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3 years ago

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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Answer:

A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)

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Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.

The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:

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That means, the equation that represents standard enthalpy of CaCO₃ is:

<em>Is the equation that has ΔH° = -1207kJ/mol</em>

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3 years ago

A 4.7g sample of ethanol is burned in a calorimeter containing 95.8 mL of water. If the initial temperature of the water is 29.1

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3 years ago

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

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3 years ago

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goblinko [34]

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3 years ago

A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,

inna [77]

Answer:

3.676 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have different values of V and T:

(V₁T₂) = (V₂T₁)

Knowing that:

V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,

V₂ = ??? L, T₂ = 40°C + 273 = 313 K,

Applying in the above equation

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.

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3 years ago

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