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bulgar [2K]
3 years ago
5

What methods were used to clean up the Chernobyl and three mile island accidents?

Chemistry
1 answer:
Assoli18 [71]3 years ago
6 0

<u>Answer</u>:

Electric evaporator has been used to clean up the Chernobyl and 3 mile island accident.

<u>Explanation</u>:

About 14 years back, there was a puff of radioactive steam from the evaporator used in the "Three-mile island" nuclear plant. It took two years for the electric evaporator to clean up the 2.2 million gallons of water that have been destroyed because of the nuclear power plant accident.

The steam that came out of the electric evaporator carried a radioactive form of hydrogen and tritium with itself. Cost of clean up was $1 billion.

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It is observed that the atoms of hydrogen in gas discharge tube emit radiations whose spectrum shows line characteristics (line
azamat

Answer: the line Spectra of hydrogen lies between the ultra-violet, visible light and infra-red of the electro magnetic spectrum

Explanation:

Electromagnetic radiation spans an wide range of wavelengths and frequencies. This range is called the electromagnetic spectrum. The electromagnetic spectrum is generally divided into seven regions, in order of decreasing wavelength and increasing energy and frequency. The 7 regions includes; radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.

lower-energy radiation, such as radio waves, is expressed as frequency while microwaves, infrared, visible and UV light are usually expressed as wavelength and finally, higher-energy radiation such as X-rays and gamma rays, is expressed in terms of energy per photon.

Therefore, hydrogen lies between the ultra-violet, visible light and infra-red region of the electro magnetic spectrum.

5 0
3 years ago
What products come from forest resources
lana66690 [7]

Some of these are products most people wouldn't think originated from trees, which only further exemplifies the value of preserving our trees and forests!

Wine Corks. ...

Natural Aspirin and Acne Medication. ...

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6 0
3 years ago
Read 2 more answers
A solution that contains a large amount of solute would be described as what
antiseptic1488 [7]

Answer:

A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.

8 0
3 years ago
The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g
ANTONII [103]

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given:  2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

<span>We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.</span>

7 0
3 years ago
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
2 years ago
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