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Nikitich [7]
3 years ago
12

Which of the highlighted carbon-oxygen bonds absorbs at a higher wavenumber in an IR spectrum?

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

1. bond in the molecule on the right

Explanation:

CH3CH2-OH

The compound above is an alcohol due to the presence of the OH bond. The wave number of the C - O bond is given as; 1050-1150 cm^-1.

CH3CH__O

The compound above is an aldehyde due to the presence of the CHO bond. The wave number of the C = O  bond is given as; 1740-1720 cm^-1

Comparing both bonds, the C = O bond absorbs at a higher wave number.

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The rising of rock layers is uplift
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Balance the following equation:<br> Sg + F2 --&gt; SF
Tanzania [10]

Answer:

S+ F2 ⇒ SF

S=1

F =2

So S +F2 ......... 2SF

2S + F2 ..........2SF this is a balance equation

S=2 F=2 in left side s=2 F = 2 in rightside

Explanation:⇆

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3 0
3 years ago
Using the standard enthalpies of formation for the chemicals involved, calculate the enthalpy change for the following reaction.
Firlakuza [10]

Answer: -134 kJ

Explanation:

The balanced chemical reaction is,

3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq)+NO(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{HNO_3}\times \Delta H_{HNO_3})+(n_{NO}\times \Delta H_{NO})]-[(n_{H_2O}\times \Delta H_{H_2O})+(n_{NO_2}\times \Delta H_{NO_2})]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(2\times -207)+(1\times 90)]-[(1\times -286)+(3\times 32)]

\Delta H=-134kJ

Therefore, the enthalpy change for this reaction is, -134 kJ

6 0
3 years ago
A 0.2 g sample of pyrolusite is analyzed for manganese content as follows. Add 50.0 mL of 0.1 M solution of ferrous ammonium sul
enyata [817]

Answer:

66.7%

Explanation:

The reaction for the titration of the excess ferrous ion is:

  • 5Fe⁺² + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

We calculate the moles of Fe⁺² from the used moles of KMnO₄:

  • 0.02 M * 15.0 mL = 0.30 mmol KMnO₄
  • 0.3 mmol KMnO₄ * \frac{5mmolFe^{+2}}{1mmolKMnO_4} = 1.5 mmol Fe⁺²

Then we substract those 0.30 mmol from the original amount used:

  • 0.1 M * 50.0 mL = 5.0 mmol Fe⁺²
  • 5.0 - 1.5 = 3.5 mmol Fe⁺²

The reaction between ferrous ammonium sulfate and MnO₂ is:

  • 2Fe⁺² + MnO₂ + 4H⁺ → 2Fe³⁺ + Mn²⁺ + 2H₂O

So we convert those 3.5 mmol Fe⁺² that were used in this reaction to MnO₂ moles:

  • 3.5 mmol Fe⁺²  * \frac{1mmolMnO_2}{2mmolFe^{+2}}= 1.75 mmol MnO₂

Then we convert MnO₂ to Mn₃O₄, using the reaction:

  • 3MnO₂ → Mn₃O₄ + O₂
  • 1.75 mmol MnO₂ * \frac{1mmolMn_3O_4}{3mmolMnO_2} = 0.583 mmol Mn₃O₄

Finally we convert Mn₃O₄ moles to grams:

  • 0.583 mmol Mn₃O₄ * 228.82 mg/mmol = 133.40 mg Mn₃O₄

And calculate the percent

  • 0.2 g = 200 mg
  • 133.40 / 200 * 100% = 66.7%
5 0
3 years ago
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