Answer:
C. Y & Z
Explanation:
V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).
Thus, Y and Z make covalent.
Answer:
D
Explanation:
The indivisibility of an atom was proved wrong an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions......
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Answer:
True
Explanation:
Atomic radius can be defined as a measure of the size (distance) of the atom of a chemical element such as hydrogen, oxygen, carbon, nitrogen etc, typically from the nucleus to the valence electrons. The atomic radius of a chemical element decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon). Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).
<em>Hence, the atomic radius of phosphorus is smaller than the atomic radius of magnesium. Basically, the atomic radius of phosphorus is 98 pm while the atomic radius of magnesium is 145 pm.</em>
A base generally releases a hydroxide ion (OH-) when dissolved in water.
There are exceptions, such as ammonia NH3, which acts as a base but does not produce OH- ions. There are three definitions of acids and bases (Arrhenius, Bronsted-Lowry, and Lewis) and each one looks at acid/base characteristics differently. OH- donation is the Arrhenius definition.
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
