4. Determine the soluble and insoluble compounds. (6 pts) (enter your answers in
ID: 1028836 • Letter: 4
Question
4. Determine the soluble and insoluble compounds. (6 pts) (enter your answers in the blank boxes) NaCI BaBr CompoundsSolubility K CO Cas(PO4) Pblz AgF S. Define acid and base. (6 pts) (enter your answers in the blank boxes) base? is strong What is weakWhat is strong What is weak acid? acid? base? Strong base: Choose one example of strong and weak acid, respectively (HF, HCI, HNO3, H2CO. H2SO3) Choose one example of strong and weak base, respectively: (NaOH, KOH Strong acid: Weak acid: Weak base: Determine the oxidation number of the following elements: (6 pts) (enter your answers in the blank boxes) 6. Compounds Your answers: Oxidation number O.N. of O: Oxidation number O.N. of Cl: HCIOs CaClz O.N. of Cl: O.N. of Ca: HSO, O.N. of s O.N. of H:Explanation / Answer
4.
NaCl – Soluble
BaBr2 – Soluble
AgF – Soluble
K2CO3 – Soluble
Ca3(PO4)3 - Soluble
PbI2 – Insoluble (or sparingly soluble)
5.
A strong acid is any acid that ionizes completely in solutions. This means it gives off the greatest number of hydrogen ions or protons when placed in a solution.
A weak acid is an acid that dissociates incompletely, releasing only some of its hydrogen atoms into the solution. Thus, it is less capable than a strong acid at donating protons. These acids have higher pKa than strong acids,
A weak base is a chemical base that does not ionize fully in an aqueous solution.
A strong base is a base that is completely dissociated in an aqueous solution.
HCl, HNO3, H2SO3 – strong acid
HF, H2CO3, – weak acid
NaOH, KOH – strong base
NH3.H2O – weak base
6.
HClO3
Oxygen always has an oxidation number of –2 within compounds except when in peroxide compounds where it has a change of –1 (which in this case, the compound is not a peroxide compound)
Hydrogen always has an oxidation number of +1 except when in metal hydrides (which in this case, the compound is not a metal hydride)
The oxidation numbers in an uncharged compound must add up to zero, so:
1 + x + (-2 x 3) = 0
1 + x – 6 = 0
x = 5
CaCl2
Ca = +2
Cl = -1
HSO3-
1 + x + (-2)3 = -1
1 + x – 6 = -1
x = - 1 + 5 = 4
H = +1
S = +4
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