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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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The pressure of the gas is expected to increase in accordance to Boyle's law.

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Which of the following solids would not decompose on heating​
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Calculate the molarity of sodium ion in a solution made
Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles =Molarity \times {Volume\ of\ the\ solution}

Sodium chloride will furnish Sodium ions as:

NaCl\rightarrow Na^{+}+Cl^-

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

Moles =0.288 \times {3.58\times 10^{-3}}\ moles

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}

<u> The final concentration of sodium anion = 0.1035 M</u>

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3 years ago
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