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3 years ago

Knowing that all three-sided figures are triangles is an example of:

Chemistry

2 answers:

pogonyaev3 years ago

5 0

Answer:

B. Concept Formation

Explanation:

All the other answer choices are not equivalent.

Hope this helps!

Brainliest??

Lerok [7]3 years ago

5 0

Answer:

Concept Formation

Explanation:

The formation of concepts is a process in which specific expressions are taken to make a generalization or a correspondence between things. The formation of concepts is very important in teaching because it is vital to acquire knowledge.

In this case, the fact that all 3-sided figures are triangles is an example of concept formation because a concept is a definition that helps us define if an object is within a certain category.

So for this example, the concept that every three-sided figure is a triangle can help us define when something can be classified or not as a triangle, thus falls into the category of concept formation.

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Identify the element: 6 energy level, metallic, 82 election

Lady bird [3.3K]

<u>Answer</u>:-

Name of element = Lead

Symbol ♾ of Lead = Pb

Atomic no. = 82

Atomic mass = 207.2 amu

No. of protons = 82

No. of electrons = 82

Yes, lead(Pb) is a metallic element and certainly it has 6 electron shells which means 6 energy level.

If you find anything unclear you can ask me...

6 0

3 years ago

What happens to isotopes that are unstable?

Lesechka [4]

<u>Explanation:</u>

Isotopes are defined as the chemical species of the same element which differs in the number of neutrons. The isotopes which are unstable are known as radioactive isotope. A radioactive (unstable )isotope can undergo 3 decay process:

1. Alpha Decay: In this decay process, a larger nuclei decays into smaller nuclei by releasing alpha particle. The particle released has a charge of +2 and a mass of 4 units.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha$

2. Beta-minus decay: In this decay process, a neutron gets converted into a proton and an electron. the particle released during this process is a beta-particle.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

3. Beta-plus decay: In this decay process, a protons gets converted into a neutron and electron-neutrino particle. The particle released during this process is a positron particle.

$_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0\beta$

Isotopes which are unstable in nature can undergo these 3 decay processes.

5 0

3 years ago

131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___

inn [45]

For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0

3 years ago

If you hit the surface of Iron with a photon of energy and find that the ejected electron has a wavelength of .75 nm, what is th

lubasha [3.4K]

Answer:

The wavelength of the incoming photon is 172.8 nm

Explanation:

The wavelength of the incoming photon can be calculated with the photoelectric equation:

$KE = h\frac{c}{\lambda_{p}} - \phi$ (1)

Where:

KE: is the kinetic energy of the electron

h: is Planck's constant = 6.62x10⁻³⁴ J.s

c: is the speed of light = 3.00x10⁸ m/s

$\lambda_{p}$ : is the wavelength of the photon =?

Φ: is the work function of the surface (Iron) = 4.5 eV

The kinetic energy of the electron is given by:

$KE = \frac{p^{2}}{2m} = \frac{(\frac{h}{\lambda_{e}})^{2}}{2m}$ (2)

Where:

p: is the linear momentum = h/λ

m: is the electron's mass = 9.1x10⁻³¹ kg

$\lambda_{e}$ : is the wavelength of the electron = 0.75 nm = 0.75x10⁻⁹ m

Hence, the wavelength of the photon is:

$\frac{(\frac{h}{\lambda_{e}})^{2}}{2m} = h\frac{c}{\lambda_{p}} - \phi$

$\lambda_{p} = \frac{hc}{\frac{h^{2}}{2m\lambda_{e}^{2}} + \phi} = \frac{6.62 \cdot 10^{-34} J.s*3.00\cdot 10^{8} m/s}{\frac{(6.62 \cdot 10^{-34} J.s)^{2}}{2*9.1 \cdot 10^{-31} kg*(0.75 \cdot 10^{-9} m)^{2}} + 4.5 eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}} = 1.728 \cdot 10^{-7} m = 172.8 nm$

Therefore, the wavelength of the incoming photon is 172.8 nm.

I hope it helps you!

3 0

3 years ago

Name the element in the second period of the periodic table with 6 valence electrons.

Rus_ich [418]

Answer:

Sulfur

Explanation:

Using this to make the answer 20 characters long

5 0

3 years ago