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Kamila [148]
3 years ago
14

Give examples of the one element from the alkaline earth metal group and one from the noble gases group. Include each element at

omic number, atomic mass number of neutrons of valence electron number of electron shells and some common characterizes of elements from it group.
Physics
1 answer:
zhuklara [117]3 years ago
8 0
Alkali metals: left column of your periodic table (not hydrogen, but anything below it). They have one valence electron, which they are happy to share in a reaction.

Halogens: second column from the right of your periodic table. They are one electron short of a full shell, so they are reactive in the opposite way that alkalis are--they want electrons.

Atomic number (number of protons) is the big number on the periodic table square. Hydrogen's is 1.

Atomic mass is a little number down below. For example, Hydrogen's is 1.008.

Neutrons are a tricky subject, because different isotopes of the same element can have different numbers of neutrons. You can't generally get this from the atomic mass, because the atomic mass is a weighted average of naturally occurring isotopes. Hydrogen can have 0,1, or 2 neutrons. To answer this, you'd have to choose a particular isotope from the table of isotopes (a completely different chart from the periodic table) which has a certain number of neutrons: n = weight - Z.

Valence electrons are the electrons in the outermost shell. (The column of the table).

<span> Number of principal shells is the row of the periodic table. </span>
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What is the energy of an electromagnetic wave that has a frequency of
sukhopar [10]

Answer:

Energy, \; E = 2.6504 * 10^{-34} \; Joules

Explanation:

Given the following data;

Frequency = 4.0 x 10⁹ Hz

Planck's constant, h = 6.626 x 10-34 J·s.

To find the energy of the electromagnetic wave;

Mathematically, the energy of an electromagnetic wave is given by the formula;

E = hf

Where;

E is the energy possessed by a wave.

h represents Planck's constant.

f is the frequency of a wave.

Substituting the values into the formula, we have;

Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34}

Energy, \; E = 2.6504 * 10^{-34} \; Joules

8 0
3 years ago
A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C
Rus_ich [418]

Answer:

1.50\ *10^{-6} }

Explanation:

Given

e=100 N/C

M=0.15 g

q=\ 22\  pC\\=\ 22\ *10^{-2}

The  ratio of the electric force on the bee to the bee's weight can be determined by the following formula

\frac{fe}{M*9.81}

\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}

=\ 1.50\ *10^{-6}

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3 years ago
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Usimov [2.4K]

Answer:

c. 0.80

Explanation:

they will choose the path that has not resistance

8 0
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Can you help me with this paper please I will give you 20 points!
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