This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
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Answer:
the answer is 70.906 g/mol
Explanation:
It is a column of elements in the periodic table of the chemical elements.
<u>Given:</u>
Concentration of Ba(OH)2 = 0.348 M
<u>To determine:</u>
pOH of the above solution
<u>Explanation:</u>
Based on the stoichiometry-
1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion
Therefore, concentration of OH- ion = 2*0.348 = 0.696 M
pOH = -log[OH-] = - log[0.696] = 0.157
Ans: pOH of 0.348M Ba(OH)2 is 0.157
Answer: Ionic substances would be good conductors since they dissolve in water and form ions.
Explanation:
Conductors can be defined as those materials which can allow the electricity to flow through them. The conductors also allow the transmission of heat and light from one to another source. They conduct electricity by allowing the electrons to flow through them easily.
The ionic substances can be good conductors because these can be dissolved in water and they will produce ions. The positively-charged ions will be attracted towards the negative electrode whereas the negatively charged ions to the positive electrode.
