According to Arrhenius concept a base is one which can donate hydroxide ion in its aqueous solution.
Here NaOH is a base as it can donate OH-.
As NaOH is a strong base it will have very high dissociation constant.
Due to high dissociation constant if will almost completely dissociate into sodium ions and hydroxide ions.
Thus in a solution of NaOH we will expect the following two ions
Na^{+} + OH^{-}[/tex]
Can't say i can answer this. :/
Answer:
3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)
Explanation:
When aqueous solutions of chromium(II) iodide and sodium phosphate are combined, solid chromium(II) phosphate and a solution of sodium iodide are formed. The molecular equation is:
3 CrI₂(aq) + 2 Na₃PO₄(aq) ⇄ Cr₃(PO₄)₂(s) + 6 NaI(aq)
The full ionic equation includes all the ions and the molecular species.
3 Cr²⁺(aq) + 6 I⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 I⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)
Answer:
sodium (Name) and Chlorine (Cl)
Explanation:
Any of the Group 17 elements (the halogens) only need one additional electron to take on the electron configuration of the noble gas just to it's right. That would be argon (Ar) in this case.
For this reaction, the needs of both elements (to reach a stable Group 17 electron configuration) is matched: Just one electron is provided (Na) and taken (Cl) and that forms a single bond
Answer:
SO3
Explanation:
Data obtained from the question include:
S = 40%
O = 59%
To obtain the empirical formula, do the following:
Divide the above by their molar mass as shown below:
S = 40/32 = 1.25
O = 59/16 = 3.69
Next, divide by the smallest as shown below:
S = 1.25/125 = 1
O = 3.69/1.25 = 3
Therefore, the empirical formula is SO3