Answer:
H₃O⁺(aq) + CO₃²⁻(aq) ⟶ HCO₃⁻(aq) + H₂O(ℓ)
Explanation:
There are three steps you must follow to get the answer.
You must write the:
- Molecular equation
- Ionic equation
- Net ionic equation
1. Molecular equation
HNO₃(aq) + Li₂CO₃(aq) ⟶ LiNO₃(aq) + LiHCO₃(aq)
2. Ionic equation
You write the molecular formulas for the weak electrolytes, and you write the ionic substances as ions.
H₃O⁺(aq) + NO₃⁻(aq) + 2Li⁺(aq) + CO₃²⁻(aq) ⟶ Li⁺(aq)(s) + NO₃⁻(aq) + Li⁺(aq) + HCO₃⁻(aq) + H₂O(ℓ)
3. Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
H₃O⁺(aq) + <u>NO₃⁻(aq)</u> + <u>2Li⁺(aq)</u> + CO₃²⁻(aq) ⟶ <u>Li⁺(aq)</u> + <u>NO₃⁻(aq</u>) +<u> Li⁺(aq)</u> + HCO₃⁻(aq) + H₂O(ℓ)
The net ionic equation is
H₃O⁺(aq) + CO₃²⁻(aq) ⟶ HCO₃⁻(aq) + H₂O(ℓ)
Answer:
C)52g KCl in 100g water at 80°C
Explanation:
A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.
A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.
From the solubility curve in the attachment below:
A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.
B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.
C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.
D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.
The balanced reaction that describes the reaction between lithium and nitrogen gas to produce lithium nitride is expressed 6Li + N2 = 2Li3N.Hence for every 6 moles of lithium used, there are 2 moles of lithium nitride produced. For this problem, 12 moles of lithium yields 4 moles of lithium nitride.
Sodium/Atomic number
11
Gold/Atomic number
79
Potassium/Atomic number
19
Silicon/Atomic number
14