Answer: The pH of a 4.4 M solution of boric acid is 4.3
Explanation:
at t=0 cM 0 0
at eqm 
So dissociation constant will be:
Give c= 4.4 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D4.4%5Ctimes%200.000011%3D4.8%5Ctimes%2010%5E%7B-5%7DM)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[4.8\times 10^{-5}]=4.3](https://tex.z-dn.net/?f=pH%3D-log%5B4.8%5Ctimes%2010%5E%7B-5%7D%5D%3D4.3)
Thus pH of a 4.4 M 
solution is 4.3
I think the correct answer from the choices listed above is option B. The reactants calcium sulfide and sodium sulfate will react and form a precipitate which is calcium sulfate since it is only slightly soluble in aqueous solution. Hope this answers the question.
Answer is: <span>the empirical formula of the hydrocarbon is CH</span>₂.<span>
Chemical reaction: C</span>ₓHₐ + O₂ → xC + a/2H₂O.<span>
m(CO</span>₂) = 33.01 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 33.01 g ÷ 44.01 g/mol.
n(CO₂) = n(C) = 0.75 mol.
m(H₂O) = 13.52 g.
n(H₂O) = 13.52 g ÷ 18 g/mol.
n(H₂O) = 0.75 mol.
n(H) = 2 · n(H₂O) = 1.5 mol.
n(C) : n(H) = 0.75 mol : 1.5 mol /0.75 mol.
n(C) : n(H) = 1 : 2.
Answer:The answer is C) The total mass of C and D only.
