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Elenna [48]
3 years ago
14

Identify the correct equation for the equilibrium constant Kc for the reaction given. C u ( s ) + 2 A g N O 3 ( a q ) − ⇀ ↽ − C

u ( N O 3 ) 2 ( a q ) + 2 A g ( s ) Cu(s)+2AgNOX3(aq)↽−−⇀Cu(NOX3)X2(aq)+2Ag(s) Select one: K c = [ Cu ( NO 3 ) 2 ] [ Ag ] 2 [ AgNO 3 ] 2 [ Cu ] Kc=[Cu(NO3)2][Ag]2[AgNO3]2[Cu] K c = [ Cu ( NO 3 ) 2 ] 2 [ Ag ] [ AgNO 3 ] [ Cu ] 2 Kc=[Cu(NO3)2]2[Ag][AgNO3][Cu]2 K c = [ Cu ( NO 3 ) 2 ] [ AgNO 3 ] 2 Kc=[Cu(NO3)2][AgNO3]2 K c = [ Cu ( NO 3 ) 2 ] [ AgNO 3 ] Kc=[Cu(NO3)2][AgNO3]
Chemistry
1 answer:
lana [24]3 years ago
3 0

Answer:

Kc = [Cu(NO₃)₂]/[2 AgNO₃]²

Explanation:

Let's consider the following balanced redox equation.

Cu(s) + 2 AgNO₃(aq) ⇄ Cu(NO₃)₂(aq) + 2 Ag(s)

The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

The concentration equilibrium constant for this reaction is:

Kc = [Cu(NO₃)₂]/[2 AgNO₃]²

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Answer:

= 1.271 J/g°C

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by  water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the solid;

= 225 g × c ×(67 -53) , where c is the specific heat capacity of the metal

= 3150 c joules

Heat absorbed by water;

= 25.6 g × 4.18 J/g°C × (53-15.6)

= 4002.0992  joules

Therefore;

3150 c joules = 4002.0992 joules

c =4002.0992/3150

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3 years ago
BaCl2(aq) + Na2CO3(aq) BaCO3(s) + NaCl(aq
Vitek1552 [10]
This is the equation balanced:

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7 0
4 years ago
Which statement is true about elements?
LiRa [457]
A or C are the best options
5 0
3 years ago
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what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

\frac{V}{T}=k

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

\frac{V1}{T1}=\frac{V2}{T2}

In this case:

  • V1= 0.730 L
  • T1= 28 °C= 301 °K (0°C= 273°K)
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Replacing:

\frac{0.730 L}{301K}=\frac{V2}{273K}

Solving:

V2=273K*\frac{0.730L}{301K}

V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

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2 years ago
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