A solution of benzene in methanol has a transmittance of 93.0 % in a 1.00 cm cell at a wavelength of 254 nm. Only the benzene absorbs light at this wavelength, not the methanol. What will the solution's transmittance be if it is placed in a 10.00 cm long pathlength cell

Answers

Answer 1

Answer:

T = 48.39%

Explanation:

In this case we need to apply the Beer law which is the following:

A = CεL  (1)

Where:

A: Absorbance of solution

C: Concentration of solution

ε: Molar Absortivity (Constant)

L: Length of the cell

Now according to the given data, we have transmittance of 93% or 0.93. We can calculate absorbance using the following expression:

A = -logT (2)

Applying this expression, let's calculate the Absorbance:

A = -log(0.93)

A = 0.03152

Now that we have the absorbance, let's calculate the concentration of the solution, using expression (1).

A = CεL

C = A / εL

Replacing:

C = 0.03152 / 1 *ε   (3)

Now, we want to know the transmittance of the solution with a length of 10 cm. so:

A = CεL

Concentration and ε are constant, so:

A = (0.03152 / ε) * ε * 10

A = 0.3152

Now that we have the new absorbance, we can calculate the new transmittace:

T = 10^(-A)

T = 0.4839 ----> 48.39%


Related Questions

a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a buffer composed of HC2H3O2 and C2H3O2 - ? Write a chemical equation for that reaction.c) What happens when HBr is added to this buffer? Write a chemical equation for that reaction.

Answers

Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The specific heat capacity of
wood is 1.716 joules/gram degree Celsius. What is the mass of the piece of wood?
ОА. 16 g
OB. 29 g
ОC. 36 g
OD. 61 g

Answers

Answer:

35.578g or 36g if you round

Explanation:

Q=mc ∆∅ where ∅ is temperature difference

1160= m x 1.716 x (42-23)

m = 1160/ 1.716 x19

m=35.578g

m = 36g to nearest whole number

Answer: C. 36 g

Explanation: I got this right on Edmentum.

Which of the following would be more reactive than magnesium (Mg)?
A. Calcium (Ca)
B. Potassium (K)
C. Argon (Ar)
D. Beryllium (Be)

Answers

Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium

Explanation:

I hope it will help you

Answer: B. Potassium(K)

Explanation:

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