Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
Answer: is C
Explanation:
Both the atomic mass and the atomic number increase from left to right
In a dilute acid solution most if not all of the molecules will split into ions.
For example HCl is a strong acid and 100% of the molecules will split into
H+ & Cl-
in a weak acid solution only a portion of the molecules will turn into ions because the ionization percentage isn't as large. Which will essentially leave a high percentage of un-reacted molecules
Answer:
0.32 M
Explanation:
Step 1: Write the balanced reaction at equilibrium
Ag₂S(s) ⇌ 2 Ag⁺(aq) + S²⁻(aq)
Step 2: Calculate the concentration of Ag⁺ at equilibrium
We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.
Keq = [Ag⁺]² × [S²⁻]
[Ag⁺] = √{Keq / [S²⁻]}
[Ag⁺] = √{2.4 × 10⁻⁴ / 0.0023} = 0.32 M