Answer:
d. the conjugate base of the weak acid
Explanation:
The strong base (BOH) is completely dissociated in water:
BOH → B⁺ + OH⁻
The resulting conjugate acid (OH⁻) is a weak acid, so it remains in solution as OH⁻ ions.
By other hand, the weak acid (HA) is only slightly dissociated in water:
HA ⇄ H⁺ + A⁻
The resulting conjugate base (A⁻) is a weak base. Thus, it reacts with H⁺ ions from water to form HA, increasing the concentration of OH⁻ ions in the solution.
Therefore, the resulting solution will have a pH > 7 (basic).
The answer is: 1.5 moles of oxygen are present.
V(O₂) = 33.6 L; volume of oxygen.
p(O₂) = 1.0 atm; pressure of oxygen.
T = 0°C; temperature.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
n(O₂) = V(O₂) ÷ Vm.
n(O₂) = 33.6 L ÷ 22.4 L/mol.
n(O₂) = 1.50 mol; amount of oxygen.
Iodine is decolorized.
The first reaction stated in the question occurs as follows;
2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)
The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.
Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.
The equation of the titration reaction is;
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
When this reaction takes place, iodine is decolorized due to its reduction to I^-.
#1- Identify a problem
#2- Collect info on your problem
#3- Make a hypothesis
#4- Design an experiment to test your hypothesis
#5- Collect data and observations
#6- Accept or reject your hypothesis
#7- Record results
Hope this helps.
Answer:
I think it would be:
NaCO3 (s)-->Na2O (s) + CO2 (g)
