Organic molecule. In particular, a carbohydrate.
The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O applies to an acid-base titration.
moles NaOH = c · V = 0.2423 mmol/mL · 32.23 mL = 7.809329 mmol
moles H2SO4 = 7.809329 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.9046645 mmol
Hence
[H2SO4]= n/V = 3.9046645 mmol / 37.21 mL = 0.1049 M
The answer to this question is [H2SO4] = 0.1049 M
Answer:
2 KOH(aq) + CuCl2(aq) = 2 KCl(s) + Cu(OH)2
Explanation:
Answer:
Kp = 8.76×10⁻³
Explanation:
We determine the carbamate decomposition in equilibrium:
NH₄CO₂NH₂ (s) ⇄ 2NH₃(g) + CO₂(g)
Let's build the expression for Kp
Kp = (Partial pressure NH₃)² . Partial pressure CO₂
We do not consider, the carbamate because it is solid and we only need the partial pressure from gases
Kp = (0.370atm)² . 0.0640 atm
Kp = 8.76×10⁻³
Remember Kp does not carry units