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2 years ago

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From gross and specific selection rules below, identify which belongs to the rotational, rotational Raman, vibrational and elect

ronic spectroscopies. Gross selection rules: The dipole moment must change during the transition Molecule must have permanent dipole moment Molecule must have anisotropic polarizability

1 answer:

guajiro [1.7K]2 years ago

8 0

Answer:

Rotational spectroscopy, the dipole moment must change during the transition.

Rotational Raman spectroscopy, molecule must have anisotropic polarizability

Vibrational and electronic spectroscopy, molecule must have permanent dipole moment.

Explanation:

For the vibration rotation spectrum to be observed, it is necessary to change the dipole moment during the vibration.

Raman scattering using an anisotropic crystal gives information about the orientation of the crystal. The polarization of Raman scattering light relative to the crystal, and the polarization of laser light, can be used to determine the orientation of the crystal, provided the crystal structure is known.

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Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at 30 °C. Calculate the pres

goldenfox [79]

The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

3 0

2 years ago

Marie and Calvin dissolved 10 grams of KNO3 in 100 grams of water at 25oC. Next they added 5 grams more. Calvin told Marie that

PSYCHO15rus [73]

Hello!

Calvin told Marie that they could continue to add solute until the reached 40 grams because the solution was still unsaturated.

Unsaturated solutions are those in which the solvent (in this case water) can still dissolve more solute (in this case KNO₃) at the given pressure and temperature. This can be seen visually when adding more solute doesn't result in the presence of grains of solids that settle in the bottom of the flask. That happens because the rate of dissolving is higher than the rate of crystallization.

Have a nice day!

8 0

2 years ago

Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

Ghella [55]

Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

$=0.0327\ moles$

Generally concentration is mathematically represented as

$concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$

$Z= \frac{0.0327}{12*10^{-3}}$

$=2.72 \ mol/L$

The dissociation reaction of $KClO_3$ is

$KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

$K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

$= 2.72^2$

$K_{sp}=7.40$

5 0

2 years ago

Read 2 more answers

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago

Which contain covalent bonds?

Ilia_Sergeevich [38]

The only compound that contains covalent bonds would be A. BCl4-.

3 0

2 years ago