Answer:
I think bread is the answer
<span>pOH = 9.2
pH = 4.8</span><span>
That's your </span>answer!!
Hope I helped!! :)
Answer:
C + D
Explanation:
C + D are the reactants because they are on the right side of the equation
(I hope this answer was quick enough!)
Answer:
0.056moles HF and 0.70M
Explanation:
When a strong acid is added to a buffer, the acid reacts with the conjugate base.
In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:
NaF + HCl → HF + NaCl
Initial moles of NaF and HF in 60.0mL of solution are:
NaF:
0.0600L × (0.80mol / L)= 0.048 moles NaF
HF:
0.0600L × (0.80mol / L)= 0.048 moles HF
Then, the added moles of HCl are:
0.0200L × (0.40mol / L) = 0.008 moles HCl.
Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:
<em>0.056moles HF</em>
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In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:
0.056mol HF / 0.0800L = <em>0.70M</em>
Answer:
Explanation:
For a general equilibrium
aA +bB ⇔ cC + dD ,
the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.
Our reasoning here should be based on the fact that Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).
So with this in mind, lets answer this question.
1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.
2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.
3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.
4. True: From our previous reasongs this is the true one.
5. False: If K is small, the equilibrium lies on the reactants side.
