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kramer
2 years ago
14

Question 8 helpppppp

Chemistry
2 answers:
Strike441 [17]2 years ago
6 0
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2><u>LE CHATELIER PRINCIPLE</u>:</h2>

Le Chatelier's principle is a principle which predicts the effect of change in condition on a the position of a equilibrium.  

Change could be in Temperature, Moles(Concentration), Pressure.

-------------------------------------------------------------------------------------------------------------

<h2>EFFECTS ON EQUILIBRIUM</h2><h3 /><h3>__________________________________________________</h3><h3>Temperature: </h3>

We could change the temperature by increasing or decreasing it, and there are two types of reactions, endothermic or exothermic, both have different outcomes by changing the condition.

__________________________________________________________

In Endothermic,

When we increase the temperature, It will favor the formation of product thus MORE PRODUCT will be formed - equilibrium will shift towards right i.e. product side.

When we decrease the temperature, It will favor the formation of reactants thus MORE reactant will be formed - equilibrium will shift towards left i.e. reactants side.

__________________________________________________________

In Exothermic,

When we increase the temperature, It will favor the formation of reactants thus MORE REACTANT will be formed - equilibrium will shift towards left i.e. reactants side.

When we decrease the temperature, It will favor the formation of product thus MORE PRODUCT will be formed - equilibrium will shift towards the right i.e. product side

<h2>_____________________________________</h2><h3>Pressure:</h3>

When we increase the pressure, The equilibrium shifts towards the fewer moles of a gas. Wherever there are less moles, it could be any, reactant or product, the increase of Pressure will favor in formation of them.

When we decrease the pressure, The equilibrium shifts towards the more moles of a gas. So whether reactants or product have the higher moles, the decrease in pressure will favor in the formation of them.

<h2>_____________________________________</h2><h3>Concentration of moles:</h3>

When we increase the concentration of moles of any substance, The equilibrium will shift away from that substance. So if we want to make more product and there are more moles of reactant present, then Increase the concentration of reactants.

When we decrease the concentration of any substance, The equilibrium will shift towards that substance. So if we want to make more product, decrease its concentration or add more reactant.

<h2>_____________________________________</h2><h3>Catalyst:</h3>

By the use of the catalyst, There is no effect on the equilibrium. The rate of  both forward and backward reaction are increased with the equal amount.

<h2>_____________________________________</h2><h2>Question:</h2>

By keeping that in mind,

A) Extra NH_{3} will shift the equilibrium towards left i.e. more reactants will be formed.

B) Extra N2 is removed, the equilibrium will shift towards the left i.e. more reactants will be formed.

C) N2 is added, the equilibrium will shift towards right i.e. more PRODUCT will be formed. (I assumed there was a typing error and it meant Addition instead of removal)

D) When we decrease the pressure the equilibrium will shift towards the left i.e. more reactant will be formed.

E) Catalyst has no effect on the system

<h2>_____________________________________</h2><h2>Best Regards</h2><h2>'Borz'</h2><h2 />
solniwko [45]2 years ago
5 0

Answer:

Explanation:

This is lechatlier's principle, which basically talks about how adding or taking away things will disrupt equilibrium and the system will act a certain way to restore equilibrium

1. adding NH3 causes the equilibrium to shift left - produce more reactants

2. removing N2 causes the equilibrium to shift left - produce more reactants

3.  same as number 2

4. decreasing pressure causes equilbrium to shift to the side with more moles so it shifts left and produces more reactants

5. catalyst doesn't change anything about equlibrium, it only affects the reaction's rate

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Why is performing extraction with several small portions of a solvent more officient than a single extraction with the same tota
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An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

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pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
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