Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
The pressure would increase. When the temperature change form cold to hot, the gas will find ways to escape from containment. Thus, if it cannot escape that pressure will keep on increasing as the temperature rises.
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Answer:</h3>
D. CH₃O₆
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Explanation:</h3>
- We are given the ratio of carbon, hydrogen and oxygen in a compound as 14: 42 : 84
We are supposed to determine the empirical formula of the compound;
- We need to know what is an empirical formula;
- An empirical formula of a compound is a formula that shows the simplest whole number ratio of elements in a compound.
In this case;
The ratio of C : H : O is 14 : 42 : 84
- Note, this is not the simplest ratio.
But we can simplify further to get the simplest whole number ratio;
- Therefore, 14 : 42 : 84 is equal to 1 : 3 : 6
- Thus, the simplest whole number ratio of C : H : O in the compound is 1 : 3 : 6
- Therefore, the empirical formula of the compound is CH₃O₆
Calcium chloride dehydrate (CaCl₂ · 2 H₂O) have a molar mass equal to 147 g/mol.
Explanation:
To calculate the molar mass of calcium chloride dehydrate (CaCl₂ · 2 H₂O) we use the following formula:
molar mass of CaCl₂ · 2 H₂O = atomic weight of Ca × 1 + atomic weight of Cl × 2 + atomic weight of H × 4 + atomic weight of O × 2
molar mass of CaCl₂ · 2 H₂O = 40 × 1 + 35.5 × 2 + 1 × 4 + 16 × 2
molar mass of CaCl₂ · 2 H₂O = 147 g/mol
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