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ss7ja [257]
3 years ago
6

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
Chemistry
1 answer:
guajiro [1.7K]3 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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Most minerals are _______ at room temperature.<br><br> solid<br> gas<br> liquid
Tomtit [17]

Answer:

solid

Explanation:

4 0
3 years ago
Read 2 more answers
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
3 years ago
Adding water to a concentrated stock solution results in the...?
SOVA2 [1]

Answer:

The solution becomes diluted.

Explanation:

When you add water to a solution, the number of moles of the solvent stays the same while the volume increases. Therefore, the molarity decreases.

Hope this helps!

5 0
3 years ago
Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1
tekilochka [14]

Answer:

3.26 % of vinegar is acetic acid

Explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid

7 0
3 years ago
In the chemical reaction:
tiny-mole [99]

Answer:

d

Explanation:

d

6 0
2 years ago
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