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2 years ago

In addition to not causing damage to the sample, what is another advantage of using a microspectrophotometer to analyze fibers?

1 answer:

5 0

Another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.

<h3>What is a microspectrophotometer?</h3>

Microspectrophotometry is a biological technique used to measure the absorption or transmission spectrum of a solid or liquid material in either transmitted or reflected light.

Microspectrophotometry can also measure the emission of light by a sample, which is usually small as the micro implies.

One advantage of microspectrophotometry is that the sample does not get damaged. However,

However, another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.

Learn more about microspectrophotometry at: brainly.com/question/5832827

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The free energy for the oxidation of glucose to CO2 and water is -686 kcal/mol, and the free energy for the reduction of NAD+ to

vitfil [10]

Answer:

Most of free energy available from oxidation of the glucose remains in pyruvate.

Explanation:

The overall reaction of the process glycolysis is:

Glucose + 2 NAD⁺ + 2 ADP + 2 Pi ⇒ 2 Pyruvate + 2 NADH + 2 H⁺ + 2ATP

Glucose is oxidized to give 2 molecules of pyruvate and 2 molecules of NADH and ATP (Energy currency).

<u>Though the free energy of oxidation of glucose is high but only 2 NADH is formed because the most of the free energy that is being released from the oxidation of glucose remains in the pyruvate which is produced in the reaction and thus only 2 molecules are formed.</u>

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3 years ago

All of the following elements will form ions by gaining

KiRa [710]

Answer:

Magnesium

Explanation:

When an atom loses are gain the electrons ions are formed.

There are two types of ions.

Anion

Cation

1 = Anion

It is formed when an atom gain the electrons. when atom gain electron negative charge is created on atom. For example.

X + e⁻ → X⁻

2= Cation

It is formed when an atom loses the electrons. when atom lose electron positive charge is created on atom. For example.

X → X⁺ + e⁻

There are seven valance electrons in iodine. To complete the octet it gain one electron and form anion.

Sulfur has six valance electrons it gain two electrons to complete the octet and form anion.

Phosphorus has five valance electrons it gain three electrons two complete the octet and form anion.

Magnesium has two valance electrons it loses two electron to complete the octet and form cation.

7 0

3 years ago

7. Which of these is not matter?<br> a. a cloud<br> c. sunshine<br> b. your hair<br> d. the sun

PilotLPTM [1.2K]

Hair ????????????????!?????

3 0

3 years ago

As methane burns, it reacts with oxygen and forms carbon dioxide and water. The balanced chemical equation below models this che

Elan Coil [88]

Explanation:

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3 0

3 years ago

Which of the following has the greatest electronegativity difference between the bonded atoms? a. A strong acid made of hydrogen

NARA [144]

Answer: Option (d) is the correct answer.

Explanation:

Electronegativity value of hydrogen is 2.2.

Electronegativity value of chlorine is 3.16.

Electronegativity value of carbon is 2.55.

Electronegativity value of oxygen is 3.44.

Electronegativity value of nitrogen is 3.04.

Electronegativity value of sodium is 0.93.

Electronegativity value of iodine is 2.66.

Therefore, calculate the electronegativity difference between the bonded atoms as follows.

Electronegativity difference of HCl = Electronegativity value of chlorine - electronegativity value of hydrogen

= 3.16 - 2.2

= 0.96

Electronegativity difference of CO = Electronegativity value of oxygen - electronegativity value of carbon

= 3.44 - 2.55

= 0.89

Electronegativity difference of $N_{2}$ = Electronegativity value of nitrogen - electronegativity value of nitrogen

= 3.04 - 3.04

= 0

Electronegativity difference of NaI = Electronegativity value of iodine - electronegativity value of sodium

= 2.66 - 0.93

= 1.73

So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.

Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.

5 0

3 years ago

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