They give a double displacement reaction where the ions switch places and give sodium nitrate (NaNO3) and silver chloride (AgCl) as the products. Silver nitrate is also very soluble in water, but silver chloride is highly insoluble in water and will precipitate out of solution as a white solid.
It is considered to be a conductor
(C) 0.1 mole of NaCl dissolved in 1,000. mL of water
<u>Explanation:</u>
The conductivity of 0.1 mole of NaCl dissolved in 1000 mL of water will be greatest as the number of ions in 0.1 mole of NaCl will be more than 0.001, 0.05 and 0.005 moles of NaCl. Greater the number of ions in the solution, greater will be the conductivity. Specific Conductivity decreases with a decrease in concentration. Since the number of ions per unit volume that carry current in a solution decrease on dilution. Hence, concentration and conductivity are directly proportional to each other.
Answer:
Option C. Energy Profile D
Explanation:
Data obtained from the question include:
Enthalpy change ΔH = 89.4 KJ/mol.
Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:
Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.
If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.
Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.
Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.
