Hello,
I think the answer is B) It is second longest river in south america
Hope this helps!!
~Girlygir101~
Answer:
Explanation:
According to Bronsted-Lowry acids or base theory , the reagent capable of giving hydrogen ion or proton will be acid and that which accepts hydrogen ion or proton will be base .
C₉H₇N + HNO₂ ⇄ C₉H₇NH⁺ + NO₂⁻
If K > 1 , reaction is proceeding from left to right .
Hence HNO₂ is giving H⁺ or proton and C₉H₇N is accepting proton to form
C₉H₇NH⁺ .
Hence HNO₂ is bronsted acid and C₉H₇N is bronsted base .
B )
when K < 1 , reaction above proceeds from right to left . That means
C₉H₇NH⁺ is giving H⁺ so it is a bronsted acid and NO₂⁻ is accepting H⁺ so it is a bronsted base .
Hence , NO₂⁻ is a bronsted base and C₉H₇NH⁺ is a bronsted acid .
We can use the dilution formula to find the volume of the diluted solution to be prepared
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
Substituting the values in the equation
15 M x 25 mL = 3 M x v2
v2 = 125 mL
The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution
Atomic size decreases in a period but the ionization energy and electronegativity increases across a period.
<h3>
Describe the trends in the atomic size, ionization energy and electronegativity?</h3>
Atomic radius decreases across a period because of nuclear charge increases whereas atomic radius of atoms generally increases from top to bottom within a group because there is again an increase in the positive nuclear charge.
Ionization energy increases when we move from left to right across an period and decreases from top to bottom.
Electronegativity also increases from left to right across a period and decreases from top to bottom.
So we can conclude that atomic size decreases in a period but the ionization energy and electronegativity increases across a period.
Learn more about Electronegativity here: brainly.com/question/24977425
#SPJ1
I believe the answer is C
