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mars1129 [50]
1 year 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]1 year 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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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec
Bess [88]

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0

Let energy change be E for 1 atom.

E=E_{\infty}-E_1=0-(-13.6  eV)=13.6 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol

E'=1,312.17 kJ/mol

The energy  required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ ,B^{4+} atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-340eV)=340 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 340eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol

E'=32,804.31 kJ/mol

The energy  required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ ,Li^{2+}atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol

E'=11,809.55 kJ/mol

The energy  required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ ,Mn^{24+}atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol

E'=820,107.88 kJ/mol

The energy  required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0
3 years ago
Mass = 35g Volume = 7cm3 What is the Density?​
olga_2 [115]

Answer:

5 g/cm^3

Explanation:√3V=1.91293cm

7 0
3 years ago
The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has
Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate law expression is:

Rate=k[NO]^2[O_2]

Now we have to determine the number of molecules of NO\text{ and }O_2

In mixture 1 : There are 5 NO and 4 O_2 molecules.

In mixture 2 : There are 7 NO and 2 O_2 molecules.

In mixture 3 : There are 3 NO and 5 O_2 molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

Rate=k[NO]^2[O_2]

Rate=k(5)^2\times (4)

Rate=k(100)

The rate law expression for mixture 2 is:

Rate=k[NO]^2[O_2]

Rate=k(7)^2\times (2)

Rate=k(98)

The rate law expression for mixture 3 is:

Rate=k[NO]^2[O_2]

Rate=k(3)^2\times (5)

Rate=k(45)

Hence, the mixture 1 has the fastest initial rate.

4 0
3 years ago
Combustion analysis of toluene, a common organic solvent, gives 3.52 mg of co2 and 0.822 mg of h2o. if the compound contains onl
IRISSAK [1]
<span>C7H8 First, lookup the atomic weight of all involved elements Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Then calculate the molar masses of CO2 and H2O Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now calculate the number of moles of each product obtained Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK. moles CO2 = 3.52 / 44.0087 = 0.079984 moles H2O = 0.822 / 18.01488 = 0.045629 Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen. moles C = 0.079984 moles H = 0.045629 * 2 = 0.091258 So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting 1: 1.140953 The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like) 1 ( 1 : 1.140953) = 1 : 1.140953 2 ( 1 : 1.140953) = 2 : 2.281906 3 ( 1 : 1.140953) = 3 : 3.422859 4 ( 1 : 1.140953) = 4 : 4.563812 5 ( 1 : 1.140953) = 5 : 5.704765 6 ( 1 : 1.140953) = 6 : 6.845718 7 ( 1 : 1.140953) = 7 : 7.986671 8 ( 1 : 1.140953) = 8 : 9.127624 That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>
7 0
3 years ago
Each of the following are descriptions of Physical properties except ____.
Naily [24]
Each of the following are descriptions of physical properties except C. Flammability
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