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1 year 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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1. Which of the following parts allow different activities of the cell to happen? A cytoplasm C vacuole B. lysosome D. vesicle 2

Anna35 [415]

Answer: A. Cytoplasm

Explanation: Cytoplasm is a jelly, like material that fills up the remaining space in the cell. It is also responsible in making the shape of the cell. Without it, other parts of the cells or the organelles cannot make its function because it serves as the hallway of the transportation of the nutrients needed by each cell.

Learn more: https://brainly.ph/question/73078

4 0

2 years ago

A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 31.μmol of copper(II) sulfate into a 150.mL volumetri

baherus [9]

Answer:

The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

Explanation:

The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.

Here's that idea written as a formula: c= n/V

where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.

You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L

Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L

Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

3 0

2 years ago

HELPPPP PLEASEEE

Igoryamba

Atomic mass is the answer!!
Explanation: An elements atomic number won’t be able to change

6 0

2 years ago

Lowering an object decrease is potential

Tresset [83]

Answer:

Lowering the object near the ground decreases its potential energy.

Explanation:

Potential Energy : Energy possessed by the object due to its shape ,Size and Position is called potential energy.

Example :

Change in shape and size : When you compress the spring , potential energy is introduced in it . So it expand quickly when you remove your hand.

Change in position : when you swing , after attaining maximum height (extreme ends) , the swing comes back on its on .This is because at maximum height ,the swing has maximum Potential energy . Hence it fall back on its on because it already has potential energy.

Potential energy(P) is given by the formula :

P = mgh

where ,

m= mass of the object

g = acceleration due to gravity

h = height of the object from the ground

If the height of the object increases from the ground , its potential energy also get increase.

On lowering the object The height of the object from the ground reduces . So potential energy also reduces.

8 0

3 years ago

A solution of H 2 SO 4 ( aq ) H2SO4(aq) with a molal concentration of 8.01 m 8.01 m has a density of 1.354 g / mL . 1.354 g/mL.

anzhelika [568]

Answer:

[H₂SO₄] = 6.07 M

Explanation:

Analyse the data given

8.01 m → 8.01 moles of solute in 1kg of solvent.

1.354 g/mL → Solution density

We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g

Mass of solvent = 1kg = 1000 g

Mass of solution = 1000g + 785.4 g = 1785.4 g

We apply density to determine the volume of solution

Density = Mass / volume → Volume = mass / density

1785.4 g / 1.354 g/mL = 1318.6 mL

We need this volume in L, in order to reach molarity:

1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L

Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M

4 0

2 years ago

Read 2 more answers