Protons, neutrons..................................
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>
<span>You must balance your equation correctly.
Here is your answer:
294gFeS2 x 1molFeS2/119.99 x 11mols O2/4mols FeS2--> 6.738mol O2
176gO2 x 1mol O2/32gO2 x 4mols FeS2/11mol FeS2--> 2mols FeS2
Now choose the molecule with the lowest amount (Limiting Reagent)
2molsFeS2 x 2molsFe2O3/4molsFeS2 x 159.7g
159.7g Fe2O3 grams produced.</span>
Answer:
Are basic:
[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M
Explanation:
A solution is basic when pH = - log [H₃O⁺] is higher than 7.
It is possible to convert [OH⁻] to [H₃O⁺] using:
[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]
a. [OH⁻] = 3.13x10⁻⁷M
[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]
[H₃O⁺] = 3.19x10⁻⁸M
pH = - log [H₃O⁺] = 7.50
[OH⁻] = 3.13x10⁻⁷M is basic
b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.
This solution is not basic
c. [H₃O⁺] = 9.55x10⁻⁹M
pH = 8.02
This solution is also basic.
