisture. Air masses can extend thousands of kilometers across the surface of the Earth, and can reach from ground level to the stratosphere/10 miles into the atmosphere. an air mass over northern Canada is a continental polar air mass and is cold and dry. One that forms over the Indian Ocean is called a maritime tropical air mass and is warm and humid.
Answer:
D
Explanation:
The high jump of ionization energy indicates that we are trying to remove electron from noble gas configuration state.
The ionization energy data specifies that the Elements are from group 1 at period 3 or greater.
Removing the first electron require 496 kJ and the second ionization energy jump significantly due to the removal of electron from the noble gas configuration which is logical because electron try to maintain the especially stable state.
Answer:
Na₂₆F₁₁
Explanation:
We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:
74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na
25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F
Divide each by smallest number of moles:
3.2255/1.3586 = 2.37
1.3586/1.3586 = 1
Multiply by common number to get a smallest whole number:
2.37*11 = 26,
1*11 = 11
The empirical formula is Na₂₆F₁₁