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Alja [10]
1 year ago
9

Atoms of what class of elements are used to absorb electrons in electron-dense stains?

Chemistry
1 answer:
lesantik [10]1 year ago
8 0

The elements of heavy metals are used to absorb electrons in electron-dense stains.

<h3>What are heavy metals?</h3>

A heavy metals are generally defined as metals with relatively high densities, atomic weight and atomic number.

They toxic and poisonous in nature. They occurs naturally and are very essential to life.

Arsenic, cadmium, lead, copper, nickel and mercury are the most commonly used heavy metals.

Electron staining means to absorb heavy metals of high scattering power to biological specimens which exhibit a small scattering power for electrons.

Thus, the elements of heavy metals are used to absorb electrons in electron-dense stains.

To learn more about  heavy metals, refer to the link below:

brainly.com/question/3937595

#SPJ4

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Why was 1990 an important year regarding air quality?
Valentin [98]

Why was 1990 an important year regarding air quality? Check all that apply.

2.Cost-effective ways to reduce pollution were emphasized.


4.Modifications and improvements were made to the Clean Air Act.


i just took the test this is 100% correct

3 0
3 years ago
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In which orbital does an electron in a bromine atom experience the greatest effective nuclear charge?
asambeis [7]

First let us determine the electronic configuration of Bromine (Br). This is written as:

Br = [Ar] 3d10 4s2 4p5

 

Then we must recall that the greatest effective nuclear charge (also referred to as shielding) greatly increases as distance of the orbital to the nucleus also increases. So therefore the electron in the farthest shell will experience the greatest nuclear charge hence the answer is:

<span>4p orbital</span>

4 0
4 years ago
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What is an extensive property that can be calculated?
lord [1]
Volume can be calculated
4 0
3 years ago
NEED HELP ASAP ITS DUE TODAY THESE QUESTIONS
beks73 [17]

1.

Density can be defined as the mass of the substance in unit volume.

Density = mass / volume

Hence, g/mL and kg/L can be used as the units of density.

Those units are interchangeable because when converting one unit into other one nothing will happen to the value.

That is because when converting the g into kg, you have to multiply the value by 1 x 10⁻³. When converting mL into L, you should again multiply the volume by 1 x 10⁻³. Then those 1 x 10⁻³ will cancel off and the original value will remain as same.

2.

Answer is 2.70 g/mL.

<em>Explanation;</em>

Mass of the block = 146 g

Volume of the block =  length x width x height

                                  = 6.0 cm x 3.0 cm x 3.0 cm

                                  = 54 cm³ = 54 mL

Density = mass / volume

            = 146 g / 54 mL

            = 2.70 g/mL

3.

Answer is  3.39 g/mL.

<em>Explanation;</em>

When immersing an object in a solution, the increased volume indicates the volume of that object.

Hence,

  Volume of the object = increased volume of water level

                                      = final volume - initial volume

                                      = 27.8 mL - 21.2 mL

                                      = 6.6 mL

Mass of the object = 22.4 g

Density = Mass / Volume

            = 22.4 g / 6.6 mL

            = 3.39 g/mL

4.

Accepted value is the value that scientists and community accept as true. This is a theoretical value.

But the measured value is the value that you obtain from doing experiments. This is the actual value.

If your measured value is more close to the accepted value, then your measured value is more precise. But, if your measured value is far away from accepted value means that your value is not precise and there may have some errors.

5.

Answers : Percent error is 9.11 %

                 The element which has 7.13 g/cm³ as density is Zinc (Zn).

<em>Explanation;</em>

Percent error can be calculated by using following formula.

% error = ( (measured value - accepted value) / accepted value ) x 100%

            = ( (7.78 g/cm³ - 7.13 g/cm³) / 7.13 g/cm³ ) x 100%

            = 9.11 %

6 0
3 years ago
1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hour
lianna [129]

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9  

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

5 0
3 years ago
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