That statement is true because to make a diluted solution, you have to add more solvent than the solute in a concentrated solution. This is extra information, but to make a diluted solution more concentrated, you have to add more solute since a concentrated solution has an excess amount of solute as stated before.
... to be called elements<span>. This lesson shows </span>you how to<span> predict the </span>numbers<span> of </span>neutrons, electrons, andprotons<span> of the isotopes they are likely to find in nature. (</span><span>cont.) ... What </span>kind<span> of </span>generalization can you make<span> about how the </span>number<span> of </span>protons<span> and </span>neutrons<span> are </span>related<span> to </span>each other<span> in the </span>elements<span>? Unit 1 • Investigation IV</span>
<span> Curium (Cm) is the answer.
Hope this helps.</span>
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>
Yes, it mixes it and has vitamins in the tea.