Answer:
this option is correct the amount of heat required to turn a mole of a solid into a liquid
Explanation:
Elements that belong to same group contains same number of valence electrons. Hence, they tend to show similar chemical properties.
When we move across a period then number of electrons get added to the same shell. As a result, there will be no increase in size of elements.
Also, metals are the elements that belong to group 1, 2 and d-block elements are also known as metals. Metallic character of elements decreases when we move left to right in a periodic table.
As most reactive metals are placed on the left side of periodic table.
Since, size of elements increases on moving down the group. So, an element is able to easily lose its valence electrons because of less force of attraction between its nucleus and valence electrons.
As a result, there will be increase in reactivity of metals on moving down the group.
Thus, we can conclude that given sentences are as follows.
- Elements in the same group have the same number of valence electrons.
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Elements in the same period have the same number of electron shells.
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Metallic elements become less reactive as you move left to right in a period.
-
Metallic elements become more reactive as you move top to bottom in a group.
C is correct. Have a good day!
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>
