Answer:
Electronegativity = 1.87.
Ionic radius = 109 pm.
Atomic radius = -39 pm
First ionization energy = 410 kJ/mol
Explanation:
Hello!
In this case, since electronegativity, ionic radius, atomic radius and first ionization energy are periodic properties that have specific trends, we can summarize it by realizing that oxygen and beryllium belong the same period 2 and differ in group, 6A and 2A respectively.
In such a way, the required comparison is written below:
Electronegativity = 3.44 (oxygen) - 1.57 (beryllium) = 1.87.
Ionic radius = 140 pm (oxygen)- 31 pm (beryllium) = 109.
Atomic radius = 73 pm (oxygen) - 112 pm (beryllium) = -39 pm
First ionization energy = 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium) = 410 kJ/mol
It means that electronegativity, ionic radius and first ionization energy increases from left to right whereas the atomic radius from right to left.
Best regards!
Answer:
The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons. They are the electrons involved in chemical bonds with other elements. Every element in the first column (group one) has one electron in its outer shell.
Explanation:
Explanation:
During a chemical reaction, the atoms of the original substances gain, lose or share their electrons with those of the substances with which they are reacting. The reaction creates new substances made up of a new combination of atoms and a different configuration of electrons.
Ionic Equation:
H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq) + Na⁺(aq) + Cl⁻(aq)
Net ionic equation:
H⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq)
The activity of the sample when it was shipped from the manufacturer is 4.54 mCi
<h3>How to determine the number of half-lives that has elapsed </h3>
From the question given above, the following data were obtained:
- Time (t) = 48 hours
- Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
- Number of half-lives (n) =?
n = t / t½
n = 48 / 342.72
n = 0.14
<h3>How to determine the activity of the sample during shipping </h3>
- Number of half-lives (n) = 0.14
- Original activity (N₀) = 5.0 mCi
- Activity remaining (N) =?
N = N₀ / 2ⁿ
N = 5 / 2^0.14
N = 4.54 mCi
Thus, the activity of the sample during shipping is 4.54 mCi
Learn more about half life:
brainly.com/question/2674699
