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Yuliya22 [10]
3 years ago
9

Nitrogen can ionically bond with an unknown element X from group 2. How many ions of element X are required to create a stable i

onic compound with nitrogen? A) 2 B) 3 C) 4 D) There is not enough information.
Chemistry
2 answers:
fredd [130]3 years ago
6 0

Answer is: B) 3.

Nitrogen (nonmetal) in this compound has oxidation number -3 and element X (metal) has oxidation number +2.

Smallest common multiple for 3 and 2 is 6.

number of nitrogen atoms = 6 ÷ 3.

number of nitrogen atoms = 2, there are two nitrogen atoms.

number of atom X: 6 ÷ 2 = 3.

For example, Mg₃N₂ is magnesium nitride. Nitrogen has -3 oxidation number and magnesium has oxidation number +2.

klemol [59]3 years ago
5 0
To make a stable ionic compound you need 8 valence electrons and since nitrogen has 5 valence electrons he would need 4 more ions of Elements to make a full 8,
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What is the reduction potential of a hydrogen electrode that is still at standard pressure, but has ph = 5.65 , relative to the
Zigmanuir [339]

Answer:

\boxed{\text{-0.275 V}}

Explanation:

We must use the Nernst equation

E = E^{\circ} - \dfrac{RT}{zF} \ln Q

1. Write the equation for the cell reaction

If you want the reduction potential, the pH 5.65 solution is the cathode, and the cell reaction is

                                                                     <u> E°/V</u>

  Anode: H₂(1 bar) ⇌ 2H⁺(1 mol·L⁻¹) + 2e⁻;     0

Cathode: <u>2H⁺(pH 5.65) + 2e⁻ ⇌ H₂(1 bar)</u>;    <u> 0 </u>

 Overall: 2H⁺ (pH 5.65) ⇌ 2H⁺(1 mol·L⁻¹);      0

Step 2. Calculate E°

(a) Data

   E° = 0

   R = 8.314 J·K⁻¹mol⁻¹

   T = 25 °C

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pH = 5.65

Calculations:  

T = 25 + 273.15 = 298.15 K

\text{H}^{+} = 10^{\text{-pH}} = 2.24 \times 10^{-5}\text{ mol/\L}\\\\Q = \dfrac{\text{[H}^{+}]_{\text{prod}}^{2}}{\text{[H}^{+}]_{\text{react}}^{2}} = \dfrac{(1.00)^{2}}{(2.24 \times 10^{-5})^{2}} =2.00 \times 10^{9}\\\\\\E = 0 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln{2.00 \times 10^{9}}\\\\= -0.01285 \times 21.41 = \textbf{-0.275 V}\\\text{The cell potential for the cell as written is }\boxed{\textbf{-0.275 V}}

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3 years ago
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insens350 [35]

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