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Anit [1.1K]
3 years ago
9

For the reaction of gaseous nitrogen with gaseous hydrogen to produce gaseous ammonia, the correct equilibrium constant expressi

on is _____
Chemistry
1 answer:
exis [7]3 years ago
7 0

The solution would be like this for this specific problem:

 

First, we need to write out the balanced reaction equation for the problem:<span>

"gaseous nitrogen with gaseous hydrogen to produce gaseous ammonia" 

gaseous nitrogen + gaseous hydrogen = gaseous ammonia 

gaseous nitrogen = N2(g) 
gaseous hydrogen = H2(g) 
gaseous ammonia = NH3(g) 

In here, I’m going to show you how to balance this from start to finish: 

N2(g) + H2(g) ↔ NH3(g) (basic, unbalanced equation that shows reactants and product) 
N2(g) + H2(g) ↔ 2NH3(g) (balances the nitrogens) 
N2(g) + 3H2(g) ↔ 2NH3(g) (balances the hydrogens) => Balanced Equation 

<span>Next, we write the Keq expression of the balanced reaction:

Keq = (∏[products]^n) /(∏[reactants]^n) 
Keq = [NH3]^2 / [[N2][H2]]^3 


Therefore, the correct equilibrium constant expression is Keq = (NH3)2 / (N2)(H2)3.</span></span>

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Which substance is an electrolyte <br> 1) CCl4 <br> 2)SiO2<br> 3)C6H12O6<br> 4) H2SO4
Elena L [17]
Electrolyte is any species which when dissolved in solvent particularly water dissociates into cations and anions. Electrolytes are conductors of electricity. In given options;

CCl₄ (Tetrachloromethane) is a covalent compound. And it doesn't dissociate to any cation or anion. So it is not electrolyte.

SiO₂ (Silicon Dioxide) is also covalent in nature and exist in giant framework. It is not electrolyte.

Glucose (C₆H₁₂O₆) is also covalent compound. And doesn't produced any ion in water, hence it is not electrolyte.

H₂SO₄ (Sulfuric acid) is Electrolyte. When it is dissolved in water it produces H⁺ and SO₄²⁻ ions as follow,

                                 H₂SO₄    →      2 H⁺ ₍aq₎    +   SO₄²⁻ ₍aq₎

Result:
               H₂SO₄ is electrolyte.
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Explanation:

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4 0
3 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
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