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Nimfa-mama [501]
1 year ago
6

The bond angles in the nitrite ion, nitrogen dioxide, and the nitronium ion (NO₂⁺) are 115°, 134°, and 180°, respectively. Expla

in these values using Lewis structures and VSEPR theory.
Chemistry
1 answer:
rewona [7]1 year ago
3 0

The bond angles in the nitrite ion , nitrogen dioxide and the Nitronium ion are 115°,134°,180° respectively .

In the nitrite ion , N-atom has sp2 hybridization so it has trigonal planner  geometry and bent shaped and has bond angle of 115° because of there is a lone pair on nitrogen atom due to which there is a repulsion occurs between the bond pair and lone pair . so the bond angle slightly deviate from 120° (general bond angle in case of sp2 hybridization ) and form a angle of 115° .

according to the Lewis structure nitrogen of nitrite ion contains 6 valance electron of oxygen and 5 valence electron of nitrogen . from which one oxygen and nitrogen bonded by double bond by sharing- pairing 4 electrons and other oxygen by single bond .

In case of Nitronium ion , N-atom has sp-hybridization , so it has linear geometry and the bond angle is 180° . also there is a  positive charge on the nitrogen atom . which cause the valence electron of nitrogen becomes 4 . and the valance electron of oxygen is 6 . so according to Lewis structure ,the two oxygen bonded to the nitrogen atom by double bond . hence the shape will be linear and bond angle will be 180° .

In case of nitrogen dioxide , it involves the sp2 hybridization . according to Lewis dot structure the nitrogen dioxide have one 2 sigma and 1 lone pair electron . it has bent geometry .due to the lone pair there is a greater repulsion occurs due to  which bond angle becomes 134° .

<h3>What is Lewis  structure ?</h3>

Lewis structure is also known as Lewis dot structure , Lewis dot diagram , electron dot structure . it shows the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule .

<h3>Hybridization :</h3>

it is the intermixing of atomic orbitals to make new hybrid orbital .

Learn more about hybridization here :

brainly.com/question/22765530

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Hitman42 [59]

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In this experiment, student groups ran repeated trials (5) and then averaged their data. ALL BUT ONE statement explains why
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A sample of lemon juice is found to have a pH of 2.3. What is the concentration of hydrogen ions in the lemon juice?
Elenna [48]

Answer: It's equal to 10^(-2.3), or 0.00501 M, or 5.01 * 10^-3 moles/Liter

Explanation:

Well, pH = - log[H+]

Or, in words, pH is equal to -1 multiplied by the logarithm (base 10) of the hydrogen ion concentration.   So you have 2.3 = -log[H+].    We want to isolate the H+, so let's start simplifying the right hand side of the equation. First, we multiply both sides by -1.   -2.3=log[H+]   Now, the definition of a logarithm says that if the log (base 10) of [H+] is -2.3, then 10 raised to the -2.3 power is [H+]   So on each side of the equation, we raise 10 to the power of that side of the equation.   10^(-2.3) = 10^(log[H+])   and because 10^log cancels out...   10^(-2.3) = [H+]   Now we've solved for [H+], the hydrogen ion concentration!

7 0
3 years ago
(3) A 10.00-mL sample of 0.1000 M KH2PO4 was titrated with 0.1000 M HCl Ka for phosphoric acid (H3PO4): Ka1= 7.50x10-3; Ka2=6.20
shtirl [24]

Answer:

The pH of this solution is 1,350

Explanation:

The phosphoric acid (H₃PO₄) has three acid dissociation constants:

HPO₄²⁻ ⇄ PO4³⁻ + H⁺        Kₐ₃ = 4,20x10⁻¹⁰  (1)

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸   (2)

H₃PO₄ ⇄ H₂PO4⁻ + H⁺       Kₐ₁ = 7,50x10⁻³   (3)

The problem says that you have 10,00 mL of KH₂PO₄ (It means H₂PO₄⁻) 0,1000 M and you add 10,00 mL of HCl (Source of H⁺) 0,1000 M. So you can see that we have the reactives of the equation (3).

We need to know what is the concentration of H⁺ for calculate the pH.

The moles of H₂PO₄⁻ are:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

The moles of H⁺ are, in the same way:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

So:

H₃PO₄   ⇄      H₂PO4⁻         +        H⁺           Kₐ₁ = 7,50x10⁻³   (3)

X mol     ⇄  (1x10⁻³-X) mol  + (1x10⁻³-X) mol                            (4)

The chemical equilibrium equation is:

Kₐ₁ = ([H₂PO4⁻] × [H⁺] / [H₃PO₄]

So:

7,50x10⁻³ = (1x10⁻³-X)² / X

Solving the equation you will obtain:

X² - 9,5x10⁻³ X + 1x10⁻⁶ = 0

Solving the quadratic formula you obtain two roots:

X = 9,393x10⁻³ ⇒ This one has no chemical logic because solving (4) you will obtain negative H₂PO4⁻ and H⁺ moles

X = 1,065x10⁻⁴

So the moles of H⁺ are : 1x10⁻³- 1,065x10⁻⁴ : 8,935x10⁻⁴ mol

The reaction volume are 20,00 mL (10,00 from both KH₂PO₄ and HCL)

Thus, the molarity of H⁺ ([H⁺]) is: 8,935x10⁻⁴ mol / 0,02000 L = 4,468x10⁻² M

pH is -log [H⁺]. So the obtained pH is 1,350

I hope it helps!

5 0
3 years ago
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