Hydroxylamine in water: HONH₂(aq) + H₂O(l) ⇄ HONH₃⁺(aq) + OH⁻(aq).
Hydroxylammonium nitrate in water: HONH₃NO₃(aq) → OHNH₃⁺(aq) + NO₃⁻(aq).
1) with positive hydrogen ions (protons) react base and gives weak conjugate acid:
H⁺(aq) + HONH₂(aq) ⇄ HONH₃⁺(aq).
2) with hydroxide anions react acid and produce weak base and weak electrolyte water:
HONH₃⁺(aq) + OH⁻(aq) ⇄ HONH₂(aq) + H₂O(l).
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
The question is incomplete, the complete question is;
Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m AlCl3 0.100 m NaCl 0.100 m MgCl2 They all have the same boiling point.
Answer:
AlCl3 0.100 m
Explanation:
Let us remember that the boiling point elevation is given by;
ΔTb = Kb m i
Where;
ΔTb = boiling point elevation
Kb = boiling point constant
m = molality of the solution
i = Van't Hoff factor
We can see from the question that all the solutions possess the same molality, ΔTb now depends on the value of the Van't Hoff factor which in turn depends on the number of particles in solution.
AlCl3 yields four particles in solution, hence ΔTb is highest for AlCl3 . The solution having the highest value of ΔTb also has the highest boiling point.
A. The pH indicates how acidic or basic a
solution is.
