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Answer:
if you want to have answers, don't take it wrong ...but put more details !
Explanation:
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
Answer:
D. Scientific notation is a way of writing large and small numbers
Answer:
Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)
Explanation:
Part 1. Volume of reactant
(a) Balanced chemical equation.

(b) Moles of CuCl₂

(c) Moles of Na₃PO₄
The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

(d) Volume of Na₃PO₄

Part 2. Net ionic equation
(a) Molecular equation

(b) Ionic equation
You write molecular formulas for the solids, and you write the soluble ionic substances as ions.
According to the solubility rules, metal phosphates are insoluble.
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>
The net ionic equation is
3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)