Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
They are both sources of power for objects that people use.
Answer:
Ionic equation:
Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)
Explanation:
Chemical equation:
NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)
Balanced chemical equation:
NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)
Ionic equation:
Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)
Net ionic equation:
OH⁻(aq) + H⁺(aq) → H₂O(l)
The Cl⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
