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Digiron [165]
3 years ago
13

Hno3, a strong acid, is added to shift the ag2co3 equilibrium (equation 7.6) to the right. explain why the shift occurs.

Chemistry
1 answer:
Deffense [45]3 years ago
7 0
Answer 1)  When a strong acid like HNO_{3} reacts with Ag_{2} CO_{3} usually the equilibrium shifts to the right because

As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change."  So, in the given reaction when HNO_{3} reacts with Ag_{2} CO_{3}  it generates carbon dioxide and water as a by product, if we are adding HNO_{3} it will remove some of the CO_{3} molecule from the reaction mixture, which then tends to shift the equilibrium towards right.

Answer 2) The same would be observed in this case, if we replace HNO_{3} with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.

As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing Ag^{+} and Cl^{-} ions from the solution.
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Answer:

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Explanation:

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7 0
3 years ago
Catalyst
Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

\tt \dfrac{4}{5}\times 0.5=0.4

volume NO at 1273 K and 1 atm

\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L

b. 15 L NH3 at STP ( 1mol = 22.4 L)

\tt \dfrac{15}{22.4}=0.67~mol

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

\tt \dfrac{6}{4}\times 0.67=1

mass H2O(MW = 18 g/mol) :

\tt mass=mol\times MW=1\times 18=18~g

c. mol NO at 1273 K and 1 atm :

\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34

mol ratio of NO : O2 = 4 : 5, so mol O2 :

\tt \dfrac{5}{4}\times 0.34=0.425

Volume O2 at STP :

\tt 0.425\times 22.4=9.52~L

5 0
3 years ago
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MArishka [77]

Answer:

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8 0
2 years ago
Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

Now put all the values in this expression, we get :

K_p=\frac{(10.24)}{(0.348)(0.441)}

K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

4 0
3 years ago
Below are shown, for five metals, reduction reactions and standard electrode potential values. which of these metals is the leas
just olya [345]
Above question is incomplete. Complete question is attached below
........................................................................................................................
Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.

In present case, reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is least reactive.

On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive. 


7 0
3 years ago
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