The net ionic equation for the reaction that occurs when solutions of lead(II) nitrate and potassium iodide are combined in a test tube is:
Pb²⁺(aq) + 2 I⁻(aq) ⇒ PbI₂(s)
Solutions of lead(II) nitrate and potassium iodide were combined in a test tube. This is a double displacement reaction. The complete ionic equation is:
Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 K⁺(aq) + 2 I⁻(aq) ⇒ 2 K⁺(aq) + 2 NO₃⁻(aq) + PbI₂(s)
Spectator ions are those that appear both in the reactants and in the products. They do not participate in the reaction, so they do not appear in the net ionic equation. The net ionic equation is:
Pb²⁺(aq) + 2 I⁻(aq) ⇒ PbI₂(s)
The net ionic equation for the reaction that occurs when solutions of lead(II) nitrate and potassium iodide are combined in a test tube is:
Pb²⁺(aq) + 2 I⁻(aq) ⇒ PbI₂(s)
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Answer:
Explanation: When solutions of potassium iodide and lead nitrate are combined?
The lead nitrate solution contains particles (ions) of lead, and the potassium iodide solution contains particles of iodide. When the solutions mix, the lead particles and iodide particles combine and create two new compounds, a yellow solid called lead iodide and a white solid called potassium nitrate. Chemical Equation Balancer Pb(NO3)2 + KI = KNO3 + PbI2. Potassium iodide and lead(II) nitrate are combined and undergo a double replacement reaction. Potassium iodide reacts with lead(II) nitrate and produces lead(II) iodide and potassium nitrate. Potassium nitrate is water soluble. The reaction is an example of a metathesis reaction, which involves the exchange of ions between the Pb(NO3)2 and KI. The Pb+2 ends up going after the I- resulting in the formation of PbI2, and the K+ ends up combining with the NO3- forming KNO3. NO3- All nitrates are soluble. ... (Many acid phosphates are soluble.)
The concentration of the resulting solution is 0.33 M. The correct option is 3. 0.33 M
From the question, we are to determine the concentration of the resulting solution
From the dilution law,
M₁V₁ = M₂V₂
Where M₁ is the initial concentration
V₁ is the initial volume
M₂ is the final concentration
V₂ is the final volume
From the given information
M₁ = 1.0 M
V₁ = 20 mL
M₂ = ?
V₂ = 60 mL
Then,
1 × 20 = M₂ × 60
20 = M₂ × 60
Therefore,
M₂ = 20 ÷ 60
M₂ = 0.33 M
Hence, the concentration of the resulting solution is 0.33 M. The correct option is 3. 0.33 M
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NH4NO3(s): ∆Hf = -365.56 kJ ∆Sf = 151.08 J/K.
NH3(g): ∆Hf = -46.11 kJ ∆Sf = 192.45 J/K.
H2O(l): ∆Hf = -285.830 kJ ∆Sf = 69.91 J/K.
O2(g): ∆Hf = 0.00 kJ ∆Sf = 205 J/K.
A) all of these measurements are needed
B) Temperature
C) Molar Mass
D) Pressure
E) Mass
Answer:
heterogeneous
Explanation: