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Igoryamba
3 years ago
9

Would it be possible to conduct an UV/Vis spectroscopy experiment on a sample of the compound C3H3O2? Explain.

Chemistry
2 answers:
ICE Princess25 [194]3 years ago
7 0

Answer: Yes, we can conduct a UV/Vis spectroscopy experiment for C_3H_3O_2 compound.

Explanation: UV/Vis Spectroscopy is used for the quantitative determination of conjugated organic compounds, biological macromolecules and transition metal ions.

We are given a compound C_3H_3O_2, in which conjugation is present and undergoes \pi-\pi^* transition. This transition means that the energy gap is low and hence the wavelength will be more.

According to the Planck's relation:

E=\frac{hc}{\lambda}

Energy and wavelength follow inverse relation.


alexandr402 [8]3 years ago
4 0
Different substances can interact differently with light. So for me, I think the answer would be yes. It is possible <span>to conduct an UV/Vis spectroscopy experiment on a sample of the compound C3H3O2. Every substance is unique and has its own light interactions.</span>
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metal salt acid hydroyon and why cuz I guessed and I haven't t learned this yet and I need points

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3 years ago
A chemistry student is given 700. mL of a clear aqueous solution at 26.° C. He is told an unknown amount of a certain compound
Alex73 [517]

Answer:

The correct answer is - yes, 4.57 g of solute per 100 ml of solution

Explanation:

The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.

Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.

Then solublity will be :

32. g solute / 700 ml solution = y / 100 ml solution

⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.

Thus, the answer is 4.57 g of solute per 100 ml of solution.

5 0
2 years ago
What is the chemical formula for Salt and hydrogen gas?
marshall27 [118]
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3 0
2 years ago
Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2
lapo4ka [179]

Answer: Concentration of CO = 0.328 M

Concentration of Cl_2 = 0.328 M

Concentration of COCl_2 = 0.532 M

Explanation:

Moles of  CO and Cl_2 = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of CO and Cl_2  =\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M

The given balanced equilibrium reaction is,

                            CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.          0.860M     0.860M           0

At eqm. conc.    (0.860-x) M  (0.860-x) M     (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

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

4.95=\frac{x}{(0.860-x)^2}

By solving the term 'x', we get :

x =  0.532 M

Thus, the concentrations of CO,Cl_2\text{ and }COCl_2 at equilibrium are :

Concentration of CO = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of Cl_2 =  (0.860-x) M  =  (0.860-0.532) M = 0.328 M

Concentration of COCl_2 = x M = 0.532 M

3 0
2 years ago
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cupoosta [38]

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7 0
2 years ago
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