Answer:
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
Explanation:
Let's consider the molecular equation between aqueous copper(II) chloride and aqueous sodium phosphate.
3 CuCl₂(aq) + 2 Na₃PO₄(aq) ⇒ 6 NaCl(aq) + Cu₃(PO₄)₂(s)
The complete ionic equation includes all the ions and insoluble species.
3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ 6 Na⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and insoluble species.
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
That is approximately .63620 grams.
The reaction between KOH and HBr is as follows ;
KOH + HBr ---> H₂O + KBr
Stoichiometry of base to acid is 1:1 molar ratio
Both are strong acid and strong base therefore complete ionization takes place
The number of KOH moles added - 0.50 M / 1000 mL/L x 22 mL = 0.011 mol
the number of HBr moles - 0.25 M /1000 mL/L x 44 mL = 0.011 mol
the number of H⁺ ions and OH⁻ ions are equal therefore the whole amount of acid has been completely neutralised by base.
No remaining acid nor base, therefore solution is neutral.
pH = 7
thats the pH value for a neutral solution
Fluorine, Chlorine, Bromine, Iodine and Astatine are representing the family of elements called : halogens.
Halogens:
- all have 7 electrons in their outer shell,
- are very reactive elements,
- form - 1 ions.
Answer: C ) They have same chemical properties.
<span>250 ml * 1.25 g/ml * 3.74 j/g-K * 9.2 K = 10.752 kJ
Pretty much, all you need to do here is multiply all of these out to get your final answer. Not all questions are this easy, but this is certainly one of them.</span>
