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
Answer:
The answer is A: 0.8 moles
Explanation:
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
Answer:
half lives passed=5
given sample=90g
sample left=2.8125g
Explanation:
no. of half lives=total time/half life
no.=19days/3.8days
no.=5 days
after 5 half lives sample left=2.8125g
Proton; neutron
Protons have a positive charge
Neutrons are neutral
Electrons have a negative charge :)