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IRISSAK [1]
3 years ago
8

Ammonia is produced commercially by the Haber reaction: N2(8)+ 3H2(8)

Chemistry
1 answer:
Shalnov [3]3 years ago
5 0

This is an incomplete question, here is a complete question.

Ammonia is produced commercially by the Haber  reaction:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)+Heat

The formation of ammonia is favored by

A) an increase in pressure

B) a decrease in pressure

C) removal of N₂(g)

D) removal of H₂(g)

Answer : The correct option is, (A) an increase in pressure.

Explanation :

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

Le-Chatelier's principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Increase the pressure :

If the pressure in the equilibrium is increased, the equilibrium will shift in the direction where fewer total molecules are shown in the balanced chemical equation.

In the given reaction, there are 4 molecules present on reactant side and 2 molecules on product side. That means, less molecules present on product side. Thus, the reaction will shift in right direction that is towards the product.

Removal the reactant molecule :

If any of reactant molecule in the equilibrium is removed, then the equilibrium will shift in left direction that is towards the reactant.

Hnece, the formation of ammonia is favored by  an increase in pressure.

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A 50.0 mL sample of 12.0 M HCl is diluted to 200 mL. What is true about the diluted solution?
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The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

  • V1 = 50mL
  • C1 = 12.0M
  • V2 = 200mL
  • C2 = ?
<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

Learn more on concentration of a solution here;

brainly.com/question/2201903

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2 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0
3 years ago
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