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Elenna [48]
2 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]2 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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Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis
almond37 [142]

Answer: a) BaCl_2

b)  BiO_3H_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = \frac{0.48}{0.48}=1

For O =\frac{0.96}{0.48}=2

The ratio of Ba: Cl= 1:2

Hence the empirical formula is BaCl_2

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles

Moles of O= \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= \frac{0.38}{0.38}=1

For O =\frac{1.15}{0.38}=3

For H=\frac{1.16}{0.38}=3

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is BiO_3H_3

6 0
2 years ago
Write the net ionic equation for the reaction of zinc metal with aqueous iron(II) nitrate. Include physical states.
kotykmax [81]

Answer:

Zn°(s) + Fe⁺²(aq)  => Zn⁺²(aq) + Fe°(s)

Explanation:

Molecular Equation:

Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)

Ionic Equation:

Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)

Net Ionic Equation: => Drop NO₃⁻ as spectator ion

Zn°(s) + Fe⁺²(aq)  => Zn⁺²(aq) + Fe°(s)

7 0
2 years ago
The molecular weight of a gas is ________ g/mol if 6.7 g of the gas occupies 6.3 l at stp.
PIT_PIT [208]
At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol. 
                               n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
                                          6.7g/0.28125 mol = 23.82 g/mol 
6 0
2 years ago
How many hydrogen atoms are in 11C2H6
tresset_1 [31]

Answer:

6

Explanation:

You will see H6 and the H stands for helium and the 6 is how many of that atom is there

5 0
2 years ago
Read 2 more answers
Identify one disadvantage to each of the following models of electron configuration:
Murrr4er [49]

Answer:

The disadvantages of each of the given model of electron configuration have been mentioned below:

1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.

2). Arrow and line diagrams make the counting of electrons and take up too much space.

3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.

6 0
2 years ago
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