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trapecia [35]
3 years ago
13

The table below shows two types of electromagnetic waves and three random applications of electromagnetic waves.

Chemistry
2 answers:
Ilia_Sergeevich [38]3 years ago
6 0

CORRECT ANSWER: GAMMA RAYS-1  AND RADIO WAVES-2

<h2>TOOK THE TEST!</h2>
Reptile [31]3 years ago
4 0
Correct Answer is 1 i.e. Gamma rays—2 and radio waves—3

Reason: 
1) In a hypernova, star<span> as similar to </span>nuclear fusion<span> converts lighter elements into heavy elements. If fusion is not capable of generating enough pressure to counteract gravity, star immediately collapses to form a </span>black hole<span>. During this process, energy will be released, along the axis of rotation to form </span>gamma-ray burst. Such gamma-ray burst was first detected using <span>Fermi Gamma-ray Space Telescope. Thus, gamma-ray is capable of providing information of gravity fields.

2) Radiowaves are capable of inducing transitions that requires less energies. These transition includes nuclear excitation and electron excitation (in rotational energy level). Depending upon the value to Jmax, it is possible to determine the temperature and </span><span>heat released by astronomical objects</span><span>
</span>
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What is the molarity of ethanol c2h5oh in an aqueous solution that is 36.4% ethanol by mass
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3 years ago
May someone assist, please...? I don't know how to do chemistry work...
EastWind [94]

Answer:

1) 90.0 mL

2) 11.25 M

3) 0.477 M

4) 144 mL

Explanation:

The main formula that will be used for all these calculations is:

                                                     C₁V₁ = C₂V₂

C stands for concentration and V stands for volume and the subscripts 1 and 2 indicate an initial concentration or volume and a final concentration or volume.

For each problem, it's best to start by figuring out what you have and what you need to find. Figure out if you're looking for an initial value or a final value.

1) We need to find the initial volume. So, take what values you have and plug them in and then solve for whatever variable:

5.00 M · V₁ = 500.0mL · 0.900 M                        - divide by 5.00

C₁ = 90.0 mL

2) This time we're finding the initial concentration:

20.0mL · C₁ = 150.0mL · 1.50 M                          - divide by 20.0mL

C₂ = 11.25 M

3) Now we're finding the final concentration:

12.00mL · 3.50 M = 88.0mL · C₂                         - divide by 88.0mL

C₂ = 0.477 M

4) Finally, we're looking for the final volume:

9.0mL · 8.0 M = 0.50 M · V₂                                - divide by 0.50 M

V₂ = 144mL

6 0
3 years ago
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