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Natasha2012 [34]
3 years ago
13

In a hurry to complete the experiment, Joseph failed to calibrate the spectrophotometer. As a result, all absorbance values for

the standard solutions that are measured and recorded are too high. How will this affect the following for the Test Solutions?
Chemistry
2 answers:
lilavasa [31]3 years ago
7 0

\boxed{{\text{Incorrect results}}} will be observed for test solutions if calibration of spectrophotometer is not done.

Further explanation:

Spectrophotometry

It is a technique that is used to measure the absorbance of light as a measure of the intensity of light when light passes through the sample of the solution. The optical instrument used for such purpose is a spectrophotometer.

Lambert-Beer’s Law

This law relates to the absorbance of the species to its concentration and the optical path length through the sample. The mathematical expression for Lambert-Beer’s law is,

\boxed{A = \varepsilon lc}  

Here,

A is the absorbance of the species.

\varepsilon is the molar absorptivity of the species.

<em>l</em> is the optical path length.

<em>c</em> is the concentration of species.

While using a spectrophotometer, calibration is an essential step in determining the absorbance of the sample. This done by using a blank solution (standard solution) and taking its absorbance value as a reference for other solutions. But if the equipment is not calibrated before measuring the absorbance for other solutions, the results will be incorrect.

In the given experiment, the absorbance values of standard solutions are recorded very high. This will lead to incorrect values of absorbance for test solutions. This is because the absorbance of the standard solution is to be used as a reference for the test solutions.

Learn more:

1. Classify the transitions as absorption or emission of light: brainly.com/question/1594022

2. What is the frequency of yellow light? brainly.com/question/5882803

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Beer-Lambert Law

Keywords: absorbance, Lambert-Beer law, calibration, standard solutions, test solutions, blank solution, spectrophotometry, spectrophotometer.

gizmo_the_mogwai [7]3 years ago
5 0
Joseph will not get the correct results for his samples. The spectophotometer will measure wrong absorbance values for the sample. It is highly advised to callibrate the instrument by first setting the absorbance of the solvent to zero. After it is done, only then one must determine the absorbance of the test solutions.
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A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
A container of a mixture of 4 gases has a total pressure of 35.7 kPa. Gas A has a partial pressure of 7.8kPa. Gas B has a partia
Maurinko [17]

Answer:

partial pressure of gas D Pd = 15.5 kPa

Explanation:

As per the Dalton's law of partial pressure, in a mixture, pressure exerted by each gas when summed gives the total partial pressure exerted by mixture.

P(Total) = P1+P2+P3.....

Given P(Total) = 35.7 kPa

Partial pressure of gas A Pa = 7.8 kPa

Partial pressure of gas B Pb = 3.7 kPa

Partial pressure of gas C Pc =  8.7 kPa

There, Partial pressure of gas D Pd = P(Total) -(Pa+Pb+Pc)

Pd = 35.7-(7.8+3.7+8.7) = 35.7-20.2 kPa = 15.5 kPa

Therefore, partial pressure of gas D Pd = 15.5 kPa

4 0
3 years ago
Which of the following is a completely elastic collision? A. a ball rebounds against a wall, reversing its direction, but at onl
Dmitry_Shevchenko [17]

Answer:

Answer E.

For a collision to be completely elastic, there must be NO LOSS in kinetic energy.

We can go through each answer choice:

A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.

B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.

C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.

D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.

E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:

Let:

m = mass of each ball

v = velocity

We have the initial kinetic energy as:

KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2

And the final as:

KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2

5 0
3 years ago
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Which is NOT a chemical reaction?
Sophie [7]
A is correct because there is no reaction involved
6 0
3 years ago
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Determine the volume of atmospheric air (at 14 lb/in.^2) needed to fill your bike tires (assuming it holds 500 mL of air) to the
Fiesta28 [93]

Boyle Law says “the pressure of fixed amount of ideal gas which is at constant temperature is inversely proportional to its volume".<span>

P = 1/V

<span>Where, P is pressure of the ideal gas and V is volume of the ideal gas.</span>

<span>For two situations, this law can be added as;
P</span>₁V₁ = P₂V₂<span>

</span><span>14 lb/in² x V₁ = 70 lb/in² x 500 mL</span><span>
                 </span><span>V₁ = 2500 mL</span><span>

Hence, the needed volume of atmospheric air = 2500 mL

<span>Here, we made two </span>assumptions. They are,
1. The atmospheric air acts as ideal gas.
2. Temperature is a constant.

<span>We didn't convert the units to SI units since converting volume and pressure are products of two numbers, they will cut off. </span></span></span>

3 0
4 years ago
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