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melamori03 [73]
3 years ago
11

18-19. If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the a

bsor- bance in a 1-cm cell. Will the absorbance of the reagent-blank solu- tion also be increased by a factor of 10?\
Chemistry
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Explanation:

Absorbance of light by a reagent of concentration c, is given as

A = εcl

A = Absorbance

ε = molar absorptivity

c = concentration of reagent.

l = length of light path or length of the solution the light passes through.

So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.

But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.

Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Hope this Helps!!!

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How many moles are there in 24.0 grams of FeF3
Sphinxa [80]

Hey there!:

Molar mass FeF3 = 112.84 g/mol

Number of moles  = mass of solute / molar mass

Number of moles = 24.0 / 112.84

Number of moles = 0.212 moles of FeF3


Hope that helps!

8 0
4 years ago
Which explains why the laws
Brums [2.3K]

Answer:

D

Explanation:

Scientific laws describe how things work with little to no exception. They do NOT provide an explanation to WHY something occurs.

8 0
3 years ago
Read 2 more answers
- Unit 1 Worksheet 4
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I think it is C or F
4 0
3 years ago
Attempt 2
Assoli18 [71]
Lets let our mass equal 3 on alletals and solve using d=m/v equation

Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium
4 0
3 years ago
Ten kilograms of R-134a fill a 1.595-m3 weighted piston-cylinder device at a temperature of −26.4°C. The container is now heated
Tju [1.3M]

Explanation:

The given data is as follows.

    Mass of refrigerant, m = 10 kg

  Volume of the refrigerant, V = 1.595 m^{3}

Formula for specific volume of the refrigerant is as follows.

        v = \frac{V}{m}

           = 0.1595 m^{3}/kg

So, at -26.4^{o}C specific volume will be within v_{f} and v_{g} and pressure is constant.

The fluid will be in super-heated state at temperature 100^{o}C and at T = -26.4^{o}C pressure 1 bar = 0.1 MPa.

According to super-heated tables, the specific volume is v = 0.30138 m^{3}/kg.

Hence, the final volume will be calculated as follows.

               V_{f} = v \times m

                         = 0.30138 m^{3}/kg \times 10 kg

                         = 3.0138 m^{3}

Thus, we can conclude that final volume of the R-134a is 3.0138 m^{3}.

7 0
3 years ago
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