Calculate the ratio by using Henderson-Hasselbalch equation:
pH = pKa + log [neutral form] / Protonated form
3.05 = 2.21 + log [neutral form] / [Protonated form]
3.05 - 2.21 = log [neutral form] / [Protonated form]
0.84 = log [neutral form] / [Protonated form]
[neutral form] / [protonated form] = anti log 0.84 = 6.91
Answer:
The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)
Explanation:
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
