Answer: A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.
Explanation:
A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.
Explanation:
The chart below shows monatomic ions formed when an atom loses or gains one or more electrons, and the ionic compounds they form. You can check your periodic table to see that the cations are monatomic ions formed from metals, and the anions are monatomic ions formed from nonmetals.
Answer:
A metalloid is a substance that has both the qualities of a metal and a non metal. Metals conduct electricity and have a high melting point.The element described does not conduct electricity until temperatures are reduced therefore it is not a metal. The fact that conduction occurs when temperatures are reduced means that it is not a non metal because non metals do not conduct electricity at all. Therefore the element is a metaloid because it exhibits some properties of a metal and others of a non mental.
Explanation:
litmus paper
Because it will just tell if the solution is acidic or basic, it won't tell the pH
Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>
1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.
(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>
