Is air a compound mixture or a element. compound
The two elements that have the same properties as sodium are potassium and rubidium .
<h3>What is periodic table?</h3>
The periodic table is an arrangement of the elements in order of increasing atomic number.
From the list, the two elements that have the same properties as sodium are potassium and rubidium .
Proton has an atomic number of 20 and a mass number of 40. The number of electrons is the same as the number of protons. Given that the atomic number is the number of protons present, we have 20 electrons and 20 protons.
Number of neutrons = Mass number - Number of protons
= 40 - 20 = 20 neutrons
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Answer:
element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.
Explanation:
- Group IIA have 2+ valency and two electrons in its valance shell.
- Its Electropositivity is high and have the tendency to donate it two electrons.
- Element of IIA form ionic with most electronegative element.
Examples:
Cu²⁺, Mg²⁺, Sr²⁺ are examples having 2+ valance electron
one of the following is examples of element that have 2+ valence electrons
MgCl₂
Atomic number of Magnesium (Mg) is 12
Electronic Configuration of Mg:
1s², 2s², 2p⁶, 3s²
or
K =2
L = 8
M = 2
So, it have to give its 2 electrons to form a stable compound.
Similarly
Chlorine atomic number is 17
Electronic Configuration of Chlorine:
1s², 2s², 2p⁶, 3s², 3p⁵
or
K =2
L = 8
M = 7
So, it have to gain one electrons to form a stable compound and complete its octet.
So,
Two chlorine atom as a molecule gain 2 electrons from Mg²⁺ atom
So one Mg²⁺ and 2 Cl⁻ atoms form an ionic bond
where in this ionic bond Mg²⁺ transfer its 2 valence electron completely and chlorine molecule accept 2 electrons.
Cl-----Mg------Cl
So the Answer is
element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.
Answer:
λ = 5.68×10⁻⁷ m
Explanation:
Given data:
Energy of photon = 3.50 ×10⁻¹⁹ J
Wavelength of photon = ?
Solution:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = 3×10⁸ m/s
Now we will put the values in formula.
3.50 ×10⁻¹⁹ J = 6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ
λ = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J
λ = 19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J
λ = 5.68×10⁻⁷ m
