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julia-pushkina [17]

3 years ago

14

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond order

s. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy.
a. CN +
b. CN
c. CN -

1 answer:

svet-max [94.6K]3 years ago

4 0

Answer:

Bond orders

CN- =3

CN+ =2.5

CN =2

CN+ is paramagnetic

The order of increasing bond energy is CN <CN+ <CN-

The order of increasing bond length is CN- <CN+ <CN

Explanation:

The molecular orbital diagram of CN- is shown in the image attached. From that image, it is easy to deduce the bond order of the other related species. Bond length is in the reverse order as bond energy because the shorter the bond, the greater the bond energy. The specie with the highest bond order has the highest bond energy.

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Volume of HCl used 25.0mL 4 l

borishaifa [10]

Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB = 25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

8 0

3 years ago

Find the concentration of H+ ions at a pH = 11 and

Black_prince [1.1K]

Explanation:

the pH of the solution defined as negatuve logarithm of $H^+$ ion concentration.

$pH=-\log[H^+]$

1. Hydrogen ion concentration when pH of the solution is 11.

$11=-\log[H^+]$

$[H^+]=1\times 10^{-11} mol/L$ ..(1)

At pH = 11, the concentration of $H^+$ ions is $1\times 10^{-11} mol/L$ .

2. Hydrogen ion concentration when the pH of the solution is 6.

$6=-\log[H^+]'$

$[H^+]'=1\times 10^{-6} mol/L$ ..(2)

At pH = 6, the concentration of $H^+$ ions is $1\times 10^{-6} mol/L$ .

3. On dividing (1) by (2).

$\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} mol/L}{1\times 10^{-6} mol/L}=1\times 10^{-5}$

The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is $1\times 10^{-5}$ .

4. Difference between the $H^+$ ions at both pH:

$1\times 10^{-6} mol/L-1\times 10^{-11} mol/L=9.99\time 10^{-7} mol/L$

This means that Hydrogen ions in a solution at pH = 7 has $9.99\time 10^{-7} mol/L$ ions fewer than in a solution at a pH = 6

6 0

3 years ago

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Which of the following compounds contains the lead (iv) ion?

Oliga [24]

Oxidation state of Pb in PbO2 is +4.
Oxidation state of Pb in PbCl2 is +2.
Oxidation state of Pb in Pb2O is +1.
Oxidation state of Pb in Pb4O3 is +6/4.

Hence option A. PbO2 is correct.
Hope this helps, have a nice day!

6 0

3 years ago

Read 2 more answers

What is the frequency of a light that has a wavelength of 3.0 x 10 -7m?

alexdok [17]

Answer:

frequency of light (f) = 1 x 10¹⁵s⁻¹

Explanation:

Given Data:

Wavelength of light λ = 3.0 x10⁻⁷m

Frequency of light: to be calculated

Formula Used to find frequency:

f = V/λ ........................... (1)

where

f is the frequency

V is the velocity

λ is wavelength

Velocity of light = 3 x 10⁸ ms⁻¹

put the values in equation (1)

f = 3 x 10⁸ ms⁻¹ / 3.0 x10⁻⁷m

f = 1 x 10¹⁵s⁻¹

So the frequency of light = 1 x 10¹⁵s⁻¹

7 0

2 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

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3 years ago