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Darya [45]
3 years ago
12

Describe two differences between Mendeleev's periodic table and the modern periodic table

Physics
1 answer:
Igoryamba3 years ago
6 0
There's a massive improvement in the layout, to begin with. Elements are frequently colour-coded to differentiate between their physical properties.

Mendeleev's periodic table has no place for isotopes, and also placed dissimilar atoms together in groups.

<span>Mendeleev also left gaps for elements he suspected would be discovered, but he did not have the equipment to discover. These were later discovered through things like electrolysis. These days these unknown elements are not so evenly spread, but are mostly found around unununium.</span>
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The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this
svlad2 [7]

Answer: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.

Explanation: On Edgenuity!!

7 0
3 years ago
A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)
natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

\cot\phi = (1+\frac{hf}{m_{e}c^{2}  })\tan\frac{\theta }{2}      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, m_{e} is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

\cot\phi = (1+\frac{h}{m_{e}c\lambda  })\tan\frac{\theta }{2}

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for m_{e}, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9}  })\tan\frac{134 }{2}

\cot\phi=2.37

\phi = 22.90°

8 0
4 years ago
Real springs have mass. How will the true period andfrequency
Ad libitum [116K]

Explanation:

An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.

Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.

4 0
4 years ago
Which of these is NOT a common building material?
Over [174]
B Quartz. Will be your answer of thia
8 0
3 years ago
Read 2 more answers
Can someone please help me with this question? ASAP thank you!
Assoli18 [71]

did you get the answer

8 0
3 years ago
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