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Vitek1552 [10]
3 years ago
6

To run a spectrophotometry experiment, begin by _______ the spectrophotometer and preparing the samples. Be sure to select the c

orrect ________, then run a measurement on the _______ solution. Follow up by running measurements on _______ solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.
1.
a) cleaning
b) warming up
c) unplugging
2.
a) wavelength
b) transmittance
c) absorbance
3.
a) blank
b) aqueous
c) sample
4.
a) blank
b) aqueous
c) sample
Chemistry
1 answer:
12345 [234]3 years ago
8 0

Answer:

b) warming up a) wavelength a) blank c) sample

Explanation:

<em>To run a spectrophotometry experiment, begin by </em><em>warming up</em><em> the spectrophotometer and preparing the samples.</em> It is important that the equipment is warmed up for at least 30 minutes before starting the measurements.

<em>Be sure to select the correct </em><em>wavelength</em><em>, then run a measurement on the </em><em>blank</em><em> solution.</em> The selected wavelength depends on the analyte of interest. The black solution contains the same matrix but it doesn´t contain the analyte.

<em>Follow up by running measurements on </em><em>sample</em><em> solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples. </em>The samples are those of unknown concentration that we want to determine.

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Compare the chemical reactivity of chlorine and bromine. Explain your answer. Thanks in advance
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Bromine vs Chlorine | Br vs Cl

 

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Bromine

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3 years ago
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Answer:

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Explanation:

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For the following reaction, 6.94 grams of water are mixed with excess sulfur dioxide . Assume that the percent yield of sulfurou
Alexxx [7]
<h3>Answer:</h3>

#a. Theoretical yield = 31.6 g

#b. Actual yield = 25.72 g

<h3>Explanation:</h3>

The equation for the reaction between sulfur dioxide and water to form sulfurous acid is given by the equation;

SO₂(g) + H₂O(l) → H₂SO₃(aq)

The percent yield of H₂SO₃ is 81.4%

Mass of water that reacted is 6.94 g

#a. To get the theoretical yield of H₂SO₃ we need to follow the following steps

Step 1: Calculate the moles of water

Molar mass of water = 18.02 g/mol

Mass of water = 6.94 g

But, moles = Mass/molar mass

Moles of water = 6.94 g ÷ 18.02 g/mol

                        = 0.385 mol

Step 2: Calculate moles of H₂SO₃

From the equation, the mole ratio of water to H₂SO₃ is 1 : 1

Therefore, moles of water = moles of H₂SO₃

Hence, moles of H₂SO₃ = 0.385 mol

Step 3: Theoretical mass of H₂SO₃

Mass = moles × Molar mass

Molar mass of H₂SO₃ = 82.08 g/mol

Number of moles of H₂SO₃ = 0.385 mol

Therefore;

Theoretical mass of H₂SO₃ = 0.385 mol ×  82.08 g/mol

                                             = 31.60 g

Thus, the theoretical yield of H₂SO₃ is 31.6 g

<h3>#b. Calculating the actual yield</h3>

We need to calculate the actual yield

Percent yield of H₂SO₃ is 81.4%

Theoretical yield is 31.60 g

But; Percent yield = (Actual yield/theoretical yield)×100

Therefore;

Actual yield = Percent yield × theoretical yield)÷ 100

                   = (81.4 % × 31.6) ÷ 100

                  = 25.72 g

The percent yield of H₂SO₃ is 25.72 g

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Katyanochek1 [597]

Answer: They always have the same functional groups.

Explanation:

use quizlet too if you have toooo

3 0
3 years ago
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