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trasher [3.6K]
1 year ago
4

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

correct ________, 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.
Chemistry
1 answer:
Strike441 [17]1 year ago
7 0

Answer:

To run a spectrophotometry experiment, begin by warming up the spectrophotometer and preparing the samples. Be sure to select the correct wavelength, then run a measurement on the blank solution. Follow up by running measurements on sample solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.

Explanation:

The spectrophotometer is a device used in laboratories to carry out analysis in experiments that need to measure and compare the amount of light absorbed or transmitted between samples, regardless of whether the samples are transparent or opaque solid. The spectrophotometer can be single-beam and double-beam. Regardless of the nature of the beam, to perform a spectrophotometry experiment, you must start by heating the spectrophotometer and preparing the samples. Make sure to select the correct wavelength and perform a measurement on the blank solution. Follow up by performing measurements on sample solutions. Once the data is collected, turn off the instrument, clean the area and discard the samples.

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Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

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2 years ago
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So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

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