Answer:
1. Molecular equation
BaCl2(aq) + 2AgNO3(aq) –> 2AgCl(s) + Ba(NO3)2 (aq)
2. Complete Ionic equation
Ba²⁺(aq) + 2Cl¯(aq) + 2Ag⁺(aq) + 2NO3¯ (aq) —> 2AgCl(s) + Ba²⁺(aq) + 2NO3¯(aq)
3. Net ionic equation
Cl¯(aq) + Ag⁺(aq) —> AgCl(s)
Explanation:
We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:
Kw = Kb * Ka
where Kw is always equal to 1x10^-14
so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:
1x10^-14 = 1.4x10^-9 * Ka
solve for Ka
Ka = 7.14x10^-6
Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.
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Answer:
B. CA, 14
Explanation:
Atoms of elements contain small particles known as electrons, neutrons, and protons. The nucleus of an atom is made up of neutrons and protons which are at the center of the atom. Electrons on the other hand surrounds the nucleus. Electron has negative charge while proton has a positive charge. The number of neutrons is equivalent to the number of protons . In addition, the number of protons is equal to mass number minus the number of electrons.
For the compound , it can be broken down into and . Its ion has a mass of 34 and 18 electrons which means it has already lost 2 electrons.
Therefore:
For the given element, the number of electrons is 18+2 = 20 electrons.
The number of protons = 34 - 20 = 14.
And the number of neutrons is 14.
Only option B has the correct answer.