Answer:
oxalate, fluoride, carbonate, phosphate.
Explanation:
It is important to consult the solubility chart in order to determine the anions.
Any compound with a low Ksp means the compound is very insoluble.
Hence, it means the formation of this compound is favoured strongly and its dissolution which would liberate calcium ions is unfavoured.
Any compound with a low Ksp would therefore "soak up" all the free calcium and will thus not be present in the solution.
Using solubility chart and noting the compounds with a low Ksp value, the anions that are probably present are: oxalate, fluoride, carbonate, phosphate.
If you are looking for a one specific electron you will have 90% chance of finding it in an area within the atom electron clouds.
According to the wave mechanical or electron cloud model of the atom, postulates that the atomic orbital is a region in space where there is a very high probability of finding an electron.
The term "electron cloud" is just another term used for "atomic orbital".
Since electrons reside in orbitals, if you are looking for a one specific electron you will have 90% chance of finding it anywhere within the electron clouds. This would make mathematical sense as the probability of finding an electron within the electron cloud can never be 100%.
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You would calculate them by dividing them and then multiplying to get the final answer
The ocean went over little spots where rocks caved In creating dents in the rock big enough for water to get in and when the wave went over them it left water and whatever else was in it
Answer:
See explanation
Explanation:
Recall that a neutralization reaction is one in which an acid reacts with a base to yield salt and water only. The equation is only balanced when the number of each atom on the reactant side is the same as the number of the same atom on the product side.
Let us now complete and balance each reaction equation;
(a) H2SO4 + 2NaOH --------> Na2SO4 + 2H2O
(b) HNO3 + KOH ---------> KNO3 + H2O
(c) 2HCl + Ca(OH)2 -------> CaCl2 + 2H2O
(d) 2H3PO4 + 3Ba(OH)2 ------> Ba3(PO4)2 + 6H2O
() CH3COOH + NaOH --------> CH3COONa + H2O
(1) 2HNO3 + Sr(OH)2 --------->Sr(NO3)2 + 2H2O
(g) 3HF + Fe(OH)3 ------> FeF3 + 3H2O
(h) 4HBr + Sn(OH)4 --------> SnBr4 + 4H2O