<span>ΔE = -RH [(1/nf2) - (1/ni2)]ΔE = -2.18 x10-18 J [(1/32) - (1/52)]ΔE = -1.55 x10-19 JE = hc/λλ = hc/Eλ = [(6.63 x10-34 J.s.) x (3.00 x1017 nm/s)] /(1.55 x10-19 J)λ = 1280nm</span>
<h3>Answer:</h3>
#1. Ca²⁺
# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
<h3>Explanation:</h3>
The question above concerns solubility of salts or ions in water.
The solution given contains Ag+, Ca2+, and Co2+ ions.
- In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
- In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
- The net ionic equation for the reaction is;
Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
- From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
- In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
- The net ionic equation for the reaction is;
3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
- According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
Thank you for posting your question here. Below is the solution:
HNO3 --> H+ + NO3-
<span>HNO3 = strong acid so 100% dissociation </span>
<span>** one doesn't need to find the molarity of water since it is the solvent </span>
<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
<span>1x10^-6M NO3- </span>
<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>
<span>1x10^-6 Ba(OH)2 = strong base, 100% dissociation </span>
<span>1x10^-6M Ba2+ </span>
<span>2x10^-6M OH- since there are 2 OH- / 1 Ba2+ </span>
<span>0M Ba(OH)2 </span>
<span>5x10^-9M H3O+</span>
2x 6.022x10^23= 1.204x10^24
