Can someone help me real quick?

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

Reptile [31]

Correct Answer is 1 i.e. Gamma rays—2 and radio waves—3

Reason:
1) In a hypernova, star as similar to nuclear fusion converts lighter elements into heavy elements. If fusion is not capable of generating enough pressure to counteract gravity, star immediately collapses to form a black hole. During this process, energy will be released, along the axis of rotation to form gamma-ray burst. Such gamma-ray burst was first detected using 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 heat released by astronomical objects

4 0

4 years ago

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Give the nuclide symbol for an atom that has mass number 64 and 35 neutrons.

azamat

The mass number represents the summation of the number of protons and the number of neutrons in the nucleus of an atom.
We are given that the mass number is 64 and that the number of neutrons is 35. Therefore:
number of protons = 64 - 35 = 29 protons

In ground state, number of protons in an atom is equal to the number of electrons. Therefore,
number of electrons = 29 electron

Using the periodic table, we will find that the element that has 29 electrons in ground state is copper.
The nuclide symbol of copper is shown in the attached image.

5 0

4 years ago

How many molecules of Ca are found in a sample with 0.2 mols?

DochEvi [55]

Answer:

$\boxed{1.2 \times 10^{23}\text{ atoms}}$

Explanation:

6.023 × 10²³ atoms of Ca are in 1 mol of Ca

$\text{No. of atoms} = \text{0.2 mol} \times \dfrac{6.023 \times 10^{23}\text{atoms }}{\text{1 mol }} = \mathbf{1.2 \times 10^{23}} \textbf{ atoms}}\\\\\text{There are }\boxed{\mathbf{1.2 \times 10^{23}} \textbf{ atoms}} \text{ atoms in 0.20 mol of Ca}$