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>
I believe it is 6ml because you do the doseage times the ml and mutiply it by 1
<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>
Air is a insulator in normal conditions,but when heated up to high temperatures it becomes a conductor. <span />
I’m so sorry if it’s wrong but I think it’s this.....
Answer:
2NaHCO3(s) = Na2CO3(s) + CO2(g) + H2O(g).
And I found it on Quora.com
By the way NaHCO3 is Sodium bicarbonate but a.k.a Baking soda
Hoped this helped :)
