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Firlakuza [10]
3 years ago
8

Infrared (IR) spectroscopy is used to identify functional groups within a molecule. Click on the peak that corresponds to the st

retching vibration for the highlighted bond.chemical bonds are given

Chemistry
1 answer:
Sergio [31]3 years ago
7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct place to click on any of the  double peak showing between  2500 and 3000cm^{-1} on the uploaded question

Explanation:

The group which is being highlighted is an Aldehyde  functional group denoted by this structure (=C-H)

This groups stretch on the infrared spectrometer gives two medium intensities peaks

Its stretch comes at  2820 - 2850 cm^{-1}  peak 1

and at  2720 - 2750 cm^{-1}  peak 2

So the correct click would be on any of the two peaks between 2500 - 3000 cm^{-1}  region

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Fog forms when a water vapor changes to the___ state of matter?
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What volume of a 1.5 M KOH solution is needed to provide 3.0 moles of KOH?
Brums [2.3K]

Answer:

  • Volume = <u>2.0 liter</u> of 1.5 M solution of KOH

Explanation:

<u>1) Data:</u>

a) Solution: KOH

b) M = 1.5 M

c) n = 3.0 mol

d) V = ?

<u>2) Formula:</u>

Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:

  • M = n / V in liter

<u>3) Calculations:</u>

  • Solve for n: M = n / V ⇒ V = n / M

  • Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter

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8 0
3 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

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