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mars1129 [50]
2 years ago
14

A chemical reaction between X2 (red) and Y2 (blue) produces XY (red-blue). All compounds are in a gaseous state and each molecul

e represents 0.10 bar of that reactant or product. The picture shown here represents the equilibrium mixture.
There are six blue-blue clusters composed of two fused blue spheres, three red-red clusters composed of two fused red spheres, and seven red-blue clusters composed of a red sphere fused with a blue sphere.

Calculate the equilibrium constant for this reaction
Chemistry
1 answer:
Katarina [22]2 years ago
4 0

The equilibrium constant for this reaction is 1.65.

Solution:

X2: 3, Y2: 6, XY: 7

X₂ + Y₂ ⇌ 2XY

Kc = [XY]² / [X₂][Y₂] = [7]²/[3][6] = 2.72

1mole of equation √(2.72) = 1.65

The equilibrium constant of a chemical reaction is the value of the reaction quotient in chemical equilibrium the state to which a dynamic chemical system approaches when, after sufficient time, its composition exhibits no further measurable tendency to change. Q is the quantity that changes as the reaction system approaches equilibrium. K is the numerical value of Q at the end of the reaction when equilibrium is reached.

Balance Example A book on a table. A car that moves at a constant speed. A chemical reaction in which the forward and reverse kinetics are equal. The specific rate constant k is a constant of proportionality that relates reaction rate to reactant concentration. The rate laws and specific rate constants of chemical reactions must be determined experimentally. The value of the rate constant is temperature dependent.

Learn more about The equilibrium constant here:-brainly.com/question/3159758

#SPJ1

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garri49 [273]
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Table sugar: 12 atoms of carbon, 22 atoms of hydrogen; 11 atoms of oxygen; 45 total atoms
Marble: 1 atom of calcium, 1 atom of carbon; 3 atoms of oxygen; 5 total atoms
Natural gas: 1 atom of carbon, 4 atoms of hydrogen; 5 total atoms
Rubbing alcohol: 3 atoms of carbon, 8 atoms of hydrogen; 1 atom of oxygen; 12 total atoms
Table sugar: 1 atom of silicon; 2 atoms of oxygen; 3 total atoms
7 0
4 years ago
What is the date of discovery of element Mercury
STALIN [3.7K]

2000 BC and was found in tubes in Egyptian tombs dated from 1500BC

hope this helps

4 0
4 years ago
Read 2 more answers
for a theoretical yield of 22 g and actual yield of 13 g, calculate the percent yield for a chemical reaction
sammy [17]

Answer:

59.09%

Explanation:

13/22*100=59.09%

8 0
3 years ago
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
4 years ago
A 0.5 mol sample of He(g) and a 0.5 mol sample of Ne(g) are placed separately in two 10.0 L rigid containers at 25°C. Each conta
KiRa [710]

Answer:

Helum (He)g will escape faster

Explanation:

the phenomemenon can be explained by the Graham's law of diffusion.

Graham's law of difussion states that the rate of difussion is inversely proportional to the square root of the molecular mass,which means the gas with lower molecular mass will escape faster.

Helium gas has a molecular mass of 4 while Neon has a molecular mass of 10.

rate of diffusion of He/rate of difussion of Ne=√4/10=√0.4=0.63

It means He(g) will move 0.63 times faster than Ne(g) under the same condition

5 0
3 years ago
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